Solucionario Cap11 Dinamica Beer Johnston

Download Solucionario Cap11 Dinamica Beer Johnston

Post on 13-Apr-2015

188 views

Category:

Documents

188 download

DESCRIPTION

COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 2. x = t3 − (t − 2) m 2 v= a= (a) Time at a = 0. dx = 3t 2 − 2 ( t − 2 ) m/s dt…

TRANSCRIPT

COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 2. x = t3 − (t − 2) m 2 v= a= (a) Time at a = 0. dx = 3t 2 − 2 ( t − 2 ) m/s dt dv = 6t − 2 m/s 2 dt 0 = 6t0 − 2 = 0 t0 = 1 3 t0 = 0.333 s W (b) Corresponding position and velocity. ⎛1⎞ ⎛1 ⎞ x = ⎜ ⎟ − ⎜ − 2 ⎟ = − 2.741 m ⎝3⎠ ⎝3 ⎠ 3 2 x = − 2.74 m W ⎛1⎞ ⎛1 ⎞ v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s ⎝3⎠ ⎝3 ⎠ 2 v = 3.67 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 3. Position: Velocity: x = 5t 4 − 4t 3 + 3t − 2 ft v= a= dx = 20t 3 − 12t 2 + 3 ft/s dt dv = 60t 2 − 24t ft/s 2 dt Acceleration: When t = 2 s, x = ( 5 )( 2 ) − ( 4 )( 2 ) − ( 3)( 2 ) − 2 v = ( 20 )( 2 ) − (12 )( 2 ) + 3 a = ( 60 )( 2 ) − ( 24 )( 2 ) 2 3 2 4 3 x = 52 ft W v = 115 ft/s W a = 192 ft/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 4. Position: Velocity: x = 6t 4 + 8t 3 − 14t 2 − 10t + 16 in. v= a= dx = 24t 3 + 24t 2 − 28t − 10 in./s dt dv = 72t 2 + 48t − 28 in./s 2 dt Acceleration: When t = 3 s, x = ( 6 )( 3) + ( 8 )( 3) − (14 )( 3) − (10 )( 3) + 16 v = ( 24 )( 3) + ( 24 )( 3) − ( 28 )( 3) − 10 a = ( 72 )( 3) + ( 48 )( 3) − 28 2 3 2 4 3 2 x = 562 in. ! v = 770 in./s ! a = 764 in./s 2 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 5. Position: Velocity: x = 500sin kt mm v= a= and k = 10 rad/s kt = (10 )( 0.05 ) = 0.5 rad x = 500sin ( 0.5 ) dx = 500k cos kt mm/s dt dv = − 500k 2 sin kt mm /s 2 dt Acceleration: When t = 0.05 s, x = 240 mm ! v = 4390 mm/s ! a = − 24.0 × 103 mm/s 2 ! v = ( 500 )(10 ) cos ( 0.5 ) a = − ( 500 )(10 ) sin ( 0.5 ) 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 6. Position: Where Let x = 50sin k1t − k2t 2 mm k1 = 1 rad/s dθ = (1 − t ) rad/s dt x = 50sin θ mm v= and k2 = 0.5 rad/s 2 ( ) θ = k1t − k2t 2 = t − 0.5t 2 rad and d 2θ = −1 rad/s 2 dt 2 Position: Velocity: Acceleration: dx dθ = 50cosθ mm/s dt dt dv a= dt a = 50cosθ d 2θ ⎛ dθ ⎞ 2 − 50sin θ ⎜ ⎟ mm/s dt dt 2 ⎝ ⎠ either cosθ = 0 t =1s 2 When v = 0, or dθ =1− t = 0 dt Over 0 ≤ t ≤ 2 s, values of cosθ are: t (s) 0 0 1.0 0.5 1.0 0.5 0.878 1.5 0.375 0.981 2.0 0 1.0 θ ( rad ) cosθ 0.375 0.931 No solutions cosθ = 0 in this range. For t = 1 s, θ = 1 − ( 0.5 )(1) = 0.5 rad x = 50sin ( 0.5 ) x = 24.0 mm W a = − 43.9 mm/s 2 W a = 50cos ( 0.5 )( −1) − 50sin ( 0.5 )( 0 ) 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 7. Given: Differentiate twice. x = t 3 − 6t 2 + 9t + 5 v= a= (a) When velocity is zero. dx = 3t 2 − 12t + 9 dt dv = 6t − 12 dt v=0 3t 2 − 12t + 9 = 3 ( t − 1)( t − 3) = 0 t = 1 s and t = 3 s W (b) Position at t = 5 s. x5 = ( 5 ) − ( 6 )( 5 ) + ( 9 )( 5 ) + 5 3 2 x5 = 25 ft W a5 = 18 ft/s 2 W Acceleration at t = 5 s. a5 = ( 6 )( 5 ) − 12 Position at t = 0. x0 = 5 ft Over 0 ≤ t < 1 s Over 1 s < t < 3 s Over 3 s < t ≤ 5 s Position at t = 1 s. x1 = (1) − ( 6 )(1) + ( 9 )(1) + 5 = 9 ft 3 2 x is increasing. x is decreasing. x is increasing. Position at t = 3 s. x3 = ( 3) − ( 6 )( 3) + ( 9 )( 3) + 5 = 5 ft 3 2 Distance traveled. At t = 1 s At t = 3 s At t = 5 s d1 = x1 − x0 = 9 − 5 = 4 ft d3 = d1 + x3 − x1 = 4 + 5 − 9 = 8 ft d5 = d3 + x5 − x3 = 8 + 25 − 5 = 28 ft d5 = 28 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 8. x = t 2 − ( t − 2 ) ft 3 v= (a) Positions at v = 0. dx 2 = 2t − 3 ( t − 2 ) ft/s dt 2t − 3 ( t − 2 ) = − 3t 2 + 14t − 12 = 0 2 t= −14 ± (14) 2 − (4)(− 3)(−12) (2)(− 3) t1 = 1.1315 s and t2 = 3.535 s At t1 = 1.1315 s, At t2 = 3.535 s, (b) Total distance traveled. At t = t0 = 0, At t = t4 = 4 s, Distances traveled. 0 to t1: t1 to t2: t2 to t4: Adding, d1 = 1.935 − 8 = 6.065 ft d 2 = 8.879 − 1.935 = 6.944 ft d3 = 8 − 8.879 = 0.879 ft d = d1 + d 2 + d3 d = 13.89 ft W x0 = 8 ft x4 = 8 ft x1 = 1.935 ft x2 = 8.879 ft x1 = 1.935 ft W x2 = 8.879 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 9. a = 3e− 0.2t ∫ 0 dv = ∫ 0 a dt v −0=∫ t 3e− 0.2t dt 0 v t 3 = e− 0.2t − 0.2 t 0 v = −15 e− 0.2t − 1 = 15 1 − e− 0.2t At t = 0.5 s, ( ) ( ) v = 1.427 ft/s W v = 15 1 − e− 0.1 ( ) t ∫ 0 dx = ∫ 0 v dt x − 0 = 15∫ 1 − e t 0 x t ( − 0.2t ) 1 − 0.2t ⎞ ⎛ dt = 15 ⎜ t + e ⎟ 0.2 ⎝ ⎠0 x = 15 t + 5e− 0.2t − 5 At t = 0.5 s, ( ) x = 15 0.5 + 5e− 0.1 − 5 ( ) x = 0.363 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 10. Given: a = − 5.4sin kt ft/s 2 , t t v0 = 1.8 ft/s, x0 = 0, 5.4 cos kt k t k = 3 rad/s v − v0 = ∫ 0 a dt = − 5.4 ∫ 0 sin kt dt = v − 1.8 = Velocity: 0 5.4 ( cos kt − 1) = 1.8cos kt − 1.8 3 v = 1.8cos kt ft/s x − x0 = ∫ 0 v dt = 1.8 ∫ 0 cos kt dt = x−0= t t 1.8 sin kt k t 0 1.8 ( sin kt − 0 ) = 0.6sin kt 3 Position: When t = 0.5 s, x = 0.6sin kt ft kt = ( 3)( 0.5 ) = 1.5 rad v = 1.8cos1.5 = 0.1273 ft/s x = 0.6sin1.5 = 0.5985 ft v = 0.1273 ft/s W x = 0.598 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 11. Given: a = − 3.24sin kt − 4.32 cos kt ft/s 2 , x0 = 0.48 ft, t k = 3 rad/s v0 = 1.08 ft/s t t v − v0 = ∫ 0 a dt = − 3.24 ∫ 0 sin kt dt − 4.32 ∫ 0 cos kt dt v − 1.08 = = 3.24 cos kt k t 0 − 4.32 sin kt k t 0 3.24 4.32 ( cos kt − 1) − ( sin kt − 0 ) 3 3 = 1.08cos kt − 1.08 − 1.44sin kt Velocity: v = 1.08cos kt − 1.44sin kt ft/s x − x0 = ∫ 0 v dt = 1.08 ∫ 0 cos kt dt − 1.44 ∫ 0 sin kt dt x − 0.48 = = 1.08 sin kt k t 0 t t t + 1.44 cos kt k t 0 1.08 1.44 ( sin kt − 0 ) + ( cos kt − 1) 3 3 = 0.36sin kt + 0.48cos kt − 0.48 Position: When t = 0.5 s, x = 0.36sin kt + 0.48cos kt ft kt = ( 3)( 0.5 ) = 1.5 rad v = 1.08cos1.5 − 1.44sin1.5 = −1.360 ft/s x = 0.36sin1.5 + 0.48cos1.5 = 0.393 ft v = −1.360 ft/s ! x = 0.393 ft ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 12. Given: At t = 0, v = 400 mm/s; a = kt mm/s 2 at t = 1 s, where k is a constant. x = 500 mm v = 370 mm/s, 1 v t t 2 ∫ 400 dv = ∫ 0 a dt = ∫ 0 kt dt = 2 kt v − 400 = 1 2 kt 2 or v = 400 + 1 2 kt 2 At t = 1 s, Thus v = 400 + 1 2 k (1) = 370, 2 k = − 60 mm/s3 v = 400 − 30t 2 mm/s v7 = 400 − ( 30 )( 7 ) 2 At t = 7 s, When v = 0, 400 − 30t 2 = 0. v7 = −1070 mm/s W Then t 2 = 13.333 s2 , v>0 v 3.651 s, x t t 2 ∫ 500 dx = ∫ 1 v dt = ∫ 1 ( 400 − 30t ) dt x − 500 = 400t − 10t 3 ( ) t 1 = 400t − 10t 3 − 390 Position: At t = 0, At t = 3.651 s, x = 400t − 10t 3 + 110 mm x = x0 = 110 mm x = xmax = ( 400 )( 3.651) − (10 )( 3.651) + 110 = 1083.7 mm x = x7 = ( 400 )( 7 ) − (10 )( 7 ) + 110 3 3 At t = 7 s, Distances traveled: Over 0 ≤ t ≤ 3.651 s, x7 = − 520 mm W d1 = xmax − x0 = 973.7 mm d 2 = xmax − x7 = 1603.7 mm d = d1 + d 2 = 2577.4 mm d = 2580 mm W Over 3.651 ≤ t ≤ 7 s, Total distance traveled: Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 13. Determine velocity. v t t ∫ − 0.15 dv = ∫ 2 a dt = ∫ 2 0.15 dt v − ( −0.15 ) = 0.15t − ( 0.15 )( 2 ) v = 0.15t − 0.45 m/s At t = 5 s, When v = 0, For 0 ≤ t ≤ 3.00 s, For 3.00 ≤ t ≤ 5 s, Determine position. v5 = ( 0.15 )( 5 ) − 0.45 0.15t − 0.45 = 0 v ≤ 0, v ≥ 0, t = 3.00 s v5 = 0.300 m/s W x is decreasing. x is increasing. x t t ∫ −10 dx = ∫ 0 v dt = ∫ 0 ( 0.15t − 0.45) dt x − ( −10 ) = 0.075t 2 − 0.45t ( ) t 0 = 0.075t 2 − 0.45t x = 0.075 t 2 − 0.45t − 10 m At t = 5 s, x5 = ( 0.075 )( 5 ) − ( 0.45 )( 5 ) − 10 = −10.375 m 2 x5 = −10.38 m W At t = 0, x0 = −10 m (given) x3 = xmin = ( 0.075 )( 3.00 ) − ( 0.45 )( 3.00 ) − 10 = −10.675 mm 2 At t = 3.00 s, Distances traveled: Over 0 ≤ t ≤ 3.00 s, Over 3.00 s < t < 5 s, Total distance traveled: d1 = x0 − xmin = 0.675 m d 2 = x5 − xmin = 0.300 m d = d1 + d 2 = 0.975 m d = 0.975 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 14. Given: Separate variables and integrate. v t 2 ∫ 0 dv = ∫ a dt = ∫ 0 ( 9 − 3t ) dt = 9 a = 9 − 3t 2 v − 0 = 9 t − t3 (a) When v is zero. t (9 − t 2 ) = 0 t = 0 and t = 3 s (2 roots) (b) Position and velocity at t = 4 s. v = t 9 − t2 ( ) t =3sW x t t 3 ∫ 5 dx = ∫ 0 v dt = ∫ 0 ( 9t − t ) dt x−5= 9 2 1 4 t − t 2 4 9 2 1 4 t − t 2 4 x=5+ At t = 4 s, ⎛9⎞ 2 ⎛1⎞ 4 x4 = 5 + ⎜ ⎟ ( 4 ) − ⎜ ⎟ ( 4 ) ⎝2⎠ ⎝4⎠ v4 = ( 4 ) 9 − 42 x4 = 13 m W v4 = − 28 m/s W ( ) (c) Distance traveled. Over 0 < t < 3 s, Over 3 s < t ≤ 4 s, At t = 3 s, At t = 3 s At t = 4 s v is positive, so x is increasing. v is negative, so x is decreasing. ⎛9⎞ 2 ⎛1⎞ 4 x3 = 5 + ⎜ ⎟ ( 3) − ⎜ ⎟ ( 3) = 25.25 m ⎝2⎠ ⎝4⎠ d3 = x3 − x0 = 25.25 − 5 = 20.25 m d 4 = d3 + x4 − x3 = 20.25 + 13 − 25.25 = 32.5 m d 4 = 32.5 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 15. Given: Separate variables Integrate using dv = kt2 dt v = –10 m/s when t = 0 and v = 10 m/s when t = 2 s. a= dv = kt 2 dt 10 2 2 ∫ −10 dv = ∫ 0 kt dt v 10 − 10 = 1 3 kt 3 1 t 0 3 ⎤ [(10) − (−10)] = 3 k ⎡ ⎢( 2 ) − 0 ⎦ ⎥ ⎣ (a) (b) Solving for k, Equations of motion. k= ( 3)( 20 ) 8 k = 7.5 m/s 4 W Using upper limit of v at t, v v −10 1 = kt 3 3 t 0 ⎛1⎞ v + 10 = ⎜ ⎟ ( 7.5 ) t 3 ⎝3⎠ v = −10 + 2.5 t 3 m/s W Then, dx = v = −10 + 2.5 t 3 dt dx = −10 + 2.5 t 3 dt Separate variables and integrate using x = 0 when t = 2 s. ( ∫ x dx 0 = ∫ −10 + 2.5 t 3 dt t 2 t 2 ( ) ) 4⎤ x−0=⎡ ⎣ −10 t + 0.625 t ⎦ ⎡( −10 )( 2 ) + ( 0.625 )( 2 )4 ⎤ =⎡ ⎥ ⎣ −10 t + 0.0625 t ⎦ − ⎢ ⎣ ⎦ 4⎤ = −10 t + 0.625 t 4 − [ −10] x = 10 − 10t + 0.625t 4 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 16. Note that a is a given function of x. a = 40 − 160 x = 160 ( 0.25 − x ) (a) Note that v is maximum when a = 0, or x = 0.25 m Use v dv = a dx = 160 ( 0.25 − x ) dx with the limits v = 0.3 m/s when x = 0.4 m and v = vmax when x = 0.25 m vmax 0.25 ∫ 0.3 v dv = ∫ 0.4 160 ( 0.25 − x ) dx 2 ( 0.25 − x ) vmax 0.32 − = −160 2 2 2 2 0.25 0.4 ⎡ ( − 0.15)2 ⎤ ⎥ = 1.8 = −160 ⎢0 − 2 ⎢ ⎥ ⎣ ⎦ 2 vmax = 3.69 m 2 /s 2 vmax = 1.921 m/s W (b) Note that x is maximum or minimum when v = 0. Use v dv = a dx = 160 ( 0.25 − x ) with the limits v = 0.3 m/s when x = 0.4 m, 0 and v = 0 when x = xm ∫ 0.3 v dv = ∫ 0.4 160 ( 0.25 − x ) dx 2 0.3) ( 0− 2 0.25 − x ) ( = −160 xm xm 2 2 = − 80 ( 0.25 − xm ) + ( 80 )( − 0.15 ) 0.4 2 2 ( 0.25 − xm )2 = 0.02306 0.25 − xm = ± 0.1519 m xm = 0.0981 m and 0.402 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 17. a is a function of x: a = 100 ( 0.25 − x ) m/s 2 Use v dv = a dx = 100 ( 0.25 − x ) dx with limits v = 0 when x = 0.2 m ∫ 0 v dv = ∫ 0.2100 ( 0.25 − x ) dx v x 1 2 1 2 v − 0 = − (100 )( 0.25 − x ) 2 2 0.2 x = − 50 ( 0.25 − x ) + 0.125 2 So v 2 = 0.25 − 100 ( 0.25 − x ) Use dx = v dt t x ∫ 0 dt = ± ∫ 0.2 2 or v = ± 0.5 1 − 400 ( 0.25 − x ) 2 or dt = dx dx dx = 2 v ± 0.5 1 − 400 ( 0.25 − x ) 2 Integrate: 0.5 1 − 400 ( 0.25 − x ) Let u = 20 ( 0.25 − x ) ; So when x = 0.2 u = 1 u 1 and du = − 20dx u 1 1  −1 π = m sin −1 u = m t = m∫  sin u −  2 10 10  2 10 1 − u 1 du sin −1 u = Solve for u. π 2 m 10t π  u = sin  m 10t  = cos ( ± 10t ) = cos10t 2  u = cos 10t = 20 ( 0.25 − x ) continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Solve for x and v. x = 0.25 − v= Evaluate at t = 0.2 s. 1 cos10t 20 1 sin10t 2 1 cos ( (10 )( 0.2 ) ) 20 x = 0.271 m W v = 0.455 m/s W x = 0.25 − v= 1 sin ( (10 )( 0.2 ) ) 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 18. Note that a is a given function of x Use Using the limits and v dv = a dx = 600 x 1 + kx 2 dx = 600 x + 600kx3 dx v = 7.5 ft/s v = 15 ft/s when x = 0, when x = 0.45 ft, 3 ( ) ( ) ∫ 7.5 v dv = ∫ 0 ⎡ v2 ⎤ ⎢ ⎥ ⎣2⎦ 15 15 0.45 ( 600x + 600kx ) dx 0.45 0 7.5 ⎡ 600 2 600 4 ⎤ =⎢ x + kx ⎥ 4 ⎣ 2 ⎦ 2 (15)2 2 − ( 7.5)2 2 = ( 300 )( 0.45 ) + (150 ) k ( 0.45 ) 4 84.375 = 60.75 + 6.1509k Solving for k , k = 3.84 ft −2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 19. Note that a is a given function of x. Use v dv = a dx = 800 x + 3200 x3 dx ( ) Using the limit v = 10 ft/s when x = 0, v x 3 ∫10 v dv = ∫ 0 ( 800 x + 3200 x ) dx v 2 (10 ) − = 400 x 2 + 800 x 4 2 2 2 v 2 = 1600 x 4 + 800 x 2 + 100 Then where u1 and u2 are the roots of Solving the quadratic equation, Let u = x 2 v 2 = 1600u 2 + 800u + 100 = 1600 ( u − u1 )( u − u2 ) , 1600u 2 + 800u + 100 = 0 u1,2 = − 800 ± (800 )2 − ( 4 )(1600 )(100 ) ( 2 )(1600 ) = − 800 ± 0 = − 0.25 ± 0 3200 u1 = u2 = − 0.25 ft 2 So Taking square roots, v 2 = 1600 ( u + 0.25 ) = 1600 x 2 + 0.52 2 ( ) 2 ft 2 /s 2 v = ± 40 x 2 + 0.52 ft/s ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Use dx = v dt or dt = dx dx =± 2 v 40 x + 0.52 ( ) t=0 40dt = ± t dx x + 0.52 2 Use limit x=0 when 40∫ 0 dt = ± ∫ 0 x dx 1 x =± tan −1 2 0.5 0.5 x + 0.5 2 40t = ± 2.0 tan −1 ( 2 x ) 2 x = ± tan ( 20t ) v= or or tan −1 ( 2 x ) = ± 20t x = ± 0.5 tan ( 20t ) dx 2 2  = ± 0.5  sec ( 20t )  ( 20 ) = ± 10 sec ( 20t ) dt At t = 0, v = ± 10 ft/s, which agrees with the given data if the minus sign is rejected. Thus, At t = 0.05 s, v = 10 sec 2 ( 20t ) ft/s, 20t = 1.0 rad v = 10sec2 (1.0 ) = x = 0.5 tan (1.0 ) and x = 0.5 tan ( 20t ) ft 10 cos 2 1.0 v = 34.3 ft/s W x = 0.779 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 20. Note that a is a given function of x. 7  a = 12 x − 28 = 12  x −  m/s 2 3  7  Use v dv = a dx = 12  x −  dx with the limits v = 8 m/s when x = 0. 3  ∫ v v dv 8 7 = 12∫ −  dx 3  x x 0  v2    2    v 8 12  7 = x −  2 3 2 x 0 2 2 v 2 82 12  7 7  − =  x −  −    2 2 2  3 3    2 2 2  7 7 4 7   v = 8 + 12  x −  −    = 12  x −  − 3 3 3 3      2 2 7 4  v = ± 12  x −  − 3 3  Reject minus sign to get v = 8 m/s at x = 0. 2 (a) Maximum value of x. v = 0 when x = xmax 7 4  12  x −  − = 0 3 3  x− 7 1 =± 3 3 2 or 7 1  x − 3 = 9   and xmax = 8 2 m=2 m 3 3 2 xmax = 2 m Now observe that the particle starts at x = 0 with v > 0 and reaches x = 2 m. At x = 2 m, v = 0 and 2 a < 0, so that v becomes negative and x decreases. Thus, x = 2 m is never reached. 3 xmax = 2 m ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (b) Velocity when total distance traveled is 3 m. The particle will have traveled total distance d = 3 m when d − xmax = xmax − x or 3 − 2 = 2 − x or x = 1 m. 7 4  Using v = − 12  x −  − , which applies when x is decreasing, we get 3 3  7 4  v = − 12 1 −  − = − 20 3 3  2 2 v = − 4.47 m/s ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 21. Note that a is a function of x. a = k 1 − e− x ( ) Use v dv = a dx = k 1 − e− x dx with the limits v = 9 m/s when x = −3 m, and v = 0 when x = 0. 0 0 −x ∫ 9 v dv = ∫ − 3 k (1 − e ) dx ( ) ⎛ v2 ⎞ ⎜ ⎜ 2⎟ ⎟ ⎝ ⎠ 0 = k x + e− x 9 ( ) 0 −3 0− (a) 92 3⎤ = k⎡ ⎣0 + 1 − ( − 3) − e ⎦ = −16.0855k 2 k = 2.5178 k = 2.52 m/s 2 W Use v dv = a dx = k 1 − e− x dx = 2.5178 1 − e− x dx with the limit v = 0 when x = 0. ( ) ( ) ∫ 0 v dv = ∫ 0 2.5178 (1 − e v x −x ) dx x 0 v2 = 2.5178 x + e− x 2 ( ) = 2.5178 x + e− x − 1 ( ) ) 1/2 v 2 = 5.0356 x + e− x − 1 (b) Letting x = −2 m, ( ) ) 1/ 2 v = ± 2.2440 x + e− x − 1 ( v = ± 2.2440 − 2 + e2 − 1 Since x begins at x = − 2 m and ends at x = 0, v > 0. Reject the minus sign. ( = ± 4.70 m/s v = 4.70 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 22. a=v v dv = 6.8 e−0.00057 x dx x −0.00057 x ∫ 0 v dv = ∫ 0 6.8 e dx x 0 v2 6.8 e−0.00057 x −0= 2 − 0.00057 = 11930 1 − e−0.00057 x ( ) When v = 30 m/s. ( 30 )2 2 = 11930 1 − e−0.00057 x ( ) 1 − e−0.00057 x = 0.03772 e−0.00057 x = 0.96228 − 0.00057 x = ln (0.96228) = − 0.03845 x = 67.5 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 23. Given: or a=v dv = − 0.4v dx dv = − 0.4 dx Separate variables and integrate using v = 75 mm/s when x = 0. ∫ 75 dv = − 0.4∫ 0 (a) Distance traveled when v = 0 0 − 75 = − 0.4 x v x v − 75 = − 0.4 x x = 187.5 mm W (b) Time to reduce velocity to 1% of initial value. v = (0.01)(75) = 0.75 t = − 2.5ln 0.75 75 t = 11.51 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 24. Given: a =v dv = − kv2 dx Separate variables and integrate using v = 9 m/s when x = 0. v x ∫ 9 v = − k ∫ 0 dx dv ln Calculate k using v = 7 m/s when x = 13 m. ln Solve for x. (a) Distance when v = 3 m/s. 7 = − ( k )(13) 9 x=− v = − kx 9 k = 19.332 × 10−3 m −1 1 v v ln = − 51.728 ln 9 9 k ⎛3⎞ x = − 51.728 ln ⎜ ⎟ ⎝9⎠ x = 56.8 m W (b) Distance when v = 0. x = − 51.728 ln ( 0 ) x=∞W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 25. v dv = a dx = −k vdx, 1 dx = − v1/2dv k x v 3/2 ∫ x0 dx = − k ∫ v0 vdv = − 3k v x0 = 0, v0 = 25 ft/s 1 2 v v0 x − x0 = 2 3/2 v0 − v3/2 3k ( ) or x= 2 ⎡ 2 ⎡ = 125 − v3/2 ⎤ ( 25)3/2 − v3/2 ⎤ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 3k 3k Noting that x = 6 ft when v = 12 ft/s, 6= Then, 2 ⎡ 55.62 125 − 123/2 ⎤ ⎦ = k 3k ⎣ or k = 9.27 ft/s3 x= 2 ⎡125 − v3/2 ⎤ = 0.071916 125 − v3/2 ⎦ ( 3)( 9.27 ) ⎣ v3/2 = 125 − 13.905 x ( ) 3/2 (a) When x = 8 ft, v3/2 = 125 − (13.905 )( 8 ) = 13.759 ( ft/s ) v = 5.74 ft/s W (b) dv = a dt = − k vdt dt = − t =− 1 dv k v1/ 2 v 1 2 1/2 ⋅ 2⎡ v1/2 ⎤ = v0 − v1/2 ⎣ ⎦ v0 k k 1/2 ( ) t = 1.079 s W At rest, v = 0 ( 2 )( 25) 2v1/2 t = 0 = 9.27 k Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 29. x as a function of v. v = 1 − e−0.00057 x 154 ⎛ v ⎞ e −0.00057 x = 1 − ⎜ ⎟ ⎝ 154 ⎠ ⎡ ⎛ v ⎞2 ⎤ − 0.00057 x = ln ⎢1 − ⎜ ⎟ ⎥ ⎢ ⎣ ⎝ 154 ⎠ ⎥ ⎦ ⎡ ⎛ v ⎞2 ⎤ x = −1754.4 ln ⎢1 − ⎜ ⎟ ⎥ 154 ⎝ ⎠ ⎥ ⎢ ⎣ ⎦ a as a function of x. v 2 = 23716 1 − e−0.00057 a=v (1) 2 ( ) dv d ⎛ v2⎞ = = 11858 )( 0.00057 ) e−0.0005 x ⎜ ⎟ ⎟ ( 2 dx dx ⎜ ⎝ ⎠ (2) ⎡ ⎛ v ⎞2 ⎤ a = 6.75906 e −0.00057 x = 6.75906 ⎢1 − ⎜ ⎟ ⎥ 154 ⎝ ⎠ ⎥ ⎢ ⎣ ⎦ (a) v = 20 m/s. From (1), From (2), (b) v = 40 m/s. From (1), From (2), x = 122.54 a = 6.30306 x = 122.5 m e a = 6.30 m/s2 e x = 29.843 a = 6.64506 x = 29.8 m e a = 6.65 m/s2 e Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 30. Given: v = 7.5 (1 − 0.04 x ) 0.3 with units km and km/h (a) Distance at t = 1 hr. Using dx = v dt , we get dt = dx dx = v 7.5(1 − 0.04 x)0.3 Integrating, using t = 0 when x = 0, t x ⋅ or [t ]t0 = 1 − 0.04 x0.7 } { ∫ 0 dt = 7.5 ∫ 0 0.3 0 ( 7.5) ( 0.7 )( 0.04 ) (1 − 0.04 ) 1 dx 1 −1 x t = 4.7619 1 − (1 − 0.04 x ) { 0.7 } (1) Solving for x, x = 25 1 − (1 − 0.210t ) { 1/0.7 } When t = 1 h, (b) Acceleration when t = 0. x = 25 1 − ⎡ ⎣1 − ( 0.210 )(1) ⎤ ⎦ { 1/0.7 } x = 7.15 km W dv = (7.5)(0.3)(− 0.04)(1 − 0.04 x)−0.7 = − 0.0900(1 − 0.04 x) − 0.7 dx When t = 0 and x = 0, v = 7.5 km/h, a=v dv − 0.0900 h −1 dx dv = (7.5)(−0.0900) = − 0.675 km/h 2 dx (0.675)(1000) m/s 2 (3600)2 a = − 52.1 × 10−6 m/s 2 W =− (c) Time to run 6 km. Using x = 6 km in equation (1), t = 4.7619 1 − ⎡ ⎣1 − ( 0.04 )( 6 ) ⎤ ⎦ { 0.7 } = 0.8323 h t = 49.9 min W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 31. The acceleration is given by v dv gR 2 =a=− 2 dr r gR 2dr r2 Then, v dv = − Integrating, using the conditions v = 0 at r = ∞, and v = vesc at r = R 0 2 ∞ ∫ vesc v dv = − gR ∫ R r 2 dr 1 2 v 2 0 vesc ⎛1⎞ = gR 2 ⎜ ⎟ ⎝r⎠ ∞ R 0− 1 2 1⎞ ⎛ vesc = gR 2 ⎜ 0 − ⎟ 2 R⎠ ⎝ vesc = 2 gR Now, R = 3960 mi = 20.909 × 106 ft and g = 32.2 ft/s2 . Then, vesc = ( 2 )( 32.2 ) ( 20.909 × 106 ) vesc = 36.7 × 103 ft/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 32. The acceleration is given by a = − 32.2 ⎡ ⎛ y ⎞⎤ 1+ ⎜ 6 ⎟⎥ ⎢ 20.9 10 × ⎠⎦ ⎣ ⎝ 2 vdv = ady = − 32.2dy ⎡ ⎛ y ⎞⎤ 1+ ⎜ 6 ⎟⎥ ⎢ 20.9 10 × ⎠⎦ ⎣ ⎝ 2 Integrate, using the conditions v = v0 at y = 0 and v = 0 at y = ymax . Also, use g = 32.2 ft/s 2 and R = 20.9 × 106 ft. dy dy 1 2 v 2 0 v0 ∫ 0 v v0 dv = − g ∫ ∞ 0 (1 + ) y R 2 = ∞ − gR 2 0 ∫ ( R + y )2 1 ⎞ = gR ⎜ ⎟ ⎝R + y⎠ 2⎛ ymax 0 0− ⎡ gRymax 1 2 1 1⎤ − ⎥=− v0 = gR 2 ⎢ R + ymax 2 ⎣ R + ymax R ⎦ 2 v0 ( R + ymax ) = 2 gRymax Solving for ymax , Using the given numerical data, ymax = 2 Rv0 2 2 gR − v0 ymax = 2 ( 2 )( 32.2 ) ( 20.9 × 106 ) − v0 6 2 2 20.9 × 106 v0 = 2 20.9 × 106 v0 2 1.34596 × 109 − v0 (a) v0 = 2400 ft/s, ymax (b) v0 = 4000 ft/s, ymax (c) v0 = 40000 ft/s, ymax ( 20.9 × 10 ) ( 2400 ) = (1.34596 × 10 ) − ( 2400) ( 20.9 × 10 ) ( 4000) = (1.34596 × 10 ) − ( 4000) ( 20.9 × 10 ) ( 40000) = (1.34596 × 10 ) − ( 40000 ) 9 2 ymax = 89.8 × 103 ft W 6 2 9 2 ymax = 251 × 103 ft W 6 2 9 2 = negative Negative value indicates that v0 is greater than the escape velocity. ymax = ∞ W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 33. (a) Given: v = v′ sin (ω nt + ϕ ) At t = 0, v = v0 = v′ sin ϕ or sin ϕ = v0 v′ (1) Let x be maximum at t = t1 when v = 0. Then, Using sin (ω nt1 + ϕ ) = 0 and or cos (ω nt1 + ϕ ) = ± 1 (2) dx =v dt dx = v dt Integrating, x=C − v′ ωn cos (ω nt + ϕ ) v′ cos ϕ v′ C = x0 + v′ cos ϕ At t = 0, x = x0 = C − v′ ωn or ωn Then, x = x0 + ωn v′ cos ϕ − ωn v′ cos (ω nt + ϕ ) using cos ω nt1 + ϕ = −1 (3) xmax = x0 + ωn cos ϕ + ω Solving for cos ϕ , With xmax = 2 x0 , Using cos ϕ = cos ϕ = ( xmax − x0 ) ω n v′ x0ω n −1 v′ −1 (4)  v0   x0ω n   v′  +  v′ − 1 = 1     2 2 sin 2 ϕ + cos 2ϕ = 1, or Solving for v′ gives v′ = (v 2 0 2 2 + x0 ωn ) 2 x0ω n (5) W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (b) Acceleration: a= dv = v′ω n cos (ω nt + ϕ ) dt Let v be maximum at t = t2 when a = 0. Then, cos (ω nt2 + ϕ ) = 0 From equation (3), the corresponding value of x is x = x0 + v′ ωn cos ϕ = x0 + v′  x0ω n v′   ′ − 1 = 2 x0 − ωn  v ωn  = 2 x0 − 2 2 2 2 ωn 3 1 v0 v0 + x0 = x0 − 2 2 x0ω n ( 2x0ω n )ω n 2  3 − x0  ( )  v0 x0ω n 2 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 34. (a ) πt ⎤ dx ⎡ = v = v0 ⎢1 − sin ⎥ dt T ⎦ ⎣ Integrating, using x = x0 = 0 when t = 0, πt ⎤ x t t ⎡ ∫ 0 dx = ∫ 0 v dt = ∫ 0 v0 ⎢1 − sin T ⎥ dt ⎣ ⎦ x x 0 vT πt ⎤ ⎡ = ⎢v0t + 0 cos ⎥ T ⎦ π ⎣ t 0 x = v0t + v0T π cos πt T − v0T π (1) When t = 3T , x = 3v0T + v0T vT ⎛ 2⎞ cos ( 3π ) − 0 = ⎜ 3 − ⎟ v0T T π π⎠ ⎝ x = 2.36v0T W a= When t = 3T , (b) Using equation (1) with t = T , πv πt dv = − 0 cos dt T T π v0 T cos 3π a= a=− π v0 T W x1 = v0T + Average velocity is vave = v0T π cos π − v0T 2⎞ ⎛ = v0T ⎜1 − ⎟ π π⎠ ⎝ Δx x1 − x0 ⎛ 2⎞ = = ⎜1 − ⎟ v0 T Δt π⎠ ⎝ vave = 0.363v0 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 35. 10 km/h = 2.7778 m/s (a) Acceleration during start test. a= dv dt 8.2 27.7778 ∫ 0 a dt = ∫ 2.7778 v dt 100 km/h = 27.7778 m/s 8.2 a = 27.7778 − 2.7778 (b) Deceleration during braking. a=v dv = dx a = 3.05 m/s 2 W 44 0 ∫ 0 a dx = ∫ 27.7778 v dv = a ( x) 44 0 1 = v2 2 ( ) 0 27.7778 44 a = − 1 ( 27.7778)2 2 deceleration = − a = 8.77 m/s 2 W a = − 8.77 m/s 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 36. 10 km/h = 2.7778 m/s (a) Distance traveled during start test. a= dv dt 100 km/h = 27.7778 m/s ∫ 0 a dt = ∫ v0 dv a= v − v0 t t v at = v − v0 a= 27.7778 − 2.7778 = 3.04878 m/s 2 8.2 v = v0 + at = 2.7778 + 3.04878 t x = ∫ 0 v dv = ∫ 0 2.7778 + 3.04878 t ) dt = ( 2.7778 )( 8.2 ) + (1.52439 )( 8.2 ) 2 t 8.2 x = 125.3 m W (b) Elapsed time for braking test. a=v dv dx ∫ 0 a dx = ∫ v0 v dv x v ax = a= v 2 v0 2 − 2 2 1 2 1 2 v − v0 = 0 − 27.77782 2x ( 2 )( 44 ) ( ) ( ) = − 8.7682 m/s 2 a= dv dt ∫ 0 a dt = ∫ v0 dv t v at = v − v0 t= v − v0 0 − 27.7778 = − 8.7682 a t = 3.17 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 37. Constant acceleration. v0 = v A = 0, v = v0 + at = at x = x0 + v0t + At point B, (a) Solving (2) for a, 1 2 1 2 at = at 2 2 and t = 30 s a = 6 ft/s 2 W vB = 180 ft/s W x0 = x A = 0 (1) (2) x = xB = 2700 ft a= 2 x ( 2 )( 2700 ) = t2 ( 30 )2 (b) Then, vB = at = ( 6 )( 30 ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 38. Constant acceleration. x0 = 0 v = v0 + at x = x0 + v0t + a= x = x0 + v0t + v= 540 = v= Then, from (3), Substituting into (1) and (2), 1 v0 2 v − v0 t 1 2 at 2 (1) (2) Solving (1) for a, (3) Then, 1 v − v0 2 1 1 t = x0 + ( v0 + v ) t = ( v0 + v ) t 2 t 2 2 x6 = 540 ft or v0 = 540 = 120 ft/s 4.5 At t = 6 s, and 1⎛ 1 ⎞ ⎜ v0 + v0 ⎟ ( 6 ) = 4.5v0 2⎝ 2 ⎠ 1 v0 = 60 ft/s 2 a= 60 − 120 60 =− ft/s 2 = − 10 ft/s 2 6 6 v = 120 − 10t x = 0 + 120t − 1 (10 ) t 2 2 At stopping, v = 0 or 120 − 10ts = 0 ts = 12 s 1 (10 )(12 )2 = 720 ft 2 Δt = 6 s W Δd = 180 ft W x = 0 + (120 )(12 ) − (a) Additional time for stopping = 12 s − 6 s (b) Additional distance for stopping = 720 ft − 540 ft Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 39. (a) During the acceleration phase x = x0 + v0t + 1 2 at 2 Using x0 = 0, and v0 = 0, and solving for a gives a= 2x t2 Noting that x = 130 m when t = 25 s, a= ( 2 )(130 ) ( 25)2 a = 0.416 m/s W (b) Final velocity is reached at t = 25 s. v f = v0 + at = 0 + ( 0.416 )( 25 ) (c) The remaining distance for the constant speed phase is Δx = 400 − 130 = 270 m For constant velocity, Total time for run: Δt = Δx 270 = = 25.96 s v 10.40 t = 51.0 s W v f = 10.40 m/s W t = 25 + 25.96 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 40. Constant acceleration. Then, Choose t = 0 at end of powered flight. y1 = 27.5 m and a = − g = − 9.81 m/s 2 t = 16 s. 1 2 1 at = y1 + v1t − gt 2 2 2 1 2 (a) When y reaches the ground, y f = 0 y f = y1 + v1t + y f − y1 + t v1 = gt 2 = 0 − 27.5 + 1 2 ( 9.81)(16 )2 16 = 76.76 m/s v1 = 76.8 m/s W (b) When the rocket reaches its maximum altitude ymax , v=0 2 2 v 2 = v1 + 2a ( y − y1 ) = v1 − 2 g ( y − y1 ) y = y1 − 2 v 2 − v1 2g ymax = 27.5 − 0 − ( 76.76 ) ( 2 )( 9.81) 2 ymax = 328 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 41. Place origin at 0. Motion of auto. ( x A )0 = 0, ( vA )0 = 0, a A = 0.75 m/s 2 x A = ( x A )0 + ( v A )0 t + x A = 0.375t 2 m Motion of bus. 1 ⎛1⎞ a At 2 = 0 + 0 + ⎜ ⎟ ( 0.75 ) t 2 2 ⎝2⎠ ( xB )0 = ?, ( vB )0 = − 6 m/s, aB = 0 xB = ( xB )0 − ( vB )0 t = ( xB )0 − 6t m At t = 20 s, xB = 0. 0 = ( xB )0 − ( 6 )( 20 ) Hence, xB = 120 − 6 t ( xB )0 = 120 m When the vehicles pass each other, xB = x A. 120 − 6t = 0.375 t 2 0.375 t 2 + 6 t − 120 = 0 t= t= Reject the negative root. Corresponding values of xA and xB. x A = ( 0.375 )(11.596 ) = 50.4 m 2 − 6 ± (6) 2 − ( 4 )( 0.375 )( −120 ) ( 2 )( 0.375) − 6 ± 14.697 = 11.596 s 0.75 and − 27.6 s t = 11.60 s W xB = 120 − ( 6 )(11.596 ) = 50.4 m x = 50.4 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 42. Place the origin at A when t = 0. Motion of A: ( x A )0 = 0, ( v A )0 = 15 km/h = 4.1667 m/s, a A = 0.6 m/s 2 v A = ( v A )0 + a At = 4.1667 + 0.6t x A = ( x A )0 + ( v A )0 t + 1 a At 2 = 4.1667 t + 0.3t 2 2 = 23 km/h = 6.3889 m/s, aB = − 0.4 m/s 2 Motion of B: ( xB )0 = 25 m, ( vB )0 vB = ( vB )0 + aBt = 6.3889 − 0.4t xB = ( xB )0 + ( vB )0 t + (a) When and where A overtakes B. 1 aBt 2 = 25 + 6.3889t − 0.2 t 2 2 x A = xB 4.1667 t + 0.3 t 2 = 25 + 6.3889 t − 0.2 t 2 0.5t 2 − 2.2222t − 25 = 0 t= 2.2222 ± 2.22222 − ( 4 )( 0.5 )( − 25 ) ( 2 )( 0.5) t = 2.2222 ± 7.4120 = 9.6343 s and − 5.19 s Reject the negative root. x A = ( 4.1667 )( 9.6343) + ( 0.3)( 9.6343) = 68.0 m xB = 25 + ( 6.3889 )( 9.6343) − ( 0.2 )( 9.6343) = 68.0 m 2 2 . t = 9.63 s W A moves 68.0 m W B moves 43.0 m W (b) Corresponding speeds. v A = 4.1667 + ( 0.6 )( 9.6343) = 9.947 m/s vB = 6.3889 − ( 0.4 )( 9.6343) = 2.535 m/s v A = 35.8 km/h W vB = 9.13 km/h W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 43. Constant acceleration ( a1 and a2 ) for horses 1 and 2. Let x = 0 and t = 0 when the horses are at point A. Then, x = v0t + a= 1 2 at 2 Solving for a, 2 ( x − v0t ) t2 Using x = 1200 ft and the initial velocities and elapsed times for each horse, a1 = 1200 − ( 20.4 )( 61.5 ) ⎤ x − v1t1 2 ⎡ ⎦ = − 0.028872 ft/s 2 = ⎣ 2 2 t1 ( 61.5) x − v2t2 2 ⎡ ⎣1200 − ( 21)( 62.0 ) ⎤ ⎦ = − 0.053070 ft/s 2 = 2 2 t2 ( 62.0 ) a2 = Calculating x1 − x2 , x1 − x2 = ( v1 − v2 ) t + 1 ( a1 − a2 ) t 2 2 1 2 ⎡( − 0.028872 ) − ( − 0.053070 ) ⎤ ⎦t 2⎣ x1 − x2 = ( 20.4 − 21) t + = − 0.6t + 0.012099 t 2 At point B, (a) x1 − x2 = 0 tB = 2 − 0.6tB + 0.012099 t B =0 0.6 = 49.59 s 0.012099 1 ( − 0.028872 )( 49.59 )2 2 1 ( −0.05307 )( 49.59 )2 = 976 ft 2 2 Calculating xB using data for either horse, Horse 1: Horse 2: xB = ( 20.4 )( 49.59 ) + xB = 976 ft W xB = ( 21)( 49.59 ) + When horse 1 crosses the finish line at t = 61.5 s, (b) x1 − x2 = − ( 0.6 )( 61.5 ) + ( 0.012099 )( 61.5 ) Δx = 8.86 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 44. Choose x positive upward. Rocket launch data: Constant acceleration a = − g Rocket A: x = 0, v = v0 , t = 0 x = 0, v = v0 , t = t B = 4 s Rocket B: Velocities: Rocket A: v A = v0 − gt Rocket B: vB = v0 − g ( t − t B ) Positions: Rocket A: x A = v0t − 1 2 gt 2 1 2 g ( t − tB ) , 2 t ≥ tB Rocket B: xB = v0 ( t − t B ) − For simultaneous explosions at x A = xB = 240 ft when t = t E , v0t E − Solving for v0 , Then, when t = t E , 1 2 1 1 2 1 2 2 gt E = v0 ( t E − t B ) − g ( t E − t B ) = v0t E − v0t B − gt E + gt E t B − gt B 2 2 2 2 v0 = gt E − gt B 2 or 2 − t Bt E − tE (1) 2xA =0 g gt  1 2  , x A =  gt E − B  t E − gt E 2  2  2 tB ± tB + ( 4 )(1) 2 xA g Solving for t E , tE = 2 ( ) = 4± 2 )( 240 ) ( 4 )2 + ( 4)(1)( 32.2 2 = 6.35 s Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (a) From equation (1), At time t E , v0 = ( 32.2 )( 6.348 ) − v A = v0 − gt E ( 32.2 )( 4 ) 2 v0 = 140.0 ft/s W vB = v0 − g ( t E − t B ) vB/ A = 128.8 ft/s W (b) vB − v A = gt B = ( 32.2 )( 4 ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 45. (a) Acceleration of A. v A = ( v A )0 + a At , ( v A )0 = 168 km/h = 46.67 m/s v A = 228 km/h = 63.33 m/s At t = 8 s, aA = v A − ( v A )0 t = 63.33 − 46.67 8 1 a At 2 2 xB = ( x B ) 0 + ( vB ) 0 t + a A = 2.08 m/s 2 W (b) x A = ( x A )0 + ( v A )0 t + 1 aBt 2 2 1 2 ⎤ x A − xB = ( x A ) 0 − ( x B ) 0 + ⎡ ⎣ ( v A ) 0 − ( vB ) 0 ⎦ t + 2 ( a A − a B ) t When t = 0, When t = 8 s, Hence, ( x A )0 − ( xB )0 = 38 m 0 = 38 + and ( vB ) 0 − ( v A ) 0 = 0 a A − aB = − 1.1875 aB = 3.27 m/s 2 W x A − xB = 0 1 ( a A − aB )(8)2 , 2 or aB = a A + 1.1875 = 2.08 + 1.1875 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 46. (a) Acceleration of A. v A = ( v A )0 + a At and and and x A = ( x A )0 + ( v A )0 t = 1 a At 2 2 Using ( v A )0 = 0 v A = a At ( xA )0 = 0 gives xA = 1 a At 2 2 When cars pass at t = t1, x A = 90 m 2 t1 = 2 x A ( 2 )( 90 ) 180 = = aA aA aA and v A = a At1 For 0 ≤ t ≤ 5 s, For t > 5 s, vB = ( vB )0 = − 96 km/h = − 26.667 m/s vB = ( vB )0 + aB ( t − 5 ) = − 26.667 + v A = − vB a At1 = 26.667 − 7 5 a At1 − a A = 26.667 6 6 1 a A ( t1 − 5 ) 6 or 7t1 − 5 = 160 aA 1 aA ( t − 5) 6 When vehicles pass, Using 1 , aA t1 = 180 7 180 160 gives −5= aA aA aA 7 180u − 5 = 160u 2 Let u = or 160u 2 − 7 180u + 5 = 0 continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Solving the quadratic equation, u = 7 180 ± ( 49 )(180 ) − ( 4 )(160 )( 5) ( 2 )(160 ) and 0.52776 and 1 = 285.2 m/s u2 = 93.915 ± 74.967 320 = 0.0592125 aA = 3.590 m/s The corresponding values for t1 are t1 = 180 = 0.794 s, 285.2 and t1 = 180 = 7.08 s 3.590 Reject 0.794 s since it is less than 5 s. Thus, (b) Time of passing. (c) Distance d. 0 ≤ t ≤ 5 s, At t = 5 s, For t > 5 s, a A = 3.59 m/s 2 W t = t1 = 7.08 s W xB = ( xB )0 − ( vB )0 t = d − 26.667t xB = d − ( 22.667 )( 5 ) = d − 133.33 xB = d − 133.33 + ( vB )0 ( t − 5 ) + xB = d − 133.33 − 26.667 ( t − 5 ) + 1 2 aB ( t − 5 ) 2 1  3.59  2 (t − 5)   2 6  When t = t1 = 7.08 s, 90 = d xB = x A = 90 ( 3.59 )( 2.08) − 133.33 − ( 26.667 )( 2.08 ) + ( 2 )( 6 ) 2 d = 90 + 133.33 + 55.47 − 1.29 d = 278 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 47. For t > 0, x A = ( x A )0 + ( v A )0 t + 1 1 a At 2 = 0 + 0 + ( 6.5 ) t 2 2 2 or x A = 3.25t 2 For t > 2 s, xB = ( x B ) 0 + ( v B ) 0 ( t − 2 ) + 2 1 1 2 2 aB ( t − 2 ) = 0 + 0 + (11.7 )( t − 2 ) 2 2 or For x A = xB , xB = 5.85 ( t − 2 ) = 5.85t 2 − 23.4t + 23.4 3.25t 2 = 5.85t 2 − 23.4t + 23.4, 2.60t 2 − 23.4t + 23.4 = 0 or Solving the quadratic equation, t = 1.1459 and t = 7.8541 s Reject the smaller value since it is less than 5 s. (a ) x A = xB = ( 3.25 )( 7.8541) 2 t = 7.85 s W x = 200 ft W v A = 51.1 ft/s W vB = 68.5 ft/s W (b) v A = ( v A )0 + a At = 0 + ( 6.5 )( 7.8541) vB = ( vB )0 + aB ( t − 2 ) = 0 + (11.7 )( 7.8541 − 2 ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 48. Let x be the position relative to point P. Then, Also, ( x A )0 = 0 and ( xB )0 = 0.62 mi = 3273.6 ft and ( vA )0 = 68 mi/h = 99.73 ft/s x A = ( x A )0 + ( v A )0 t + aA = ( vB )0 = − 39 mi/h = − 57.2 ft/s ⎤ 2⎡ ⎣ x A − ( x A ) 0 − ( v A )0 t ⎦ t2 (a) Uniform accelerations. 1 a At 2 2 2 or aA = 2⎡ ⎣3273.6 − 0 − ( 99.73)( 40 ) ⎤ ⎦ ( 40 ) = − 0.895 ft/s 2 a A = 0.895 ft/s 2 W xB = ( x B ) 0 + ( v B ) 0 + aB = 1 aBt 2 2 2 or aB = ⎤ 2⎡ ⎣ x B − ( x B ) 0 − ( vB ) 0 t ⎦ t2 2⎡ ⎣0 − 3273.6 − ( − 57.2 )( 42 ) ⎤ ⎦ ( 42 ) = − 0.988 ft/s 2 aB = 0.988 ft/s 2 W (b) When vehicles pass each other x A = xB . ( x A )0 + ( v A )0 t + 0 + 99.73t + 1 1 a At 2 = ( xB )0 + ( vB )0 t + aBt 2 2 2 1 1 ( − 0.895) t 2 = 3273.6 − 57.2t + ( − 0.988) t 2 2 2 −0.0465t 2 − 156.93t + 3273.6 = 0 Solving the quadratic equation, Reject the negative value. Then, (c) Speed of B. vB = ( vB )0 + aBt = − 57.2 + ( − 0.988 )( 20.7 ) = − 77.7 ft/s vB = 77.7 ft/s W t = 20.7 s and − 3390 s t = 20.7 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 49. Let x be positive downward for all blocks and for point D. v A = 1 m/s Constraint of cable supporting A: x A + ( x A − xB ) = constant 2v A − vB = 0 or vB = 2v A = ( 2 )(1) = 2 m/s Constraint of cable supporting B: 2 xB + xC = constant vC + 2vB = 0 (a) (b) (c) vB/ A = vB − v A = 2 − 1 xD + xC = constant, vD + vC = 0 vD = − vC = 4 m/s vD/ A = vD − v A = 4 − 1 or vC = − 2vB = − ( 2 )( 2 ) = − 4 m/s vC = 4 m/s v B/ A = 1 m/s W W v D/ A = 3 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 50. Let x be positive downward for all blocks. Constraint of cable supporting A: x A + ( x A − xB ) = constant 2v A − vB = 0 or vB = 2v A and aB = 2a A Constraint of cable supporting B: 2 xB + xC = constant 2vB + vC = 0, or vC = − 2vB , and aC = − 2aB = − 4a A Since vC and aC are down, v A and a A are up, i.e. negative. 2 2 ⎤ vA − ( v A )0 = 2a A ⎡ ⎣ x A − ( x A )0 ⎦ 2 vA − ( v A )0 2 (a ) a A = ⎤ 2⎡ ⎣ x A − ( x A )0 ⎦ = ( 0.2 )2 − 0 = − 0.04 m/s2 ( 2 )( − 0.5) aC = − 4a A a A = 0.04 m/s 2 aC = 0.16 m/s 2 W W (b) aB = 2a A = ( 2 )( − 0.04 ) = − 0.08 m/s 2 ΔvB = aBt = ( − 0.08 )( 2 ) = − 0.16 m/s ΔxB = 1 1 2 aBt 2 = ( −0.08 )( 2 ) = − 0.16 m 2 2 ΔvB = 0.16 m/s ΔxB = 0.16 m W W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 51. Let xA, xB, xC, and xD be the displacements of blocks A, B, C, and D relative to the upper supports, increasing downward. Constraint of cable AB: v A + vB = 0 Constraint of cable BED: vB + 2vD = 0 Constraint of cable BCD: 2vC − vB − vD = 0 (a) Velocity of block A. 1 v A = − 2vC = − (2)(4) 2 (b) Velocity of block D. vD = 1 v A = − 4 ft/s 2 vD = 4 ft/s W x A + xB = constant vB = − v A xB + 2 xD = constant or 1 1 vD = − v B = v A 2 2 ( xC − xB ) + ( xC − xD ) = constant or 2vC + v A − 1 vA = 0 2 v A = − 8 ft/s v A = 8 ft/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 52. Let xA, xB, xC, xD, and xE be the displacements of blocks A, B, C, and D and cable point E relative to the upper supports, increasing downward. Constraint of cable AB: v A + vB = 0 a A + aB = 0 x A + xB = constant vB = − v A aB = − a A xB + 2 xD = constant Constraint of cable BED: vB + 2vD = 0 a B + 2aD = 0 1 1 v D = − vB = v A 2 2 1 1 aD = − a A = a A 2 2 ( xC − xB ) + ( xC − xD ) = constant Constraint of cable BCD: 2vC − vB − vD = 0 2aC − aB − aD = 0 2vC + v A = 0 2aC + 1 aA = 0 2 1 aC = − a A 4 Since block C moves downward, vC and aC are positive. Then, vA and aA are negative, i.e. upward. Also, vD and aD are negative. Relative motion: v A/D = v A − vD = a A/D = a A − aD = 1 vA 2 1 aA 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (a) Acceleration of block C. a A = 2a A/D = 2v A/D t = (2)(8) = 3.2 ft/s 2 5 aC = 0.8 ft/s 2 a A = − 3.2 ft/s 2 1 aC = − a A = 0.8 ft/s 2 4 W Constraint of cable portion BE: vB + vE = 0 xB + xE = constant aB + aE = 0 (b) Acceleration of point E. aE = − aB = a A = − 3.2 ft/s 2 aE = 3.2 ft/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 53. Let x be position relative to the right supports, increasing to the left. Constraint of entire cable: 2 x A + xB + ( xB − x A ) = constant 2vB + v A = 0 Constraint of point C of cable: v A = − 2vB 2 x A + xC = constant 2v A + vC = 0 vC = − 2v A (a) Velocity of collar A. v A = − 2vB = − ( 2 )( 300 ) = − 600 mm/s (b) Velocity of point C of cable. vC = − 2v A = − ( 2 )( −600 ) = 1200 mm/s (c) Velocity of point C relative to collar B. vC/B = vC − vB = 1200 − 300 = 900 mm/s vC/B = 900 mm/s W v A = 600 mm/s W vC = 1200 mm/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 54. Let x be position relative to the right supports, increasing to the left. Constraint of entire cable: 2vB + v A = 0, (a) Accelerations of A and B. 1 vB/A = vB − v A = − v A − v A 2 vA = − v A − ( v A )0 = a At , or 2 v A = − v B/A 3 or 2 x A + xB + ( xB − x A ) = constant, 1 vB = − v A , 2 and 1 aB = − a A 2 2 ( 610 ) = − 406.67 mm/s 3 aA = v A − ( v A )0 t = − 406.67 − 0 = − 50.8 mm/s2 8 a A = 50.8 mm/s 2 1 1 aB = − a A = − ( −50.8 ) 2 2 aB = 25.4 mm/s 2 W W (b) Velocity and change in position of B after 6 s. vB = ( vB )0 + aBt = 0 + ( 25.4 )( 6 ) xB − ( xB )0 = ( vB )0t + 1 1 2 aBt 2 = ( 25.4 )( 6 ) 2 2 vB = 152.5 mm/s ΔxB = 458 mm W W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 55. Let x be position relative to left anchor. At the right anchor, x = d . Constraint of cable: 2vB − 3v A = 0 Constraint of point D of cable: v A + vD = 0 (a) Accelerations of A and B. or or xB + ( xB − x A ) + 2 ( d − x A ) = constant vA = 2 vB 3 and aA = 2 aB 3 (d vD = − v A 2 3 − x A + d − xD ) = constant and aD = − a A ( vB )0 = 6 in./s ( vA )0 = ( 6 ) = 4 in./s 2 2 ⎤ vA − ( v A ) 0 = 2a A ⎡ ⎣ x A − ( x A )0 ⎦ 2 vA − ( v A )0 2 aA = aB = (b) Acceleration of point D. ⎤ 2⎡ ⎣ x A − ( x A )0 ⎦ = ( 2.4 )2 − ( 4 )2 ( 2 )(10 ) = − 0.512 in./s 2 a A = 0.512 in./s 2 aB = 0.768 in./s 2 aD = 0.512 in./s 2 W 3 3 a A = ( 0.512 ) = − 0.768 in./s 2 2 2 aD = − a A = − ( − 0.512 ) vB = ( vB )0 + aBt = 6 + ( − 0.768 )( 4 ) W W (c) Velocity of block B after 4 s. vB = 2.93 in./s W Change in position of block B. xB − ( xB ) 0 = ( vB ) 0 t + 1 1 2 aBt 2 = ( 6 )( 4 ) + ( − 0.768 )( 4 ) 2 2 ΔxB = 17.86 in. W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 56. Let x be position relative to left anchor. At right anchor x = d . Constraint of entire cable: (a) Velocity of A: xB + ( xB − x A ) + 2 ( d − x A ) = constant vA = 2 2 vB = (12 ) 3 3 xB + xB − xC = constant vC = 2vB = 2 (12 ) d − x A + d − xC = constant vD = − v A = − 8.00 in./s vC/ A = vC − v A = 24 − 8 v A + vD = 0, vD = 8.00 in./s vC/ A = 16.00 in./s W W 2vB − 3v A = 0 v A = 8.00 in./s W Constraint of point C of cable: (b) Velocity of C: Constraint of point D of cable: (c) Velocity of D: (d) Relative velocity. 2vB − vC = 0 vC = 24 in./s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 57. Let x be position relative to the anchor, positive to the right. Constraint of cable: − xB + ( xC − xB ) + 3 ( xC − x A ) = constant 4vC − 2vB − 3v A = 0 When t = 0, (a) vB = − 50 mm/s and 4aC − 2aB − 3a A = 0 (1, 2) ( va )0 = 100 mm/s ( vC )0 = 50 mm/s W ( vC )0 = 1 1 ⎡ 2vB + 3 ( v A ) ⎤ = ⎡ ( 2 )( − 50 ) + ( 3)(100 )⎤ ⎦ 0⎦ ⎣ 4 4⎣ Constraint of point D: ( xD − xA ) + ( xC − x A ) + ( xC − xB ) − xB = constant vD + 2vC − 2v A − 2vB = 0 (b) ( vD )0 = 2 ( v A )0 + 2vB − 2 ( vC )0 = ( 2 )(100 ) + ( 2 )( − 50 ) − ( 2 )( 50 ) xC − ( xC )0 = ( vC )0 t + 1 aC t 2 2 = ⎡ 40 − ( 50 )( 2 ) ⎦ ⎤ 2⎣ ( vD ) 0 = 0 W (c) aC = ⎤ 2⎡ ⎣ xC − ( xC )0 − ( vC )0 t ⎦ t 2 ( 2) 2 = − 30 mm/s 2 aC = 30 mm/s 2 W Solving (2) for aA aA = 1 1 2 ( 4aC − 2aB ) = ⎡ ( 4 )( −30 ) − ( 2 )( 0 )⎤ ⎦ = − 40 mm/s 3 3⎣ a A = 40 mm/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 58. Let x be position relative to the anchor, positive to the right. Constraint of cable: − xB + ( xC − xB ) + 3 ( xC − x A ) = constant 4vC − 2vB − 3v A = 0 (a) Accelerations of B and C. At t = 2 s, vC = v A = 420 mm/s and vB = − 30 mm/s and 4aC − 2aB − 3a A = 0 1 1 ( 2vB + 3v A ) = ⎡ ( 2 )( −30 ) + ( 3)( 420 )⎤ ⎦ = 300 mm/s 4 4⎣ ( vC )0 = 0 vC = ( vC )0 + aC t aB = aC = vC − ( vC )0 t = 300 − 0 2 aC = 150 mm/s 2 W 1 1 2 ( 4 )(150 ) − ( 3)( 270 )⎤ ( 4aC − 3a A ) = ⎡ ⎦ = −105 mm/s 2 2⎣ aB = 105 mm/s 2 W (b) Initial velocities of A and B. v A = ( v A )0 − a At ( vA )0 = v A − a At = 420 − ( 270 )( 2 ) = −120 mm/s ( vA )0 = 120 mm/s W W v B = ( vB ) 0 − a B t Constraint of point E: ( vB )0 = vB − aBt = − 30 − ( −105)( 2 ) vE − 3v A + 2vC = 0 ( vB )0 = 180 mm/s 2 ( xC − x A ) + ( xE − x A ) = constant (c) ( vE )0 = 3 ( vA )0 − 2 ( vC )0 = ( 3)( −120 ) − ( 2 )( 0 ) = − 360 mm/s ( vE )0 = 360 mm/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 59. Define positions as positive downward from a fixed level. Constraint of cable. ( xB − xA ) + ( xC − x A ) + 2 ( xC − xB ) = constant 3xC − xB − 2 x A = constant 3vC − vB − 2v A = 0 3aC − aB − 2a A = 0 Motion of block C. ( v A )0 = 0, a A = − 3.6 in./s 2 , vB = ( vB )0 = 18 in./s, aB = 0 ( vC )0 = aC = 1 ( vB ) + 2 ( v A )  = 6 in./s 0 0 3 1 1 2 0 + ( 2 )( − 3.6 )  ( a B + 2a A ) =   = − 2.4 in./s 3 3 vC = ( vC )0 + aC t = 6 − 1.2t xC − ( xC )0 = ( vC )0 t + 1 aC t 2 = 6t − 0.6t 2 2 (a) Time at vC = 0. 0 = 6 − 2.4t t = 2.5 s W (b) Corresponding position of block C. 2 1 xC − ( xC )0 = ( 6 )( 2.5 ) +   ( −2.4 )( 2.5 ) 2 xC − ( xC )0 = 7.5 in. W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 60. Define positions as positive downward from a fixed level. Constraint of cable: ( xB − xA ) + ( xC − x A ) + 2 ( xC − xB ) = constant 3xC − xB − 2 x A = constant 3vC − vB − 2v A = 0 3aC − aB − 2a A = 0 Motion of block C. ( vA )0 = 0, a A = − 2.5t in./s 2 , 1 ⎡ ( vB ) + 2 ( v A ) ⎤ = 0 0 0⎦ 3⎣ ( vB )0 = 0, aB = 15 in./s 2 ( vC )0 = aC = 1 1 ( aB + 2a A ) = (15 − 5t ) in./s2 3 3 t vC = ( vC )0 + ∫ 0 aC dt =0+ xC − ( xC )0 = (a) Time at vC = 0 1 15t − 2.5t 2 in./s 3 ( ) 1 7.5t 2 − 0.83333t 3 in. 3 ( ) 1 15t − 2.5t 2 = 0 t = 0 and t = 6 s 3 (b) Corresponding position of block C. 1 xC − ( xC )0 = 0 + ⎡ ( 7.5)( 6 )2 − ( 0.83333)( 6 )3 ⎤ ⎥ ⎣ ⎦ 3⎢ 0+ ( ) t =6sW xC − ( xC )0 = 30 in. W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 61. Let x be position relative to the support taken positive if downward. Constraint of cable connecting blocks A, B, and C: 2x A + 2 xB + xC = constant, 2v A + 2vB + vC = 0 2a A + 2aB + aC = 0 Constraint of cable supporting block D: (1) ( xD − xA ) + ( xD − xB ) = constant, 2aD − aB − a A = 0 Given: Given: aC/B = aC − aB = −120 aD/ A = aD − a A = 220 2a A + 2aB + ( aB − 120 ) = 0 2 ( a A + 220 ) − a A − aB = 0 Solving (5) and (6) simultaneously, a A = − 240 mm/s 2 From (3) and (4), (a) Velocity of C after 6 s. vC = ( vC )0 + aC t = 0 + ( 80 )( 6 ) (b) Change in position of D after 10 s. xD − ( x D ) 0 = ( v D ) 0 t + 2vD − v A − vB = 0 (2) aC = aB − 120 aD = a A + 220 (3) (4) or or Substituting (3) and (4) into (1) and (2), or or 2a A + 3aB = 120 a A − aB = − 440 (5) (6) and and aB = 200 mm/s 2 aD = − 20 mm/s 2 aC = 80 mm/s 2 vC = 480 mm/s W 1 1 2 aDt 2 = 0 + ( − 20 )(10 ) = −1000 mm 2 2 ΔxD = 1.000 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 62. Let x be position relative to the support taken positive if downward. Constraint of cable connecting blocks A, B, and C: 2 x A + 2 xB + xC = constant, 2v A + 2vB + vC = 0, 2a A + 2aB + aC = 0 ( vA )0 = ( vB )0 = ( vC )0 = 0, 2 2 ( xA )0 = ( xB )0 = ( xC )0 , ( xB/A )0 = 0, ( vB/A )0 = 0 ⎤ x − x ( vB/A ) − ( vB/A )0 = 2aP/A ⎡ ⎣ B/ A ( B/ A )0 ⎦ 2 vB / A − 0 = 2aB/ A ( xB − x A − 0 ) aB/ A = 2 ( xB − x A ) 2 vB /A = 402 = 10 mm/s 2 2 (160 − 80 ) xB/ A = xB/ A t2 = ( 1 aB/ At 2 = 0 + 0 + aB/ At 2 )0 + ( vB/A )0 t + 1 2 2 , or t= 2 ( xB − x A ) aB/ A = 2 (160 − 80 ) 10 =4s 2xB/ A aB/ A x A − ( x A )0 = ( v A )0 t + (a) aA = ⎤ 2⎡ ⎣ x A − ( x A )0 − ( v A )0 t ⎦ t 2 1 a At 2 2 2 = 2 ( 80 − 0 ) ( 4) a A = 10 mm/s 2 W aB = 20 mm/s 2 W aB = a A + aB/ A = 10 + 10 aC = − ( 2aB + 2a A ) = − ⎡ ⎣( 2 )( 20 ) + ( 2 )(10 ) ⎤ ⎦ = − 60 mm/s vC = ( vC )0 + aC t Constraint of cable supporting block D: t= vC − ( vC )0 aC = − 300 − 0 =5s − 60 ( xD − xA ) + ( xD − xB ) = constant, 2aD − a A − aB = 0, (b) xD − ( x D ) 0 aD = 2vD − v A − vB = 0 1 1 ( a A + aB ) = (10 + 20 ) = 15 mm/s 2 2 1 1 2 = ( vD )0 t + aDt 2 = 0 + (15 )( 5 ) ΔxD = 187.5 mm W 2 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 63. a−t curve A1 = −12 m/s, A2 = 8 m/s (a) v−t curve v6 = − 4 m/s v0 = v6 − A1 = − 4 − ( −12 ) = 8 m/s v10 = − 4 m/s (b) v14 = v10 + A2 = − 4 + 8 A3 = 16 m, A4 = − 4 m A5 = −16 m, A6 = − 4 m A7 = 4 m x−t curve x0 = 0 x4 = x0 + A3 = 16 m x6 = x4 + A4 = 12 m x10 = x6 + A5 = − 4 m x12 = x10 + A6 = − 8 m v14 = 4 m/s W (a) (b) Distance traveled: 0 ≤ t ≤ 4 s, x14 = x12 + A7 d1 = 16 − 0 = 16 m x14 = − 4 m W 4 s ≤ t ≤ 12 s, 12 s ≤ t ≤ 14 s, Total distance traveled: d 2 = − 8 − 16 = 24 m d3 = − 4 − ( − 8 ) = 4 m d = 16 + 24 + 4 d = 44 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 64. (a) Construction of the curves. a−t curve A1 = −12 m/s, A2 = 8 m/s v−t curve v0 = 8 m/s v6 = v0 + A1 = 8 + ( −12 ) = − 4 m/s v10 = v6 = − 4 m/s v14 = v10 + A2 = − 4 + 8 = 4 m/s A3 = 16 m, A4 = − 4 m A5 = −16 m, A6 = − 4 m A7 = 4 m x−t curve x0 = 0 x4 = x0 + A3 = 16 m x6 = x4 + A4 = 12 m x10 = x6 + A5 = − 4 m x12 = x10 + A6 = − 8 m x14 = x12 + A7 = − 4 m (b) Time for x > 8 m. From the x−t diagram, this is time interval t1 to t2. Over 0 < t < 6 s, dx = v = 8 − 2t dt continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Integrating, using limits x = 0 when t = 0 and x = 8 m when t = t1 x or 8 0 2 = 8t − t  t or 0 2 8 = 8t1 − t1 t12 − 8t1 + 8 = 0 Solving the quadratic equation, t1 = 8± (8)2 − ( 4 )(1)(8) ( 2 )(1) = 4 ± 2.828 = 1.172 s and 6.828 s The larger root is out of range, thus t1 = 1.172 s Over 6 < t < 10, Setting x = 8, Required time interval: x = 12 − 4 ( t − 6 ) = 36 − 4t 8 = 36 − 4t2 or t2 = 7 s ( t2 − t1 ) = 5.83 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 65. The a–t curve is just the slope of the v–t curve. 0 < t < 10 s, 10 s < t < 18 s, 18 s < t < 30 s, 30 s < t < 40 s a= a= a=0W 18 − 6 = 1.5 ft/s 2 W 18 − 10 −18 − 18 = − 3 ft/s 2 W 30 − 18 a=0W Points on the x–t curve may be calculated using areas of the v–t curve. A1 = (10)(6) = 60 ft A2 = A3 = A4 = 1 (6 + 18)(18 − 10) = 96 ft 2 1 (18)(24 − 18) = 54 ft 2 1 (−18)(30 − 24) = − 54 ft 2 x0 = − 48 ft W x10 = x0 + A1 = 12 ft W x18 = x10 + A2 = 108 ft W x24 = x18 + A3 = 162 ft W x30 = x24 + A4 = 108 ft W x40 = x30 + A5 = − 72 ft W continued A5 = (−18)(40 − 30) = −180 ft Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (a) Maximum value of x. Maximum value of x occurs When v = 0, i.e. t = 24 s. xmax = 162 ft W (b) Time s when x = 108 ft. From the x–t curve, t = 18 s and t = 30 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 66. Data from problem 11.65: 0 < t < 10 s, 10 s < t < 18 s, a= a= x0 = − 48 ft The a–t curve is just the slope of the v–t curve. a=0! 18 − 6 = 1.5 ft/s 2 ! 18 − 10 18 s < t < 30 s, 30 s < t < 40 s, −18 − 18 = − 3 ft/s 2 ! 30 − 18 a=0! ! Points on the x–t curve may be calculated using areas of the v–t curve. A1 = (10)(6) = 60 ft A2 = A3 = A4 = 1 (6 + 18)(18 − 10) = 96 ft 2 1 (18)(24 − 18) = 54 ft 2 1 (−18)(30 − 24) = − 54 ft 2 A5 = (−18)(40 − 30) = −180 ft x0 = − 48 ft ! x10 = x0 + A1 = 12 ft ! x18 = x10 + A2 = 108 ft ! x24 = x18 + A3 = 162 ft ! x30 = x24 + A4 = 108 ft ! x40 = x30 + A5 = − 72 ft ! continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (a) Total distance traveled during 0 ≤ t ≤ 30 s . For 0 ≤ t ≤ 24 s For 24 s ≤ t ≤ 30 s Total distance. d1 = x24 − x0 = 210 ft d 2 = x30 − x24 = 54 ft d = d1 + d 2 d = 264 ft ! (b) Values of t for which x = 0. In the range Set x = 0. 0 ≤ t ≤ 10 s x = x0 + v0t = − 48 + 6t − 48 + 6t1 = 0 t1 = 8 s ! In the range 30 s < t < 40 s, x = x30 + v30 (t − 30) = 108 + (−18)(t − 30) = 648 − 18t Set x = 0. 648 − 18t2 = 0 t2 = 36 s ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 67. Sketch v − t curve as shown. Label areas A1, A2 , and A3 A1 = ( 3)( 20 ) = 60 in. Δv = at1 = 2t1 in./s A2 = 1 ( Δv ) t1 = t12 in. 2 A3 = ( Δv )( 20 − t1 ) = 2t1 ( 20 − t1 ) in. Distance traveled: Δx = 12 ft = 144 in. 2 Δx = total area, 144 = 60 + t1 + 2t1 ( 20 − t1 ) or t1 = Reject the larger root. 2 t1 − 40t1 + 84 = 0 40 ± 402 − ( 4 )(1)( 84 ) ( 2 )(1) = 2.224 s t1 = 2.224 s and 37.8 s Δv = 2t1 = 4.45 in./s vmax = 3 + Δv = 3 + 4.45 vmax = 7.45 in./s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 68. Let x be the altitude. Then v is negative for decent and a is positive for deceleration. Sketch the v−t and x−t curves using times t1, t2 and t3 as shown. Use constant slopes in the v−t curve for the constant acceleration stages. Areas of v−t curve: A1 = − 1 (180 + 44 ) t1 = −112t1 ft 2 A2 = − 44t2 A3 = 1 ( − 44 ) t3 = − 22t3 2 Δx1 = 1800 − 1900 = −100 ft Δx2 = 100 − 1800 = −1700 ft Δx3 = 0 − 100 = −100 ft Using Δxi = Ai gives t1 = t2 = t3 = (a) Total time: (b) Initial acceleration. a= −100 = 0.893 s −112 −1700 = 38.64 s − 44 −100 = 4.55 s − 22 t1 + t2 + t3 = 44.1 s W Δv ( − 44 ) − ( −180 ) = Δt 0.893 a = 152.3 ft/s 2 W Changes in position: Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 69. Sketch the v−t curve Data: v0 = 64 km/h = 17.778 m/s x2 = 4.8 km = 4.8 × 103 m v1 = 32 km/hr = 8.889 m/s x1 = 4.8 × 103 − 800 = 4.0 × 103 m t2 = 450 s (a) Time t1 to travel first 4 km. x1 = 4.0 × 103 = A1 = 1 1 ( v0 + v1 ) t1 = (17.778 + 8.889 ) t1 2 2 t1 = 300 s W (b) Velocity v2. x2 − x1 = 800 = A2 = 1 1 ( v1 + v2 )( t2 − t1 ) = ( v1 + v2 )( 450 − 300 ) 2 2 v2 + v1 = 10.667 m v2 = 10.667 − 8.889 (c) Final deceleration. a12 = v2 − v1 1.778 − 8.889 = = − 0.0474 m/s 2 t2 − t1 450 − 300 a12 = 0.0474 m/s 2 W v2 = 1.778 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 70. 10 min 20 s = 10 20 + = 0.1722 h 60 3600 Sketch the v−t curve ta = tb = tc = 60 a 25 a 35 a A1 = 60t1 − But 1 1 1 1 ( 60 ) ( ta ) − ( 25) tb = 60 t1 − 1800 − 312.5 2 2 a a A1 = 5 mi 60t1 − 2112.5 1 =5 a 1 a (1) A2 = 35 ( 0.1722 − t1 ) − 35tc = 6.0278 − 35t1 − 612.5 But A2 = 8 − 5 = 3 mi 35t1 + 612.5 Solving equations (1) and (2) for t1 and 1 , a 1 = 3.0278 a (2) t1 = 85.45 × 10−3 h = 5.13 min 1 = 60.23 × 10−6 h 2 /mi a a = 16.616 × 103 mi/h 2 = (16.616 × 10 ) (5280) 3 ( 3600 ) 2 a = 6.77 ft/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 71. Sketch the a−t curve as shown ) v0 = 20 ft/s, v1 = − 6 ft/s A1 = − 6t1 A2 = − 1 ( 40 − 6 ) t1 = −17t1 2 v1 = v0 + A1 + A2 6 = 20 − 6t1 − 17t1 t1 = 0.6087 s t2 = 1.4 s t2 − t1 = 0.7913 s t1 = 0.609 s W (a) A1 + A3 = − ( 6 )(1.4 ) = − 8.4 ft/s A2 = − (17 )( 0.6087 ) = − 10.348 ft/s v2 = v0 + A1 + A3 + A2 = 20 − 8.4 − 10.348 v2 = 1.252 ft/s W (b) x2 = x0 + v0t2 + ( A1 + A3 ) x13 + A2 x2 by moment-area method 1 ⎞ ⎛1 ⎞ ⎛ = 0 + v0t2 + ( A1 + A3 ) ⎜ t2 ⎟ + A2 ⎜ t2 − t1 ⎟ 2 3 ⎠ ⎝ ⎠ ⎝ 0.6087 ⎞ ⎛1⎞ ⎛ = 0 + ( 20 )(1.4 ) − ( 8.4 ) ⎜ ⎟ (1.4 ) − (10.348 ) ⎜ 1.4 − 3 ⎟ ⎝2⎠ ⎝ ⎠ x2 = 9.73 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 72. Note that 1 5280 mile = = 660 ft 8 8 Sketch v−t curve for first 660 ft. Runner A: t1 = 4 s, t2 = 25 − 4 = 21 s A1 = 1 ( 4 )( v A )max = 2 ( vA )max 2 A2 = 21( v A )max A1 + A2 = ∆x = 5280 ft = 660 ft 8 or 23 ( v A )max = 660 ( vA )max = 28.696 ft/s Runner B: A1 = t1 = 5 s, t2 = 25.2 − 5 = 20.2 s 1 ( 5)( vB )max = 2.5 ( vB )max 2 A2 = 20.2 ( vB )max A1 + A2 = ∆x = 660 ft 22.7 ( vB )max = 660 or ( vB )max or = 29.075 ft/s Sketch v−t curve for second 660 ft. A3 = vmaxt3 − t3 = vmax ± ∆v = a t3 = 0.3t3 1 ∆vt3 = 660 2 2 0.15t3 − vmaxt3 + 660 = 0 Runner A: ( vmax ) A = 28.696, ( vmax )2 − ( 4 )( 0.15)( 660 ) ( 2 )( 0.15) ( t3 ) A = 164.57 s and (a)  = 3.3333  vmax ±  26.736 s ( vmax )2 − 396    Reject the larger root. Then total time t A = 25 + 26.736 = 51.736 s t A = 51.7 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Runner B: ( vmax ) B = 29.075, ( t3 ) B = 167.58 s and 26.257 s t B = 25.2 + 26.257 = 51.457 s t B = 51.5 s W Reject the larger root. Then total time Velocity of A at t = 51.457 s: v1 = 28.696 − ( 0.3)( 51.457 − 25 ) = 20.759 ft/s Velocity of A at t = 51.736 s: v2 = 28.696 − ( 0.3)( 51.736 − 25 ) = 20.675 ft/s Over 51.457 s ≤ t ≤ 51.736 s, runner A covers a distance ∆x (b) ∆x = vave ( ∆t ) = 1 ( 20.759 + 20.675)( 51.736 − 51.457 ) 2 ∆x = 5.78 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 73. Sketch the v−t curves. At t = 12 min = 720 s, xtruck = (19.44 )( 720 ) = 14000 m xbus = 14000 + 1200 = 15200 m xbus = area under v−t curve 1 ( t1 − 120 )( 27.78) + ( 720 − t1 )( 27.78) = 15200 2 t1 = 225.8 s (a) When xbus = xtruck , areas under the v−t curves are equal. 1 ( 27.78)( t1 − 120 ) + 27.78 ( t2 − t1 ) = 19.44t2 2 With t1 = 225.8 s, xtruck = (19.44 )( 576 ) = 11200 m (b) abus = v − v0 27.78 − 0 = t1 − 120 225.8 − 120 t2 = 576 s W xtruck = 11.20 km W abus = 0.262 m/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 74. ( vA )0 = 32 km/h = 8.889 m/s vB = 24 km/h = 6.667 m/s Sketch the v−t curves. A1 = ( 6.667 )( 45 ) = 300 m A2 = 1 1 ( 2.222 )( 45) + vA/B ( 45) 2 2 = 50 + 22.5v A/B x A = ( x A )0 + A1 + A2 xB = ( xB )0 + A1 xB/ A = xB/ A (b) ( )0 − A2 v A/B = 0.444 m/s W 0 = 60 − 50 − 22.5v A/B v A = vB + v A/B = 6.667 + 0.444 = 7.111 m/s (a) aA = v A − ( v A )0 t = 7.111 − 8.889 45 a A = − 0.0395 m/s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 75. ( vA )0 = 22 mi/h = 32.267 ft/s ( vB )0 = 13 mi/h = 19.067 ft/s Sketch the v−t curves. Slope of v−t curve for car A. a=− t1 = 13.2 = − 0.14 ft/s 2 t1 13.2 = 94.29 s 0.14 1 A2 = (13.2 )( 94.29 ) = 622.3 m 2 xB = ( xB )0 + A1 x A = ( x A )0 + A1 + A2 xB/ A = xB − x A = ( xB )0 − ( x A )0 − A2 , d = A2 or 0 = d − A2 d = 622 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 76. Construct the a−t curves for the elevator and the ball. Limit on A1 is 24 ft/s. Using A1 = 4t 4t2 = 24 Motion of elevator. For 0 ≤ t1 ≤ 6 s, t2 = 6 s ( xE )0 = 0 ( vE )0 = 0 4t1 t1 2 = 2t1 2 Moment of A1 about t = t1 : 2 2 xE = ( xE )0 + ( vE )0 t1 + 2t1 = 2t1 Motion of ball. At t = 2, For t1 > 2 s, Moment of A2 about t = t2 : ( xB )0 = 40 ft ( vB )0 = 64 ft/s A2 = − 32.2 ( t1 − 2 ) ft/s 2 ⎛t − 2⎞ − 32.2 ( t1 − 2 ) ⎜ 1 ⎟ = −16.1( t1 − 2 ) ⎝ 2 ⎠ 2 xB = ( xB )0 + ( vB )0 ( t1 − 2 ) − 16.1( t1 − 2 ) = 40 + 64 ( t1 − 2 ) − 16.1( t1 − 2 ) When ball hits elevator, xB = x E 2 2 2 40 + 64 ( t1 − 2 ) − 16.1( t1 − 2 ) = 2t1 2 18.1t1 − 128.4t1 + 152.4 = 0 or Solving the quadratic equation, t1 = 1.507 s and 5.59 s t1 = 5.59 s W The smaller root is out of range, hence Since this is less than 6 s, the solution is within range. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 77. Let x be the position of the front end of the car relative to the front end of the truck. Let v= dx dt and a= dv . dt The motion of the car relative to the truck occurs in 3 phases, lasting t1, t2, and t3 seconds, respectively. Phase 1, acceleration. Phase 2, constant speed. a1 = 2 m/s 2 v2 = 90 km/h − 54 km/h = 36 km/h = 10 m/s a3 = − 8 m/s 2 t1 = t3 = v2 − 0 10 − 0 = =5s a1 2 0 − v2 0 − 10 = = 1.25 s a2 8 Phase 3, deceleration. Time of phase 1. Time of phase 3. Sketch the a−t curve. Areas: A1 = t1v2 = 10 m/s A3 = t3v = −10 m/s Initial and final positions. x0 = − 30 − 16 = − 46 m x f = 30 + 5 = 35 m Initial velocity. v0 = 0 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Final time. t f = t1 + t2 + t3 x f = x0 + v0t f + t1 = t f − ∑ Ai ti 1 t1 2 = 5 + t2 + 1.25 − 2.5 = 3.75 + t2 t2 = 1 t3 = 0.625 s 2 35 = − 46 + 0 + (10 )( 3.75 + t2 ) + ( −10 )( 0.625 ) t2 = 49.75 = 4.975 s 10 t f = t1 + t2 + t3 = 11.225 s Total time. t1 + t2 = 9.975 s t f = 11.23 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 78. Let x be the position of the front end of the car relative to the front end of the truck. Let v= dx dt and a= dv . dt The motion of the car relative to the truck occurs in two phases, lasting t1 and t2 seconds, respectively. Phase 1, acceleration. Phase 2, deceleration. Sketch the a–t curve. Areas: Initial and final positions x0 = − 30 − 16 = − 46 m x f = 30 + 5 = 35 m A1 = 2t1 A2 = − 8t2 a1 = 2 m/s 2 a2 = − 8 m/s 2 Initial and final velocities. v0 = v f = 0 v f = v0 + A1 + A2 0 = 0 + 2t1 − 8t2 t1 = 4t2 x f = x0 + v0t f + t1 = t2 + t2 = ∑ Ai ti 1 t1 = 3t2 2 1 t2 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System 1  35 = − 46 + 0 + 2 ( 4t2 )( 3t2 ) + ( −8t2 )  t2  2  2 81 = 20 t2 t2 = 2.0125 s t1 = 8.05 s t f = t1 + t2 = 10.0625 s. Maximum relative velocity. vm = a1 t1 = ( 2 )( 8.05 ) = 16.10 m/s vm = 60.0 km/h Maximum velocity relative to ground. vmax = vT + v = 54 + 60.0 vmax = 112.0 km/h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 79. Sketch acceleration curve. Let Then, j = jerk = amax = j ( Δt ) da dt A1 = 1 amax ( 2Δt ) = amax ( Δt ) 2 2 = j ( Δt ) v f = v0 + A1 − A2 0 = 0 + A1 − A2 A2 = A1 Δx = v0 ( 4Δt ) + ( A1 )( 3Δt ) − A2 ( Δt ) = 0 + 3 j ( Δt ) − j ( Δt ) = 2 j ( Δt ) Δt = (a) Shortest time: (b) Maximum velocity: 3 3 3 3 Δx = 2j 3 ( 2 )(1.5) 0.36 = 0.4932 4Δt = ( 4 )( 0.4932 ) = 1.973 s W vmax = v0 + A1 = 0 + j ( Δt ) 2 2 = (1.5 )( 0.4932 ) = 0.365 m/s W Average velocity: vave = 0.36 Δx = = 0.1825 m/s W 4Δt 1.973 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 80. Sketch the a−t curve. From the jerk limit, j ( Δt1 ) = amax A1 = or ( Δt1 ) = amax 1.25 = = 5 s. j 0.25 1 ( 5)(1.25) = 3.125 m/s 2 vmax = 32 km/hr = 8.889 m/s = 2 A1 + A2 A2 = vmax − 2 A1 = 8.889 − ( 2 )( 3.125 ) = 2.639 m/s Δt2 = A2 2.639 = = 2.111 s amax 1.25 Total distance is 5 km = 5000 m. Use moment-area formula. 1 1 ⎛ ⎞ ⎛ ⎞ x f = x0 + v0t f + ( 2 A1 + A2 ) ⎜ t f − Δt1 − Δt2 ⎟ − ( 2 A1 + A2 ) ⎜ Δt1 + Δt2 ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ = 0 + 0 + vmax t f − 2Δt1 − Δt2 ( ) 5000 = 10 + 2.111 + 562.5 = 575 s 8.889 t f = 9.58 min W (a) t f = 2Δt1 + Δt2 + xf vmax = ( 2 )( 5 ) + 2.111 + (b) vave = xf tf = 5000 = 8.70 m/s 575 vave = 31.3 km/h W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 81. Indicate areas A1 and A2 on the a − t curve. A1 = A2 = 1 T ( 0.6 ) = 0.1T m/s 2 3 1 2T ( 0.6 ) = 0.2T m/s 2 3 By moment-area formula, ⎛7 ⎞ ⎛4 ⎞ x = v0t + ( A1 ) ⎜ T ⎟ + A2 ⎜ T ⎟ ⎝9 ⎠ ⎝9 ⎠ 7 2 8 2 15 2 1 2 40 = 0 + T + T = T = T 90 90 90 6 T 2 = ( 40 )( 6 ) = 240 s 2 (a) vmax = v0 + A1 + A2 = 0 + 0.1T + 0.2T = 0.3T (b) T = 15.49 s W vmax = 4.65 m/s W Indicate area A3 and A4 on the a − t curve. A1 = 0.1T A4 = (c) By moment-area formula, x = v0 T ⎛ T 2T ⎞ ⎛2 T ⎞ ⎛1 T ⎞ + A1 ⎜ − ⎟ + A3 ⎜ ⋅ ⎟ + A4 ⎜ ⋅ ⎟ 2 9 ⎠ ⎝2 ⎝3 6⎠ ⎝3 6⎠ A3 = 1 T ( 0.6 ) = 0.05T 2 6 1 T ( 0.45) = 0.0375T 2 6 v = 2.90 m/s W v = v0 + A1 + A3 + A4 = 0.1875T T ⎛ 5T ⎞ ⎛T ⎞ 2 = 0 + ( 0.1T ) ⎜ ⎟ + ( 0.05T ) ⎜ ⎟ + ( 0.0375T ) = 0.035417T 18 ⎝ 18 ⎠ ⎝9⎠ = ( 0.035417 )(15.49 ) 2 x = 8.50 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 82. Divide the area of the a−t curve into the four areas A1, A2 , A3 and A4. 2 ( 3)( 0.2 ) = 0.4 m/s 3 A2 = ( 5 )( 0.2 ) = 1 m/s A1 = 1 ( 5 + 2.5)( 0.1) = 0.375 m/s 2 1 A4 = ( 2.5 )( 0.1) = 0.125 m/s 2 A3 = (a) Velocities: v0 = 0 v0.2 = v0 + A1 + A2 v0.3 = v0.2 + A3 v0.4 = v0.3 + A4 v0.2 = 1.400 m/s W v0.3 = 1.775 m/s W v0.4 = 1.900 m/s W Sketch the v − t curve and divide its area into A5 , A6 , and A7 as shown. 0.3 0.4 ∫ x dx = 0.3 − x = ∫ t vdt or x = 0.3 − ∫ t vdt 0.4 At t = 0.3 s, (b) With At t = 0.2 s, With A5 + A6 = A5 = x0.3 = 0.3 − A5 − (1.775 )( 0.1) 2 ( 0.125)( 0.1) = 0.00833 m 3 x0.2 = 0.3 − ( A5 + A6 ) − A7 2 ( 0.5)( 0.2 ) = 0.06667 m 3 A7 = (1.400 )( 0.2 ) = 0.28 m x0.2 = − 0.0467 m W x0.3 = 0.1142 m W and x0.2 = 0.3 − 0.06667 − 0.28 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 83. Approximate the a−t curve by a series of rectangles of height ai , each with its centroid at t = ti . When equal widths of Δt = 0.25 s are used, the values of ti and ai are those shown in the first two columns of the table below. ti ai 2 − ti 2 ai ( 2 − ti ) (s ) 0.125 0.375 ( ft/s ) −3.215 −1.915 (s) 1.875 1.625 ( ft/s ) −6.028 −3.112 −1.547 −0.759 −0.341 −0.128 −0.036 −0.004 0.625 0.875 1.125 −1.125 −0.675 −0.390 1.375 1.125 0.875 1.375 1.625 1.875 Σ −0.205 −0.095 −0.030 0.625 0.375 0.125 2 −7.650 ft/s v = v0 + 2 ( ) −11.955 ( ft/s ) At t = 2 s, ∫ 0 adt ≈ v0 + Σai ( Δt ) ≈ v0 + ( Σai ) ( Δt ) (a) Using moment-area formula, 0 ≈ v0 − ( 7.650 )( 0.25 ) x = x0 + v0t + ∫ 0 ai ( t − ti ) dt ≈ x0 + v0t + Σai ( 2 − ti ) ( Δt ) 2 v0 = 1.913 ft/s W ≈ x0 + v0t + ( Σai ( 2 − ti ) ) ( Δt ) (b) ≈ 0 + (1.913)( 2 ) − (11.955 )( 0.25 ) x = 0.836 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 84. Approximate the a−t curve by a series of rectangles of height ai , each with its centroid at t = ti . When equal widths of Δt = 2 s are used, the values of ti and ai are those shown in the first two columns of table below. 20 − ti 2 ti ai ai ( 20 − ti ) (s ) 1 3 ( ft/s ) 17.58 13.41 (s) 19 17 ( ft/s ) 334.0 228.0 5 7 9 11 13 15 10.14 7.74 6.18 15 13 11 152.1 100.6 68.0 5.13 4.26 3.69 9 7 5 46.2 29.8 18.5 17 19 Σ 3.30 3.00 3 1 9.9 3.0 990.1( ft/s ) (a) At t = 8 s, v8 = v0 + ∫ 0 adt ≈ 0 + Σai ( Δt ) 8 = ( Σai ) ( Δt ) Since t = 8 s, only the first four values in the second column are summed: Σai = 17.58 + 13.41 + 10.14 + 7.74 = 48.87 ft/s 2 v8 = ( 48.87 )( 2 ) v8 = 97.7 ft/s W (b) At t = 20 s, x20 = vot + ∫ 0 a ( 20 − t ) dt = 0 + Σai ( 20 − t )( Δt ) 20 = ( 990.1)( 2 ) x20 = 1980 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 85. The given curve is approximated by a series of uniformly accelerated motions. For uniformly accelerated motion, 2 2 − v1 = 2a ( x2 − x1 ) v2 or Δx = Δt = 2 2 v2 − v1 2a v2 − v1 = a ( t2 − t1 ) or v2 − v1 a For the regions shown above, Region 1 2 3 4 v1 ( m/s ) v2 ( m/s ) a m/s 2 −3 ( ) Δx ( m ) Δt ( s ) 32 30 20.67 0.667 30 25 20 25 20 10 −8 −11.5 −13 17.19 9.78 11.54 0.625 0.435 0.769 5 Σ 10 0 −14.5 3.45 62.63 0.690 3.186 t = Σ ( Δt ) = 3.19 s W (a) (b) Assuming x0 = 0, x = x0 + Σ ( Δx ) = 62.6 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 86. Use a = v dv dv = slope of the given curve. noting that dx dx Slope is calculated by drawing a tangent line at the required point, and using two points on this line to dv Δv = . determine Δx and Δv. Then, dx Δx (a) When x = 0.25, v = 1.4 m/s Δv = 1m/s from the curve and Δx = 0.25m from the tangent line a = (1.4 )( 4 ) a = 5.6 m/s 2 W dv 1 = = 4 s −1 dx 2.5 (b) When v = 2.0 m/s, Δv = 1 m/s x = 0.5 m from the curve. and Δx = 0.6 m from the tangent line. a = ( 2 )(1.667 ) a = 3.33 m/s 2 W dv 1 = = 1.667s−1, dx 0.6 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 87. The a−t curve for uniformly accelerated motion is shown. The area of the rectangle is A = at. Its centroid lies at By moment-area formula, t = 1 t. 2 ⎛1 ⎞ x = x0 + v0 + A ( t − t ) = x0 + v0t + ( at ) ⎜ t ⎟ ⎝2 ⎠ = x0 + v0t + 1 2 at 2 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 88. From the a−t curve, A1 = ( −2 )( 6 ) = − 12 m/s A2 = ( 2 )( 2 ) = 4 m/s Over 6 s < t < 10 s, v = v0 + A1, v = − 4 m/s or − 4 = v0 − 12, or v0 = 8 m/s By moment-area formula, x12 = x0 + v0t + moment of shaded area about t = 12s x12 = 0 + ( 8 )(12 ) + ( − 12 )(12 − 3) + ( 4 )(12 − 11) x12 = − 8 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 89. (a) T = 0.2s. A1 = 2 ( − 24 )( 0.2 ) = − 3.2 ft/s 3 A2 = ( − 24 )( t1 − 0.2 ) = − 24t1 + 4.8 v f = v0 + ΣA 0 = 90 − 3.2 − 24t1 + 4.8 t1 = 3.8167 s A2 = − 86.80 ft/s t1 − T = 3.6167 s By moment-area formula, x1 = x0 + v0t1 + moment of area ⎡⎛ 3 ⎞ ⎛ 3.6167 ⎞ x1 = 0 + ( 90 )( 3.8167 ) + ( − 3.2 ) ⎢⎜ ( 0.2 ) + 3.6167 ⎟ + ( − 86.80 ) ⎜ ⎟ ⎠ ⎝ 2 ⎠ ⎣⎝ 8 x1 = 174.7 ft W (b) T = 0.8 s. A1 = 2 ( − 24 )( 0.8) = −12.8 ft/s, 3 A2 = ( − 24 )( t1 − 0.8 ) = − 24t1 + 19.2 v f = v0 + ΣA or 0 = 90 − 12.8 − 24t1 + 19.2, A2 = − 77.2 ft/s t1 = 4.0167 s t1 − T = 3.2167s By moment-area formula, ⎡3 ⎤ ⎛ 3.2167 ⎞ x1 = 0 + ( 90 )( 4.0167 ) + ( −12.8 ) ⎢ ( 0.8 ) + 3.2167 ⎥ + ( −77.2 ) ⎜ ⎟ ⎣8 ⎦ ⎝ 2 ⎠ x1 = 192.3 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 90. Data from Prob. 65 x0 = − 48 ft, v0 = 6 ft/s The a – t curve is just the slope of the v – t curve. 0 < t < 10 s, 10 s < t < 18 s, 18 s < t < 30 s, 30 s < t < 40 s x = x0 + v0t + ∑ Ai ti a= a=0! 18 − 6 = 1.5 ft/s ! 18 − 10 −18 − 18 a= = − 3 ft/s ! 30 − 18 a=0 ! (a) Position when t = 20 s. A1 = (18 − 10 )(1.5 ) = 12 ft/s t1 = 20 − 14 = 6s A2 = ( 2 )( − 3) = − 6 ft/s t2 = 20 − 19 = 1 s x20 = −48 + ( 6 )( 20 ) + (12 )( 6 ) + ( − 6 )(1) x20 = 138 ft ! (b) Maximum value of position coordinate. x is maximum where v = 0. From velocity diagram, tm = 24 s A1 = (18 − 10 )(1.5 ) = 12 ft/s t1 = ( 24 − 14 ) = 10 s A2 = ( 24 − 18 )( − 3) = −18 ft/s t2 = ( 24 − 21) = 3 s xm = −48 + ( 6 )( 24 ) + (12 )(10 ) + ( −18 )( 3) xm = 162 ft ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 91. x = ( t + 1) 2 y = 4 ( t + 1) −2 −3 −4 & = 2 ( t + 1) vx = x &x = 2 ax = v & = − 8 ( t + 1) vy = y & y = 24 ( t + 1) ay = v Solve for (t + 1)2 from expression for x. Substitute into expression for y. Then, y= 4 x (t + 1)2 = x xy = 4 vx = 2 m/s, v= v y = − 8 m/s = 8.25 m/s This is the equation of a rectangular hyperbola. (a) t = 0. ( 2 ) 2 + ( − 8 )2 θ = tan −1   −8   = − 76.0°  2 v = 8.25 m/s ax = 2 m/s 2 , a= a y = 24 m/s 2 = 24.1 m/s 2 76.0° W ( 2 )2 + ( 24 )2  24  θ = tan −1   = 85.2°  2  a = 24.1 m/s 2 85.2°W 1 s. 2 (b) t = vx = 3 m/s, v= v y = − 2.37 m/s = 3.82 m/s ( 3)2 + (2.37)2 θ = tan −1   −2.37   = − 38.3°  3  v = 3.82 m/s 38.3° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System ax = 2 m/s, a y = 4.74 m/s 2 a = 22 + 4.742 = 5.15 m/s 2 θ = tan −1   = 67.2°  2  a = 5.15 m/s 2 67.2°W  4.74  Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 92. Let u = t t 2 − 9t + 18 = t 3 − 9t 2 + 18t du d 2u = 3t 2 − 18t + 18, and = 6t − 18 dt dt 2 x = 6 − 0.8u m y = − 4 + 0.6u m dx du = −0.8 dt dt dy = dx dy dt dx dt ( ) Then, dy du = + 0.6 dx dt 0.6 = − 0.75 = constant 0.8 =− Since (a) dy does not change, the path is straight. dx At t = 2 s, du d 2u = − 6, and = − 6. dt dt 2 dx vx = = ( − 0.8)( − 6 ) = 4.8 m/s, dt ax = d 2x = ( − 0.8 )( − 6 ) = 4.8 m/s 2 , dt 2 d u =0 dt 2 v y = ( 0.6 )( −9 ) = −5.4 m/s ay = 0 2 vy = dy = ( 0.6 )( − 6 ) = − 3.6 m/s dt a y = ( 0.6 )( − 6 ) = − 3.6 m/s 2 v = 6.0 m/s 36.9°, a = 6.0 m/s 2 36.9° W (b) At t = 3 s, du = −9, dt ax = 0, du = −6, dt and vx = ( −0.8 )( −9 ) = 7.2 m/s, v = 9.0 m/s (c) At t = 4 s, and d 2u =6 dt 2 v y = ( 0.6 )( −6 ) = −3.6 m/s a y = ( 0.6 )( 6 ) = 3.6 m/s 2 36.9°, a = 0 W vx = ( −0.8 )( −6 ) = 4.8 m/s, ax = ( −0.8 )( 6 ) = −4.8 m/s 2 , v = 6.0 m/s 36.9°, a = 6.0 m/s 2 36.9° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 93. Substitute the given expressions for x and y into the given equation of the ellipse, and note that the equation is satisfied. 16 cos 2 π t − 16 cos π t + 4 9sin 2 π t x2 y2 + = + 2 2 4 3 4 ( 2 − cos π t ) 3 ( 2 − cos π t ) = 4 cos 2 π t − 4 cos π t + 1 + 3sin 2 π t ( ) ( 2 − cos π t )2 = 4 − 4 cos π t + cos 2 π t ( 2 − cos π t )2 =1 & and y & by differentiation. Calculate x & = x & = y ( 4cos π t − 2 )(π sin π t ) = −6π sin π t −4π sin π t − ( 2 − cos π t ) ( 2 − cos π t )2 ( 2 − cos π t )2 3sin π t (π sin π t ) 3π ( 2cos π t − 1) 3π cos π t − = ( 2 − cos π t ) ( 2 − cos π t )2 ( 2 − cos π t )2 & =0 x & = x (a) When t = 0 s, 1 (b) When t = s, 3 (c) When t = 1 s, and 3 2 & = 3π , y & =0 3, y v = 9.42 m/s v = 7.26 m/s W W −6π (2 − 1 2) ( ) = − 4π − 2 3 & =0 x and & = y 3π ( −3) ( 3)2 = −π , v = 3.14 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 94. Sketch the path of the particle, i.e. plot of y versus x. Using x = 6t − sin t , and y = 6 − 3cos t obtain the values in the table below. Plot as shown. t ( s) 0 x (m) 0 y (m) 3 π 2 6.42 18.85 31.27 37.70 6 9 6 3 3 π π 2 2π (a) Differentiate with respect to t to obtain velocity components. dx dy = 6 − 3cos t and = 3sin t vx = vy = dt dx 2 2 + v2 v 2 = vx y = ( 6 − 3cos t ) + 9sin t = 45 − 36 cos t ( m/s ) 2 2 d (v ) = 36sin t = 0 dt 2 t = 0, π , and 2π in the range 0 ≤ t ≤ 2π . When t = 0 or 2π , When t = π , cos t = 1, cos t = −1, and and v 2 is minimum. v 2 is maximum. 2 (v ) 2 min = 45 − 36 = 9 ( m/s ) , vmin = 3 m/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System (v ) 2 max = 45 + 36 = 81 ( m/s ) , vmax = 9 m/s W (b) t = 0, x = 0, y = 3 m, vx = 3 m/s, v y = 0 t =0W r = (3 m) j W tan θ = vy vx =0 θ =0W t = 2π s, x = 12π m, y = 3 m, vx = 3 m/s, vy = 0 t = 2π s W r = (12π m ) i + ( 3 m ) j W tan θ = t = π s, x = 6π m, y = 9 m, vy vx vy = 0 t =π sW θ =0W vx = 9 m/s, r = ( 6π m ) i + ( 9 m ) j W tan θ = vy vx θ = 0° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 95. Given: r = A ( cos t + t sin t ) i + A ( sin t − t cos t ) j v= dr = A ( − sin t + sin t + t cos t ) i + A ( cos t − cos t + t sin t ) j dt = A ( t cos t ) i + A ( t sin t ) j dv = A ( cos t − t sin t ) i + A ( sin t + t cos t ) j dt a= (a) When r and a are perpendicular, r ⋅ a = 0 A⎡ ⎣( cos t + t sin t ) i + ( sin t − t cos t ) j⎤ ⎦ ⋅ A⎡ ⎣( cos t − t sin t ) i + ( sin t + t cos t ) j⎤ ⎦ =0 A2 ⎡ ⎣( cos t + t sin t )( cos t − t sin t ) + ( sin t − t cos t )( sin t + t cos t ) ⎤ ⎦=0 ( cos (b) 2 t − t 2 sin 2 t + sin 2 t − t 2 cos 2 t = 0 1 − t2 = 0 ) ( ) t = 1s W When r and a are parallel, r × a = 0 A⎡ ⎣( cos t + t sin t ) i + ( sin t − t cos t ) j⎤ ⎦ × A⎡ ⎣( cos t − t sin t ) i + ( sin t + t cos t ) j⎤ ⎦ =0 A2 ⎡ ⎣( cos t + t sin t )( sin t + t cos t ) − ( sin t − t cos t )( cos t − t sin t ) ⎤ ⎦k = 0 (sin t cos t + t sin 2 t + t cos 2 t + t 2 sin t cos t − sin t cos t − t cos2 t − t sin 2 t + t 2 sin t cos t = 0 2t = 0 ) ( ) t =0 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 96. Given: Differentiating to obtain v and a, v= 1 ⎤ ⎡ r = 30 ⎢1 − i + 20 e−π t/2 cos 2π t j t + 1⎥ ⎣ ⎦ ( ) dr 1 ⎛ π ⎞ = 30 + 20 ⎜ − e−π t/2 cos 2π t − 2π e−π t/2 sin 2π t ⎟ j 2 dt ⎝ 2 ⎠ ( t + 1) = a= ⎡ ⎛1 ⎞⎤ i − 20π ⎢e−π t/2 ⎜ cos 2π t + 2sin 2π t ⎟ ⎥ j 2 ⎝ ⎠⎦ ⎣ ( t + 1) 30 2 ⎡ π ⎤ dv 2 ⎛1 ⎞ i − 20π ⎢ − e−π t/2 ⎜ cos 2π t + 2sin 2π t ⎟ + e−π t/2 ( −π sin 2π t + 4π cos 2π t ) ⎥ j = −30 3 dt ⎝2 ⎠ ⎣ 2 ⎦ ( t + 1) = −60 ( t + 1) 3 i − 10π 2e −π t/2 ( 4sin 2π t − 7.5cos 2π t ) j (a) At t = 0, 1⎞ ⎛ r = 30 ⎜1 − ⎟ i + 20 (1) j 1⎠ ⎝ ⎡ ⎛1 ⎛1⎞ ⎞⎤ v = 30 ⎜ ⎟ i − 20π ⎢(1) ⎜ + 0 ⎟ ⎥ j 1 2 ⎝ ⎠ ⎠⎦ ⎣ ⎝ r = 20 in. v = 43.4 in./s W 46.3° W 85.4° W a=− (b) At t = 1.5 s, 60 i − 10π 2 (1)( 0 − 7.5 ) j 1 a = 743 in./s 2 1 ⎞ ⎛ −0.25π r = 30 ⎜1 − cos 3π j ⎟ i + 20e 2.5 ⎝ ⎠ = (18 in.) i + ( −1.8956 in.) j v= 30 r = 18.10 in. 6.0° W ( 2.5) 2 ⎛1 ⎞ i − 20π e −0.75π ⎜ cos 3π + 0 ⎟ j ⎝2 ⎠ = ( 4.80 in./s ) i + ( 2.9778 in./s ) j v = 5.65 in./s 31.8° W a=− 60 = −3.84 in./s 2 i + 70.1582 in./s 2 j ( ( 2.5) 3 i + 10π 2e−0.75π ( 0 − 7.5cos 3π ) j ) ( ) a = 70.3 in./s2 86.9° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 97. Given: Differentiating to obtain v and a. v= a= r = ( Rt cos ω nt ) i + ctj + ( Rt sin ω nt ) k dr = R ( cos ω nt − ω nt sin ω nt ) i + cj + R ( sin ω nt + ω nt cos ω nt ) k dt dv = R −ω n sin ω nt − ω n sin ω nt − ω n 2t cos ω nt i + R ω n cos ω nt + ω n cos ω nt − ω n 2t sin ω nt k dt = R ⎡ −2ω n sin ω nt − ω n 2t cos ω nt i + 2ω n cos ω nt − ω n 2t sin ω nt ⎤ k ⎣ ⎦ ( ) ( ) ( ) ( ) Magnitudes of v and a. 2 v 2 = vx 2 + v 2 y + vz =⎡ ⎣ R ( cos ω nt − ω nt sin ω nt ) ⎤ ⎦ + (c) + ⎡ ⎣ R ( sin ω nt + ω nt cos ω nt ) ⎤ ⎦ 2 2 2 2 2 ⎤ = R2 ⎡ ⎣cos ω nt − 2ω nt sin ω nt cos ω nt + ω n t sin ω nt ⎦ + c 2 2 2 2 ⎤ + R2 ⎡ ⎣sin ω nt + 2ω nt sin ω nt cos ω nt + ω n t cos ω nt ⎦ 2 2 = R2 1 + ωn t + c2 2 2 2 ( ) v= 2 2 R2 1 + ωn t + c2 W ( ) 2 2 a 2 = ax + a2 y + az ⎡ 2 t cos ω nt = R 2 ⎢ −2ω n sin ω nt − ω n ⎣ ( ) + ( 2ω 2 n cos ω nt 2⎤ 2 t sin ω nt ⎥ − ωn ⎦ ) 2 2 3 4 2 2 2 2 = R2 ⎡ ⎣ 4ω n sin ω nt + 4ω nt sin ω nt cos ω nt + ω n t cos ω nt + 4ω n cos ω nt 3 4 2 t sin ω nt cos ω nt + ω n t sin 2 ω nt ⎤ − 4ω n ⎦ 2 4 2 = R 2 4ω n + ωn t ( ) 2 2 a = Rω n 4 + ω n t W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 98. r = ( At cos t ) i + A t 2 + 1 j + ( Bt sin t ) k Given: ( ) from which x = At cos t , cos t = x At 2 y = A t 2 + 1, sin t = z Bt 2 z = Bt sin t  y t2 =   − 1  A or x  y t2 =   +    A B 2 2 2 2  x   z  cos 2 t + sin 2 t = 1 ⇒   +   = 1  At   At  Then, For A = 3 and  y x z  A −1 =  A +  B        B = 1, 2 2 2 r = ( 3t cos t ) i + 3 t 2 + 1 j + ( t sin t ) k ( )  y x z  A −  A −  B  = 1!       2 2 Differentiating to obtain v and a. dr t v= = 3 ( cos t − t sin t ) i + 3 j + ( sin t + t cos t ) k 2 dt t +1 dv 1 a= = 3 ( −2sin t − t cos t ) i + 3 j + ( 2cos t − t sin t ) k 3 dt t2 + 1 2 ( ) (a) At t = 0, And Then, v = 3 (1 − 0 ) i + ( 0 ) j + ( 0 ) k a = −3 ( 0 ) i + 3 (1) j + ( 2 − 0 ) h a 2 = ( 3) + ( 2 ) = 13 2 2 v = 3 ft/s ! a = 3.61 ft/s 2 ! (b) If r and v are perpendicular, r ⋅ v = 0 ( 3t cos t )  3 ( cos t − t sin t )  + or (3 t + 1)  2  + t sin t )( sin t + t cos t ) = 0   ( t2 + 1  3t (9t cos t − 9t 2 2 sin t cos t + ( 9t ) + t sin 2 t + t 2 sin t cos t = 0 continued ) ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System With t ≠ 0, 9cos 2 t − 8t sin t cos t + 9 + sin 2 t = 0 10 − 8t sin t cos t + 8cos 2 t = 0 or The smallest root is The next root is 7 + 2cos 2t − 2t sin 2t = 0 2t = 7.631 s t = 4.38 s t = 3.82 s ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 99. (a) At the landing point, Horizontal motion: Vertical motion: from which y = − x tan 30° x = x0 + ( vx )0 t = v0t y = y0 + v y t2 = − 1 gt 2 = − gt 2 ( )0 t − 1 2 2 2y 2 x tan 30° 2v0t tan 30° = = g g g 2v0 tan 30° ( 2 )( 25 ) tan 30° = g 9.81 t = 2.94 s W Rejecting the t = 0 solution gives t = (b) Landing distance: (c) Vertical distance: or d = ( 25)( 2.94 ) x v0t = = cos 30° cos 30° cos 30° 1 2 gt 2 d = 84.9 m W h = x tan 30° + y h = v0t tan 30° − Differentiating and setting equal to zero, dh = v0 tan 30° − gt = 0 dt Then, or t = vo tan 30° g 2 hmax = ( v0 )( v0 tan 30°) tan 30° − 1 g ⎛ v0 tan 30° ⎞ g ⎜ 2 ⎝ 2 g ⎟ ⎠ v 2 tan 2 30° ( 25 ) ( tan 30° ) = 0 = 2g ( 2 )( 9.81) 2 hmax = 10.62 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 100. Horizontal motion: x = x0 + ( vx )0 t = v0t , 1 gt 2 = y0 − gt 2 ( )0 t − 1 2 2 so that y0 = or t = x v0 y = y0 − gx 2 2 2v0 Vertical motion: y = y0 + v y or At ground level, y = 0, gx 2 2 2v0 At x = 50 m, y0 = ( 9.81)( 50 )2 ( 2 )( 30 )2 = 13.625 m h = y0 − 13 = 0.625 m At x = 53 m, y0 = ( 9.81)( 53)2 ( 2 )( 30 )2 = 15.31 m h = y0 − 13 = 2.31 m Range to avoid: 0.625 m < h < 2.31 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 101. Horizontal motion. Vertical motion. Eliminate t. vx = v0 y=h− t= x v0 gx 2 2(h − y) x = v0t 1 2 gt 2 y=h− gx 2 2 2v0 Solve for v0. Data: h = 3 ft, g = 32.2 ft/s2 v0 = (a) To strike corner C. v0 = x = d = 15 ft, y = 0 v0 = 34.7 ft/s W x = 15 ft, y = 1 ft v0 = 42.6 ft/s x = 15 − 1 = 14 ft, y = 0 v0 = 32.4 ft/s 32.4 ft/s < v0 < 42.6 ft/s W ( 32.2 )(15)2 ( 2 )( 3 − 0 ) ( 32.2 )(15)2 ( 2 )( 3 − 1) ( 32.2 )(14 )2 ( 2 )( 3 − 0 ) To strike point B. v0 = To strike point D. v0 = (b) Range to strike corner BCD. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 102. Place origin of coordinates at point A. Horizontal motion: ( vx )0 = 90 mi/h = 132 ft/s x = x0 + ( vx )0 t = 0 + 132t ft At point B where t B = 6.5 s, xB = (132 )( 6.5 ) = 858 ft (a) Distance AB. From geometry Vertical motion: At point B − xB tan 10° = h + 0 − (b) Initial height. 1 ( 32.2 )( 6.5)2 2 h = 529 ft W d = 858 cos 10° gt 2 ( )0 t − 1 2 d = 871 ft W y = y0 + v y Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 103. Data: Horizontal motion. Vertical motion. Eliminate t. v0 = 25 ft/s, α = 90° − 55° = 35°, g = 32.2 ft/s 2 x = ( v0 cos α ) t y = h + ( v0 sin α ) t − t= x v0 cos α gx 2 2 2v0 cos 2 α 1 2 gt 2 y = h + x tan α − Solve for h. To hit point B. h = y − x tan α + x = 20 ft, y = 0 h = 0 − 20 tan 35° + gx 2 2 2v0 cos 2 α ( 32.2 )( 20 )2 ( 2 )( 25cos 35°)2 ( 32.2 )( 24 )2 ( 2 )( 25cos 35°)2 = 1.352 ft To hit point C. x = 24 ft, y = 0 h = 0 − 24 tan 35° + = 5.31 ft 1.352 ft < h < 5.31 ft W Range of values of h. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 104. Place the origin at A. Let β be the direction of the discharge velocity measured counterclockwise from the x-axis Horizontal motion. ( vx )0 = v0 cos β t = x v0 cos β x = ( v0 cos β ) t Solve for t. Vertical motion. ( vy )0 = v0 sin β y = ( v0 sin β ) t − = x tan β − 1 2 gt 2 gx 2 2 2v0 cos 2 β Geometry. At points B and C Hence, y = x tan α x tan α = x tan β − x= gx 2 2 2v0 cos 2 β Solve for x. To water point B. xB 2 2v0 cos 2 β ( tan β − tan α ) g β = 90° − φ0 = 90° − 40° = 50° 2 2 )( 24 ) cos 2 50° ( = 32.2 ( tan 50° − tan10°) = 15.01 ft d B = 15.01 ft W To water point C. xC β = 90° + φ0 = 90° + 40° = 130° 2 2 )( 24 ) cos 2 130° ( = 32.2 ( tan130° − tan10°) = − 20.2 ft dC = − xC = 20.2 ft W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 105. v0 = 13 m/s, α = 33°, x0 = 0, y0 = 0.6 m Vertical motion: v y = v0 sin α − gt y = y0 + ( v0 sin α ) t − At maximum height, vy = 0 t = or t = 1 2 gt 2 v0 sin α g (a) 13sin 33° = 0.7217 s 9.81 1 ( 9.81)( 0.7217 )2 2 ymax = 3.16 m W yes W t = x − x0 v0 cos α ymax = 0.6 + (13sin 33° )( 0.7217 ) − 1.8 m < 3.16 m < 3.7 m Horizontal motion: x = x0 + ( v0 cos α ) t t = 15.2 − 0 = 1.3941 s 13cos 33° 1 ( 9.81)(1.3941)2 2 or At x = 15.2 m, (b) Corresponding value of y : y = 0.6 + (13sin 33° )(1.3941) − y = 0.937 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 136. Velocities: v A/B = v A − v B = 1 m/s Accelerations: a A/B = a A − a B = 0.25 m/s 2 (a) aA = aB = a A/B = 2 vA ρA ρB 2 vB = = 2 vA 100 ( vA − 1) 96 2 2 ( v − 1) = 0.25 vA − A 100 96 2 2 vA − 50v A + 625 = 0 v A = ± 25 v A = 25 m/s ! vB = 24 m/s ! (b) vB = 25 − 1 = 24 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 137. an = v2 ρ , at = 0, 2 = ρ an vmax 2 vmax = ( 25)( 3g ) = ( 25 )( 3)( 9.81) = 735.35 m 2 /s 2 vmax = 27.125 m/s vmax = 97.6 km/h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 138. v2 ( ac )  = c , nA  ρA v2 ( ac )  = c nA  ρA 2    vc = ρA  ( ac )n  A = ρ B ( ac )n  B  ρB  ( ac )n  A 0.66 = = = 0.09706 ρ A ( ac )  6.8 n  B ρ B = 0.09706 ρ A = ( 0.09706 )( 60 ) = 5.8235 mm d B = 2 ρ B = 11.65 mm ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 139. Initial speed. Tangential acceleration. (a) Total acceleration at t = 0. an = v0 2 = v0 = 72 km/h = 20 m/s at = −1.25 m/s 2 ρ ( 20 )2 350 = 1.14286 m/s 2 2 a = at2 + an = ( −1.25)2 + (1.14286 )2 a = 1.694 m/s 2 ! (b) Total acceleration at t = 4 s. v = v0 + at t = 20 + ( −1.25 )( 4 ) = 15 m/s an = v2 = ρ (15)2 350 = 0.6426 m/s 2 2 a = at2 + an = ( −1.25)2 + ( 0.6426 )2 a = 1.406 m/s 2 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 140. Length of run. Radius of circle. L = π D = 130 π meters (1) ρ= 1 D = 65m 2 Tangential acceleration of starting portion of run. vm = at t1 = ( at ) ( 4 ) = 4 at m/s s1 = Constant speed portion of run. 1 1 2 2 at t1 = ( at ) ( 4 ) = 8 at m 2 2 v = vm s = s1 + vm ( t − t1 ) Substituting (1), (2) and (3) into (4) 130 π = 8 at + 4 at ( 54 − 4 ) Solving for at . From (2) at = 130 π = 1.9635 m/s 2 8 + 200 (4) (2) (3) vm = ( 4 )(1.9635 ) = 7.854 m/s 2 7.854 ) ( = Normal acceleration during constant speed portion of run. an = 2 vm ρ 65 = 0.9490 m/s 2 Maximum total acceleration. 2 a = at2 + an = (1.9635)2 + ( 0.9490 )2 a = 2.18 m/s 2 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 141. For uniformly decelerated motion: At t = 9 s, v = v0 + at t 0 = 150 − at ( 9 ) , or at = −16.667 ft/s 2 2 a 2 = at2 + an Total acceleration: 2 2 an =   a − at  1/2 2 2 = 130 ) − ( −16.667 )  (     1/2 = 128.93 ft/s 2 Normal acceleration: an = v2 ρ , where ρ= 1 5 diameter = ft 2 12  5 v 2 = ρ an =   (128.93) = 53.72 ft 2 /s 2 ,  12  Time: t= v − v0 7.329 − 150 = at −16.667 v = 7.329 ft/s t = 8.56 s ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 142. Speeds: Distance: Tangential component of acceleration: v0 = 0 s= v1 = 65 mi/h = 95.33 ft/s π 2 ( 450 ) + 300 = 1006.86 ft 2 2 v1 = v0 + 2at s at = 2 2 ( 95.33) + 0 = 4.5133 ft/s2 v1 − v0 = 2s ( 2 )(1006.86 ) 2 At point B, 2 2 vB = v0 + 2at sB where sB = π 2 ( 450 ) = 706.86 ft 2 vB = 0 + ( 2 )( 4.5133)( 706.86 ) = 6380.5 ft 2 /s 2 (a) At t = 15 s, vB = 79.88 ft/s v = v0 + at t = 0 + ( 4.5133)(15 ) = 67.70 ft/s vB = 54.5 mi/h ! Since v < vB , the car is still on the curve. Normal component of acceleration: an = v2 2 67.70 ) ( = ρ = 450 ft = 10.185 ft/s 2 ρ 450 (b) Magnitude of total acceleration: 2 a = at2 + an = ( 4.5133)2 + (10.185)2 a = 11.14 ft/s 2 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 143. (a) v A = 420 km/h v B = v A + v B/ A or , v B = 520 km/h 60° v B/ A = v B − v A = v B + ( − v A ) Sketch the vector addition as shown. 2 2 2 vB / A = v A + vB − 2v AvB cos 60° = ( 420 ) + ( 520 ) − ( 2 )( 420 )( 520 ) cos 60° 2 2 or vB/ A = 477.9 km/h sin α sin 60° = 520 477.9 or α = 70.4° v B/ A = 478 km/h 70.4° ! (b) a A = 6 m/s 2 ( aB )t = 2 m/s 2 60° vB = 520 km/h = 144.44 m/s ( a B )n = 2 vB ρ = (144.44 )2 200 = 104.32 m/s 2 30° a B/ A = a B − a A = ( a B )t + ( a B )n − a A = [2 60° ] + [ 104.32 30° ] − [ 6 ] = 2 ( − cos 60°i + sin 60° j) + 104.32 ( − cos 30°i − sin 30° j) − 6i = − 97.34 m/s 2 i − 50.43 m/s2 j aB/ A = 109.6 m/s 2 27.4° ! ( ) ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 144. (a) v A = 180 km/h = 50 m/s 30°, v B = 162 km/h = 45 m/s 45° v B/ A = v B − v A = 45 ( cos 45°i − sin 45° j) − 50 ( cos120°i + sin120° j) = 56.82i − 75.12 j = 94.2 m/s 52.9° vB/ A = 339 km/h 52.9° ! (b) ( a A )t = 8 m/s 2 2 vA 60°, ( a B )t = 3 m/s 2 45° ( a A )n = ( a B )n = ρA 2 vB 2 50 ) ( = 400 = 6.25 m/s 2 30° ρB = ( 45)2 300 = 6.75 m/s 2 45° a B/ A = a B − a A = ( a B )t + ( a B )n − ( a A )t − ( a A )n = 3 ( cos 45°i − sin 45° j) + 6.75 ( cos 45°i + sin 45° j) − 8 ( cos 60°i − sin 60° j) − 6.25 ( − cos 30°i − sin 30° j) = 8.31 m/s 2 i + 12.07 m/s 2 j or a B/ A = 15.18 m/s 2 ( ) ( ) 56.8° ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 145. (a) As water leaves nozzle. v = 8 m/s an = g sin 55° = 9.81 sin 55° = 8.04 m/s 2 an = v2 ρ 2 v 2 (8) ρ= = an 8.04 ρ = 7.96 m ! (b) At maximum height of stream. v = ( vx )0 = 8 sin 55° = 6.55 m/s an = g = 9.81 m/s 2 an = v2 ρ v 2 ( 6.55 ) = an 9.81 2 ρ= ρ = 4.38 m ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 146. Horizontal motion. Vertical motion. vx = v0 cos α v y = v0 sin α − gt x = v0 t cos α y = y0 + v0 t sin α − y = y0 + x tan α − 1 2 gt 2 (1) Eliminate t. gx 2 2 2 v0 cos 2 α Solving (1) for v0 and applying result at point B v0 = gx 2 = 2 ( y0 + x tan α − y ) cos 2 α ( 2 )(1.5 + 6 tan 3° − 0.97 ) ( cos 2 3° ) v0 = 14.48 m/s ! ( 9.81)( 6 )2 (a) (b) Magnitude of initial velocity. Minimum radius of curvature of trajectory. an = g = v2 ρ ρ= v2 v2 = an g cosθ (2) where θ is the slope angle of the trajectory. The minimum value of ρ occurs at the highest point of the trajectory where cos θ = 1 and v = vx = v0 cos α Then 2 v 2 cos 2 α (14.48 ) cos 3° = 0 = 9.81 g 2 ρ min ρ min = 21.3 m ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 147. (a) At point A, v = v0 = 120 ft/s v = 120 ft/s 60° aA = g = 32.2 ft/s 2 2 vA ( a A )n = g sin 30° = ρA = (b) ρA 2 2 (120 ) vA = g sin 30° 32.2sin 30° ρ A = 894 ft ! At the point where velocity is parallel to incline, vx = v0 sin 30° = 120 sin 30° = 60 ft/s v y = vx tan 30° = 60 tan 30° = 34.64 ft/s v= ( 60 )2 + ( 34.64 )2 2 vB = 69.282 ft/s an = g sin 60° = ρB 2 2 ( 69.282 ) vB ρB = = g sin 60° 32.2sin 60° ρ B = 172.1 ft ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 148. Compute x- and y-components of velocity and acceleration. x= 2cos π t − 1 , 2 − cos π t − 3π 2 cos π t + &= x − 3π sin π t ( 2 − cos π t )2 , && x= ( 2 − cos π t ) 1.5sin π t , 2 − cos π t 2 6π sin π t (π sin π t ) ( 2 − cos π t )3 y = &= y 1.5π ( 2cos π t − 1) ( 2 − cos π t )2 , && y= − 3π 2 sin π t ( 2 − cos π t ) x = 1, 2 − 3π ( 2cos π t − 1)(π sin π t ) ( 2 − cos π t )3 & = 1.5π , y (a) t = 0, v = && y = 1.5π , 1 , 3 &=− v=−x 2π , 3 y = 0, & = 0, x && x = − 3π 2 , 2 an = − && x = 3π 2 , 3 , 2 ρ= v 2 (1.5π ) = an 3π 2 2π , 3 ρ = 0.75 ft ! 2π 2 , 3 (b) t= x = 0, y= &=− x & = 0, y && y=− an = − && y= 2π 2 , 3 ρ= v 2 4π 2 3 = an 3.2π 2 ρ = 1.155 ft ! π2 3 (c) t = 1, x = −1, y = 0, & = 0, x &=− y π 2 , && x= , &= v = −y π 2 , an = && x= π2 3 , ρ= v2 π 2 3 = ⋅ an 4 π2 ρ = 0.75 ft ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 149. Given: Differentiating twice x= ( t − 4 )3 6 + t2 m y= t 3 ( t − 1) − m 6 4 2 vx 2 − 2) ( &= =x 2 + ( 2 )( 2 ) = 6 m/s t 2 ( t − 1) − m/s 2 2 1 && y = t − m/s 2 2 &= y &= x ( t − 4 )2 2 + 2t m/s && x = t − 4 + 2 = t − 2 m/s 2 At t = 2 s. vy 2 2 ax = && x=2−2=0 a y = && y=2− 2 2) ( &= =y − (1) = 1.5 m/s 1 = 1.5 m/s 2 2 (a) (b) Acceleration. Radius of curvature of path. tan θ = vy vx = 1.5 6 a = 1.5 m/s 2 j ! ( ) θ = 14.036° 2 2 2 v 2 = vx + v2 y = 6 + 1.5 = 38.25 m 2 /s 2 an = a cosθ = 1.5 cos14.036° = 1.45522 m/s 2 an = v2 ρ v2 38.25 = an 1.45522 ρ= ρ = 26.3 m ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 150. vx = v A At point B ( vB ) x = v A vB = ( vB ) x cosθ = vA cosθ cosθ = vA vB an = aB cosθ = g cosθ =g vA vB ρB = 2 vB v2 v = B B an gv A ρB = 3 vB ! gv A Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 151. Let θ be the slope angle of the trajectory at an arbitrary point C. Then, ( aC )n = gcosθ = ρC 2 vC , or ρC = 2 vC gcosθ But, the horizontal component of velocity is constant, ( vC ) x = ( v A ) x where ( vA ) x = v0 cosα v0 cos α = vC cosθ vC = cos α v0 cosθ ( vC ) x = vC cosθ Then, or so that (a) 1 ρC = gcosθ  cos α  v 2 cos 2 α v0  = 0 3  gcos θ  cosθ  2 Since v0 , α, and g are constants, ρC is a minimum at point B where cos θ is a maximum or θ = 0. Then, (b) ρ min = ρ B = 2 v0 cos 2 α g Q.E.D. ! ρC = or 1 cos3 θ 2  v0 cos 2 α      g   ρC = ρ min cos3 θ Q.E.D. ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 152. Let θ be the slope angle of the trajectory at an arbitrary point C. Then, ( aC )n = g cosθ = ρC 2 vC or ρC = 2 vC gcosθ But the horizontal component of velocity is constant, ( vC ) x = ( v A ) x ( vA ) x = ( vC ) x where Then, so that x = ( v0 ) x t = ( v0 cos α ) t and or t= x v0 cos α (1) ( vA ) x = v0 cosα ( vC ) x = v0 cosθ v0 cos α = vC cosθ ρC = 3 vC gv0 cos α (2) The vertical motion is uniformly accelerated ( vC ) y = ( v0 ) y − gt = v0 sin α − But 2 vC gx v0 cos α 2 (3) 2 =( 2 v0 x ) +( 2 v0 y ) = ( v0 cos α )  x  +  v0 sin α − g  v0 cos α   2 gx tan α g 2 x2  2 = v0 − + 1   2 4  cos 2 α  v0 v0   or 3 vC = 3 v0  1 2 gx tan α g 2 x2  − +  4 v0 2 v0 cos 2 α    3/2 (4) Finally, substituting (4) into (2) gives 2  v0 2 gx tan α g 2 x2  − + ρ= 1   2 4 g cos α  v0 v0 cos 2 α    3/2 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 153. Given: r = ( Rt cos ω nt ) i + ctj + ( Rt sin ω nt ) k Differentiating to obtain v and a, dr v= = R ( cos ω nt − ω nt sin ω nt ) i + cj + R ( sin ω nt + ω nt cos ω nt ) k dt dv 2 2 a= t cos ω nt i + R ω n cos ω nt + ω n cos ω nt − ω n t sin ω nt k = R − ω n sin ω nt − ω n sin ω nt − ω n dt ( ) ( ) 2 2 t cos ω nt  i + 2ω n cos ω nt − ω n t sin ω nt k = R  − 2ω n sin ω nt − ω n   Magnitudes of v and a. 2 2 v 2 = vx + v2 y + vz ( ) ( ) =   R ( cos ωnt − ωnt sin ωnt )   + (c) +   R ( sin ωnt + ωnt cos ωnt )   2 2 2 2 2  = R2  cos ωnt − 2ωnt sin ωnt cos ωnt + ωn t sin ωnt  + c 2 2 2 2  + R2  sin ωnt + 2ωnt sin ωnt cos ωnt + ωn t cos ωnt  2 2 = R 2 1 + ωn t + c2 2 2 a 2 = ax + a2 y + az 2 2 2 ( ) or v= 2 2 R 2 1 + ωn t + c2 ( )  2 = R 2  −2ωn sin ωnt − ωn t cos ωnt  ( ) + ( 2ω 2 n cos ωnt 2 2 − ωn t sin ωnt   ) 2 2 3 4 2 2 = R2   4ωn sin ωnt + 4ωnt sin ωnt cos ωnt + ωn t cos ωnt 2 3 4 2 cos 2 ωnt − 4ωn + 4ωn t sin ωnt cos ωnt + ωn t sin 2 ωnt   2 4 2 + ωn t = R 2 4ωn ( ) or 2 2 a = Rωn 4 + ωn t Tangential component of acceleration: At t = 0, v2 = R2 + c2 , at = a = 2 Rω n , dv R 2ω n 2t = 1/2 dt  2 2 2 R 1 + ωn t + c2    ( ) at = 0 Normal component of acceleration: But an = v2 an = a 2 − at2 = 2Rω n ρ v2 an or ρ = ρ = R2 + c2 ! 2 Rω n Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 154. With A = 3 and B = 1, the position vector is r = ( 3t cos t ) i + 3 t 2 + 1 j + ( t sin t ) k Differentiating to obtain v and a, ( ) v=  3t  dr  j + ( sin t + t cos t ) k = 3 ( cos t − t sin t ) i +   t2 + 1  dt      t  t2 + 1 − t    t2 + 1    dv   j a= = 3 ( − sin t − sin t − t cos t ) i + 3  dt t2 + 1         + ( cos t + cos t − t sin t ) k = −3 ( 2sin t + t cos t ) i + Magnitude of v 2. 2 2 v 2 = vx + v2 y + vz = 9 ( cos t − t sin t ) + 2 3 (t 2 +1 ) 3/2 j + ( 2cos t − t sin t ) k 9t 2 2 + ( sin t + t cos t ) 2 t +1 Differentiating, 2v dv 18t = 18 ( cos t − t sin t )( −2sin t − t cos t ) + dt 1 + t2 + 2 ( sin t + t cos t )( 2cos t − t sin t ) ( ) 2 When t = 0, a = 3j + 2k , v 2 = 9, 2v dv =0 dt a 2 = 32 + 22 = 13 Tangential acceleration: Normal acceleration: But an = v2 at = dv =0 dt or 9 13 an = 13 v2 = an an 2 = a 2 − at 2 = 13 ρ or ρ = ρ = 2.50 ft ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 155. For the sun, and g = 274 m/s 2 , R= 1 1 D =   1.39 × 109 = 0.695 ×109 m 2 2 ( ) Given that an = gR 2 v2 and that for a circular orbit a = n r r2 r= gR 2 v2 Eliminating an and solving for r, For the planet Earth, v = 107 × 106 m/h = 29.72 × 103 m/s Then, ( 274 ) ( 0.695 × 109 ) r= ( 29.72 )2 2 = 149.8 × 109 m r = 149.8 Gm ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 156. For the sun, and g = 274 m/s 2 R= 1 1 D =   1.39 × 109 = 0.695 × 109 m 2 2 ( ) Given that an = gR 2 v2 and that for a circular orbit: a = n r r2 r= gR 2 v2 Eliminating an and solving for r, For the planet Saturn, v = 34.7 × 106 m/h = 9.639 × 103 m/s Then, ( 274 ) ( 0.695 × 109 ) r= ( 9.639 )2 2 = 1.425 × 1012 m r = 1425 Gm ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 157. From Problems 11.155 and 11.156, an = an = gR 2 r2 v2 r g r For a circular orbit, Eliminating an and solving for v, For Venus, v= R g = 29.20 ft/s 2 R = 3761 mi = 19.858 × 106 ft. r = 3761 + 100 = 3861 mi = 20.386 × 106 ft Then, v = 19.858 × 106 29.20 = 23.766 × 103 ft/s 20.386 × 106 v = 16200 mi/h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 158. From Problems 11.155 and 11.156, an = an = gR 2 r2 v2 r g r For a circular orbit, Eliminating an and solving for v, For Mars, v=R g = 12.24 ft/s 2 R = 2070 mi = 10.930 × 106 ft r = 2070 + 100 = 2170 mi = 11.458 × 103 ft Then, v = 10.930 × 106 12.24 = 11.297 × 103 ft/s 11.458 × 106 v = 7700 mi/h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 159. From Problems 11.155 and 11.156, an = an = gR 2 r2 v2 r g r For a circular orbit, Eliminating an and solving for v, For Jupiter, g = 75.35 ft/s 2 v= R R = 44432 mi = 234.60 × 106 ft r = 44432 + 100 = 44532 mi = 235.13 × 106 ft Then, v = 234.60 × 106 ( ) 75.35 = 132.8 × 103 ft/s 235.13 × 106 v = 90600 mi/h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 160. Radius of Earth Radius of orbit Normal acceleration R = ( 3960 mi )( 5280 ft/mi ) = 20.908 × 106 ft r = ( 3960 + 10900 )( 5280 ) = 78.4608 × 106 ft an = gR 2 r2 and an = v2 = 2 v2 r gR 2 r Thus, v 2 gR 2 = 2 r r v2 = or ( 32.2 ) ( 20.908 × 106 ) 78.4608 × 106 = 179.40 × 106 ft 2 /s 2 v = 13.3941 × 103 ft/s Time T for one orbit. vT = 2π r 6 2π r 2π 78.4608 × 10 T = = = 36.806 × 103 s v 13.3941 × 103 ( ) T = 10.22 h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 161. Normal acceleration. Solve for v2. Data: an = gR 2 r2 gR 2 r and an = v2 ρ = v2 r v 2 = ran = g = 9.81 m/s 2 , R = 6370 km = 6.370 × 106 m r = 384 × 103 km = 384 × 106 m v2 = ( 9.81) ( 6.370 × 106 ) 384 × 106 2 = 1.0366 × 106 m 2 /s 2 v = 3670 km/h ! v = 1.018 m/s Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 162. From Problems 155 through 156, For a circular orbit, Eliminating an and solving for v, For one orbit the distance traveled is 2π r; hence, the time is or an = an = gR 2 r2 v2 r g r v=R t= 2π r v t= tA = 2π r 3 2 Rg1 2 2π rA3 2 Rg1 2 and For satellites A and B, tB = 2π rB 3 2 Rg1 2 Let n = number of orbits of B. For the next alignment, ( n + 1) t A = nt B 1  rB  =  n  rA  32 or −1 n + 1 t B  rB  = =  n t A  rA  32 Data: R = 6370 km = 6.370 × 103 m rA = 6370 + 190 = 6560 km = 6.560 × 103 m rB = 6370 + 320 = 6690 km = 6.690 × 103 m Then, 1  6.690 × 103  = 3  n   6.560 × 10  3/2 −1 = 0.02987 or n = 33.475 continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Time for orbit of satellite B is tB = 2π 6.690 × 106 ( 6.370 × 106 ( 9.81) ( ) ) 32 12 = 5.449 × 103 s = 1.5137 h Time for next alignment is nt B = ( 33.475 )(1.5137 ) nt B = 50.7 h ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 163. Differentiate the expressions for r and θ with respect to time. r = 1 + 2t − 6t 2 + 8t 3 & = 2 − 12t + 24t 2 r && r = −12 + 48t θ = 0.5e−0.8t sin 3π t θ& = − 0.4e−0.8t sin 3π t + 1.5π e−0.8t cos 3π t && = 0.32e−0.8t sin 3π t − 1.2π e−0.8t cos 3π t θ −1.2π e −0.8t cos 3π t − 4.5π 2e−0.8t sin 3π t At t = 0.5 s, e −0.8t = 0.67032, r = 1.5 ft, & = 2.00 ft/s, r && r = 12 ft/s 2 , θ = − 0.33516 rad, (a) Velocity of the collar. sin 3π t = −1, cos 3π t = 0 && = 29.56 rad/s2 θ& = 0.26812 rad/s, θ &e r + rθ&eθ v=r v = ( 2.00 ft/s ) er + ( 0.402 ft/s ) eθ ! vr = 2 ft/s, vθ = 0.402 ft/s ! (b) Acceleration of the collar. & 2 e + rθ && + 2r & e =ae +ae &θ a = && r − rθ r θ r r θ θ ar = 12 − (1.5 )( 0.26812 ) 2 ( ) ( ) ar = 11.89 ft/s 2 ! aθ = 45.41 ft/s 2 ! a = 11.89 ft/s 2 er + 45.41 ft/s 2 eθ ! aθ = (1.5 )( 29.56 ) + ( 2 )( 2 )( 0.26812 ) ( ) ( ) (c) Acceleration of the collar relative to the rod. && rer = 12 ft/s 2 er ! ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 164. Differentiate the expressions for r and θ with respect to time. r= 10 mm, t+6 4 &=− r 10 (t + 6) 2 mm/s, && r= 20 ( t + 6 )3 mm/s 2 θ = At t = 1 s, π sin π t rad, r= 10 mm; 7 θ& = 4cos π t rad/s &=− r && = 4π sin π t rad/s 2 θ && r= && = 0 θ 10 mm/s, 49 20 mm/s 2 343 θ = 0, (a) Velocity of the collar. θ& = − 4 rad/s, & = 0.204 mm/s, vr = r & = − 5.71 mm/s vθ = rθ v B = ( 0.204 mm/s ) er − ( 5.71 mm/s ) eθ ! (b) Acceleration of the collar. &2 = ar = && r − rθ 20  10  2 −   ( −4 ) = − 22.8 mm/s 2 343  7  && + 2r & =  10  ( 0 ) + ( 2 )  − 10  ( −4 ) = 1.633 mm/s 2 &θ aθ = rθ      7   49  a B = − 22.8 mm/s 2 er + 1.633 mm/s 2 eθ ! ( ) ( ) (c) Acceleration of the collar relative to the rod. a B/OA = && rer = 20 er 343 a B/OA = 0.0583 mm/s 2 er ! ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 165. Given Differentiating twice r = 2 B cos ( At/2B ) & = − A sin ( At/2B ) r θ = At/2B θ& = A/2 B && = 0 θ && r = − A2 /2 B cos ( At/2B ) ( ) Components and magnitude of velocity. & = − A sin ( At/2B ) = − A sin θ vr = r & = 2 B cos ( At/2B )  ( A/2 B ) = A cosθ vθ = rθ   (a) 2 2 v = vr + vθ = A2 sin 2 θ + A2 cos 2 θ = A 2 v=A! Components and magnitude of acceleration. & 2 = − A2 /2B cos ( At/2 ) + 2 B cos ( At/2 B )  [ A/2 B ] ar = && r − rθ   ( ) = − A2 /B cosθ && + 2r & = 0 + (2)  − A sin ( At/2 B )  ( A/2B ) &θ aθ = rθ   ( ) = − A2 /B sin θ 2 2 a = ar + aθ = ( A /B ) cos θ + ( A /B ) sin θ 4 2 2 4 2 2 = A2 /B From the figure a is perpendicular to v Thus, an = v2 a = A2 /B ! an = a = A2 /B ρ ρ= ρ= v2 an A2 2 (b) ( A /B ) =B ρ=B! Since ρ is constant, the path is a circle of radius B. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 166. Differentiate the expressions for r and θ with respect to time. r = b ( 2 + cos π t ) , & = − π b sin π t , r && r = − π 2b cos π t θ = π t, θ& = π , && = 0 θ (a) At t = 2 s, r = 3b, & = 0, vr = r sinπ t = 0, & = 0, r cosπ t = 1 && r = − π 2b, θ = 2π rad, θ& = π rad/s v = 3π beθ W & = 3π b, vθ = rθ & 2 = − π 2b − ( 3b ) π 2 = − 4π 2b ar = && r − rθ && + 2r & = 0, &θ aθ = rθ a = − 4π 2be r W (b) Values of θ for which v is maximum. & = − π b sin π t vr = r & = − b ( 2 + cos π t ) π v = rθ θ 2 2 + vθ 2 = π 2b2 ⎡ v 2 = vr sin 2 π t + ( 2 + cos π t ) ⎤ ⎢ ⎥ ⎣ ⎦ 2 2 ⎤ = π 2b 2 ⎡ ⎣sin π t + 4 + 4cos π t + cos π t ⎦ = π 2b 2 ( 5 + 4cos π t ) v 2 is maximum when cos π t = 1 or hence π t = 0, 2π , 4π , 6π , etc But θ = π t, θ = 2Nπ , N = 0, 1, 2, K W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 167. Differentiate the expressions for r and θ with respect to time. r = 6t 1 + 4t 2 , & = 6 1 + 4t 2 + 24t 2 1 + 4t 2 r && r = 72t 1 + 4t 2 ( ) −1 2 ( ) −1 2 − 96t 3 1 + 4t 2 ( ) −3 2 , θ = arctan 2t θ& = 2 1 + 4t 2 ( ) −1 , && = − 16t 1 + 4t 2 θ ( ) −2 (a) At t = 0, r = 0, & = 6 ft/s, r && r =0 && = 0 θ θ = 0, & = 6 ft/s, vr = r & 2 = 0, ar = && r − rθ θ& = 2 rad/s, & = 0, vθ = rθ v = ( 6 ft/s ) e r W && + 2r & = 24 ft/s 2 , &θ aθ = rθ & = 9 2 ft/s, r = 3 2 ft, r && r = 15 2 ft/s 2 a = 24 ft/s 2 eθ W ( ) (b) At t = 0.5 s, θ = π 4 rad, && = − 2 rad/s 2 θ& = 1 rad/s, θ & = 4.243 ft/s vθ = rθ & = 12.73 ft/s, vr = r v = (12.73 ft/s ) er + ( 4.24 ft/s ) eθ W & 2 = 15 2 − 3 2 (1)2 = 16.97 ft/s 2 ar = && r − rθ && + 2r & = 3 2 ( −2 ) + ( 2 ) 9 2 (1) = 16.97 ft/s 2 &θ aθ = rθ ( ) a = 16.97 ft/s 2 er + 16.97 ft/s 2 eθ W ( ) ( ) Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 168. Change to rectangular coordinates. cosθ = r= x r and sin θ = y r Equation of the path: 3 3 3r = = y x sin θ − cosθ y −x − r r or y = x + 3. from which Also, y−x=3 tan θ = x = 3t 2 & = 6t , vx = x y x+3 3 1 = =1+ =1+ 2 x x x t and y = 3t 2 + 1 from which Differentiating, & = 6t vy = y ax = && x = 6, (a) Magnitudes: a y = && y=6 v = 6 2 t ft/s W 2 v = vx + v2 y 2 a = ax + a2 y a = 6 2 ft/s 2 W (b) y = x + 3 is the equation of a straight line. Hence, ρ =∞W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 169. Sketch the directions of the vectors v and eθ. vθ = v ⋅ eθ = − v cosθ But Hence, & vθ = rθ & = − v cosθ rθ But from geometry, & bθ = − v cosθ cosθ r= b cosθ v=− & bθ 2 cos θ or Speed is the absolute value of v. v= & bθ W 2 cos θ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 170. From geometry, Differentiating with respect to time, Transverse component of acceleration && + 2r &= &θ aθ = rθ r= b cosθ &= r & b sin θθ 2 cos θ && &2 bθ 2b sin θθ + 2 cosθ cos θ (1) Sketch the directions of the vectors a and eθ. aθ = a ⋅ eθ = − a cosθ Matching from (1) and (2) and solving for a, && &2 bθ 2b sin θθ − cos 2 θ cos3 θ b && + 2 tan θθ &2 =− θ 2 cos θ (2) a=− ( ) Since magnitude of a is sought, |a| = b && θ + 2 tan θθ& 2 W cos 2 θ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 171. Sketch the geometry. θ + (180° − β ) + α = 180° α = β −θ r d = sin (180° − β ) sin α r= Sketch the velocity vectors. vθ = v ⋅ eθ = v cos ( 90° − α ) = v sin α But or & vθ = rθ d sin β sin α or v sin α = d sin β & θ, sin α v= d sin β θ& W sin 2 ( β − θ ) v= d sin β & θ sin 2 α Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 172. & = 0, Looking at d and β as polar coordinates with d &=0 & = dω , v = dβ v =d β d &β && + 2d & = 0, aβ = d β && − d β & 2 = − dω 2 ad = d Geometry analysis: r = d 3 for angles shown. (a) Velocity analysis: Sketch the directions of v, er and eθ. & = v ⋅ er = dω cos120° vr = r 1 & = − dω W r 2 & = v ⋅ e = dω cos 30° vθ = rθ θ 3 dω cos 30° dω 2 = r d 3 Sketch the directions of a, er and eθ. θ& = θ& = ω W 1 2 (b) Acceleration analysis: ar = a ⋅ er = a cos150° = − & 2 = − 3 dω 2 && r − rθ 2 3 dω 2 2 && r=− 3 & 2 = − 3 dω 2 + d 3 ⎛ 1 ω ⎞ dω 2 + rθ ⎜ ⎟ 2 2 ⎝2 ⎠ 2 && r=− 3 dω 2 W 4 1 aθ = a ⋅ eθ = dω 2 cos120° = − dω 2 2 && & &θ aθ = rθ + 2r && = θ 1 & = &θ aθ − 2r r ( ) 1 3d ⎡ 1 ⎛ 1 ⎞⎛ 1 ⎞ ⎤ 2 ⎢ − 2 dω − ( 2 ) ⎜ − 2 dω ⎟⎜ 2 ω ⎟ ⎥ ⎝ ⎠⎝ ⎠⎦ ⎣ && = 0 W θ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 173. Rate of change of θ . Δθ = 48.0° − 47.0° = 1.0° = 17.453 × 10−3 rad Δt = 0.5 s θ& ≈ Let r be a polar coordinate with origin at A. b = 4 km = 4 × 103 m r= b 4 × 103 = = 5.921 × 103 m cosθ cos 47.5° Δθ 17.453 × 10−3 = = 34.907 × 10−3 rad/s Δt 0.5 & = 5.921 × 103 34.907 × 10−3 = 206.68 m/s vθ = rθ From geometry, v= vθ 206.68 = cosθ cos 47.5° v = 306 m/s W Alternate solution. x = b tan θ ( )( ) & & = bθ & = b sec2θθ v=x cos 2θ v= ( 4 × 10 )(34.907 × 10 ) = 306 m/s 3 −3 cos 2 47.5° Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 174. Changes in values over the interval Δr = 13600 − 12600 = 1000 ft Δθ = 28.3° − 31.2° = − 2.9° = − 5.0615 × 10−3 rad Δt = 2s Rates of change. &= r Δr 1000 = = 500 ft/s Δt 2 Δθ − 5.0615 × 10−3 = = − 2.5307 × 10−3 rad/s Δt 2 12600 + 13600 = 13100 ft 2 31.2° + 28.3° = 29.75° 2 θ& = Mean values. r= θ = Velocity components. & = 500 ft/s vr = r & = (13100 ) −2.5307 × 10−3 = − 331.53 ft/s vθ = rθ 2 2 v = vr + vθ = ( ) ( 500 )2 + ( −331.53)2 = 600 ft/s v = 409 mi/h W vx = vr cosθ − vθ sin θ = 500cos 29.75° − ( −331.53) sin 29.75° = 598.61 ft/s v y = vr sin θ + vθ cosθ = 500sin 29.75° + ( ( −331.53) cos 29.75° ) = − 39.73 ft/s tan α = −v y vx = 39.73 = 0.06636 598.61 α = 3.80° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 175. r = be1/2θ , 2 2 & & = be1/2θ θθ r 2 &, & = be1/2θ θθ vr = r 2 2 v 2 = vr + vθ = be1/2θ ( 2 2 ) (θ & = be1/2θ 2θ & vθ = rθ 2 &2 +1 θ v = be1 2θ θ 2 + 1 2 ) ( ) 1/2 θ& W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 176. r= b θ 2 , &=− r 2b & θ, 3 θ3 2b & θ &=− vr = r θ &= vθ = rθ θ2 b & θ 2 2 v 2 = vr + vθ = 4b 2 & 2 b3 & 2 b 2 θ + 4 θ = 6 4 + θ 2 θ& 2 6 θ θ θ ( ) v= b θ 3 (4 + θ ) 2 12 θ& W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 177. r = be1/2θ , 2 2 &, & = be1/2θ θθ r 2 && r = be1/2θ ⎡ θθ& ⎢ ⎣ ( ) 2 & 2 + θθ &&⎤ +θ ⎥ ⎦ & 2 = be1/2θ 2 ⎡ θθ & ar = && r − rθ ⎢ ⎣ ( ) 2 & 2 + θθ && − θ & 2 ⎤ = be1/2θ 2 ⎡ θθ & +θ ⎥ ⎢ ⎦ ⎣ ( ) 2 &&⎤ + θθ ⎥ ⎦ && + 2r & = be1/2θ 2θ && + 2be1/2θ 2θθ & 2 = be1/2θ 2 ⎡θ && &2 ⎤ &θ aθ = rθ ⎣ + 2θθ ⎦ But 2 θ& = ω ar = be1/2θ (θω ) 2 and and 2 2 && = 0 θ aθ = be1/2θ 2 2 2 a 2 = ar + aθ = be1/2θ ( ) (θ ( 2θω ) 2 4 + 4θ 2 ω 4 ) a = be1/2θ θ θ 2 + 4 2 ( ) 1/2 ω2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 178. r= b θ 2 , &=− r θ 2b & θ, 3 && r=− θ 2b && 6b & 2 θ + 4θ 3 θ &2 = − ar = && r − rθ θ 2b & 6b & 2 b b && + 6θ & 2 − θ 2θ &2 θ + 4 θ − 2 θ& 2 = 4 −2θθ 3 θ θ θ ( ) b && && + 2r &= bθ && = ( 2 ) ⎛ − 2b ⎞θ &2 &2 &θ aθ = rθ ⎜ 3 ⎟ = 3 θθ − 4θ θ2 θ θ ⎝ ⎠ ( ) But ar = b θ& = ω and && = 0 θ θ 4 ( 6 − θ )ω 2 2 and aθ = − 2 4b θ3 ω2 ω2 2 2 a 2 = ar + aθ = b2 θ8 2 ( 36 − 12θ + θ 4 ω2 + θ 4 ω2 + ) 16b 2 θ6 = b2 θ8 (36 + 4θ ) a= b θ 4 (36 + 4θ 2 +θ4 ) 12 ω2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 179. Sketch the geometry. Law of cosines: r 2 = d 2 + h 2 − 2dh cos ϕ Differentiating with respect to time and noting that d and h are constant, & = 2dh sin ϕϕ & 2rr &= r Law of sines: so that dh sin ϕ & ϕ r sin ϕ sin θ = r d & = h sin θϕ & r Q.E.D W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 180. Given: Differentiating with respect to time, &=− R && = R (a) t = 0. A R= A , t +1 θ = Bt , z= Ct t +1 ( t + 1) 2A 3 2 , θ& = B, && = 0, θ &= z &=− z C ( t + 1) − Ct ( t + 1) 2 = C (1 + t )2 ( t + 1) , 2C (1 + t )3 z=0 &=C z && z = − 2C &=C vz = z R = A, & = − A, R && = 2 A, R & = − A, vR = R θ = 0, && = B, θ && = 0, θ & = AB, vθ = Rθ 2 2 2 v 2 = vR + vθ + vz = A2 + A2 B 2 + C 2 v= A2 + A2 B 2 + C 2 W & 2 = 2 A − AB 2 && − Rθ aR = R && + 2R & = 0 − 2 AB &θ aθ = Rθ 2 aR = 4 A2 − 4 A2 B 2 + A2 B 4 2 aθ = 4 A2 B 2 az = 4C 2 az = && z = − 2c 2 2 2 a 2 = aR + aθ + az = 4 A2 + A2 B 4 + 4C 2 a = 4 A2 + A2 B 4 + 4C 2 W (b) t = ∞. R = 0, θ = ∞, z = C, && = 0, R & = 0, R && = 0, θ θ& = B, && z=0 & = 0, vz = z & = 0, z & = 0, vr = R & 2 = 0, && − Rθ ar = R & = 0, vθ = Rθ v=0W az = && z = 0, a=0W && − Rθ & 2 = 0, aθ = Rθ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 182. From problem 11.97, the position vector is r = ( Rt cos ω nt ) i + ctj + ( Rt sin ω nt ) k. Differentiating to obtain v and a, dr v= = R ( cos ω nt − ω nt sin ω nt ) i + cj + R ( sin ω nt + ω nt cos ω nt ) k dt dv 2 2 = R − ω n sin ω nt − ω n sin ω nt − ω n a= t cos ω nt i + R ω n cos ω nt + ω n cos ω nt − ω n t sin ω nt k dt ( ) ( ) 2 2 = R  − 2ω n sin ω nt − ω n t cos ω nt i + 2ω n cos ω nt − ω n t sin ω nt k    ( ) ( ) ) i j k v × a = vx v y vz = v y a z − vz a y i + ( vz a x − vx az ) j + v x a y − v y a x k ax a y az ( ( ) 2 2 = cR 2ω n cos ω nt − ω n t sin ω nt  i +  R 2 ( sin ω nt + ω nt cos ω nt ) − 2ω n sin ω nt − ω n t cos ω nt    2 − R 2 ( cos ω nt − ω nt sin ω nt ) 2ω n cos ω nt − ω n t sin ω nt  j  2 +  − cR −2ω n sin ω nt − ω n t cos ω nt  k   2 2 = cRω n ( 2 cos ω nt − ω nt sin ω nt ) i − R 2ω n 2 + ω n t j + cRω n ( 2sin ω nt + ω nt cos ω nt ) k 2  2 2 2 2 2 2  | v × a | = c 2 R 2ω n 4 + ωn τ + R 4ω n 2 + ωn t    ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1/2 The binormal unit vector eb is given by Let α be the angle between the y-axis and the binormal. cos α = eb ⋅ j = eb = v×a | v×a | ( v × a) ⋅ j = |v×a | c 2 R 2ω 2 4 + ω 2t 2 + R 4ω 2 2 + ω 2t 2  n n n n   ( R 2ω n ( 2 + ω nt ) ) ( ) 12 Let 2 2 A = R 2ω n 2 + ω n t , ( ) A as C shown in the sketch. The angle that the osculating plane makes with the y-axis is the angle β. 2 2 B = cRω n 4 + ω n t 12 ( ) , C= A2 + B 2 so that cos α = 2 2 R 2 + ωn t A tan β = = B c 4 + ω 2t 2 1 2 n ( ( ) ) β = tan −1 2 2 R 2 + ωn t 2 2 c 4 + ωn t ( ( ) ) 12 ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 183. For A = 3 and B = 1, r = ( 3t cos t ) i + 3 t 2 + 1 j + ( t sin t ) k Differentiating to obtain v and a. v= a= dr t = 3 ( cos t − t sin t ) i + 3 j + ( sin t + t cos t ) k 2 dt t +1 dv 1 = 3 ( − 2sin t − t cos t ) i + 3 dt t2 + 1 ( ) ( ) 3/2 j + ( 2cot t − t sin t ) k (a) At t = 0, v = 3 (1 − 0 ) i + ( 0 ) j + ( 0 ) k = 3i a = − 3(0)i + 3(1) j + ( 2 − 0 ) k = 3j + 2k i j k v × a = 3 0 0 = − 6 j + 9k 0 3 2 | v × a | = 62 + 92 = 10.817 eb = v×a = − 0.55470 j + 0.83205k | v×a | cosθ 2 = 0.83205 cosθ x = 0, cosθ y = − 0.55470, θ x = 90°, (b) At t = θ y = 123.7°, θ z = 33.7° ! π 2 s, v = − 4.71239i + 2.53069 j + k a = − 6i + 0.46464 j − 1.5708k i j k 1 v × a = − 4.71239 2.53069 0.46464 1.5708 −6 = − 4.43985 i − 13.4022 j + 12.9946 k continued Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System 2 2 2 | v×a|=  ( 4.43985 ) + (13.4022 ) + (12.9946 )    1/2 = 19.1883 eb = v×a = − 0.23138i − 0.69846 j + 0.67721k | v×a | cosθ x = − 0.23138, cosθ y = − 0.69846, cosθ z = 0.67721 θ x = 103.4°, θ y = 134.3°, θ z = 47.4° ! Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 184. Given: a = kt 2 ft/s 2 , t x0 = 36 ft, t x9 = 144 ft, 1 3 kt 3 v9 = 27 ft/s v − v0 = ∫ 0 a dt = ∫ 0 kt 2 dt = Velocity: t v = v0 + x − x0 = ∫ 0 v dt = v0t + 1 3 kt 3 1 4 kt 12 Position: x = x0 + v0t + x = 144 ft 1 4 1 4 kt = 36 + v0t + kt 12 12 and v = 27 ft/s When t = 9 s, 36 + v0 ( 9 ) + 1 4 k ( 9 ) = 144 12 or 9v0 + 546.75k = 108 v0 + 1 3 k ( 9 ) = 27 3 (1) v0 + 243k = 27 Solving equations (1) and (2) simultaneously yields: v0 = 7 ft/s Then, (2) and k = 0.082305 f t/s 4 x = 36 + 7t + 0.00686t 4 ft W v = 7 + 0.0274t 3 ft/s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 185. (a) Determination of k. From dv = a dt , dt = dv dv = a 0.6 (1 − kv ) Integrating, using the condition v = 0 when t = 0, t v ∫ 0 dt = ∫ 0 0.6 1 − kv ( ) dv or t t 0 =− v 1  ln (1 − kv )   0 0.6k t=− 1 ln (1 − kv ) 0.6k (1) Using t = 20 s when v = 6 mm/s, Solving by trial, 20 = − 1 ln (1 − 6k ) 0.6k k = 0.1328 s/m W (b) Position when v = 7.5 m/s. From v dv = a dx, dx = v dv v dv = a 0.6 (1 − kv ) Integrating, using the condition x = 6 m when v = 0, 1 v 1  1 x−6= −1 +  dv = ∫ 0 0.6k  1 − kv  0.6k x=6− 1 0.6k 1   v + k ln (1 − kv )    ∫ 6 dx = ∫ 0 0.6 1 − kv ( ) 1    −v − k ln (1 − kv )  0 v x v v dv Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Using v = 7.5 m/s and the determined value of k: x=6− 1 1   7.5 + ln (1 − ( 0.1328 )( 7.5 ) )   0.6 0.1328 0.1328 ( )( )  vmax = 1 1 = k 0.1328 x = 434 m W vmax = 7.53 m/s W (c) Maximum velocity occurs when a = 0. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 186. Constant acceleration. v0 = 25 mi/h = 36.667 ft/s v f = 65 mi/h = 95.333 ft/s x0 = 0 and x f = 0.1 mi = 528 ft 2 v2 f = v0 + 2a x f − x0 2 v2 f − v0 ( ) (a) Acceleration. a= 2 x f − x0 ( ) = 95.3332 − 36.667 2 = 7.3333 ft/s 2 2 ( 528 − 0 ) a = 7.33 ft/s 2 W (b) Time to reach 65 mph. v f = v0 + at f tf = v f − v0 a = 95.333 − 36.667 7.3333 t f = 8.00 s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 187. Let x be position relative to the fixed supports, taken positive if downward. Constraint of cable on left: 2v A + 3vB = 0, Constraint of cable on right: vB + 2vC = 0, or vC = − or vB = − 2 x A + 3xB = constant 2 vA, 3 and aB = − 2 aA 3 xB + 2 xC = constant 1 1 vB = v A , 2 3 and aC = 1 aA 3 Block C moves downward; hence, block A also moves downward. (a) Accelerations. v A = ( v A )0 + a AT or aA = v A − ( v A )0 t = 456 − 0 = 38.0 mm/s2 12 a A = 38.0 mm/s 2 W W W aB = − aC = (b) 2 ⎛2⎞ a A = − ⎜ ⎟ ( 38.0 ) = − 25.3 mm/s 2 3 ⎝3⎠ a B = 25.3 mm/s 2 aC = 12.67 mm/s 2 1 ⎛1⎞ a A = ⎜ ⎟ ( 38.0 ) = 12.67 mm/s 2 3 ⎝3⎠ vB = ( vB )0 + aBt = 0 + ( − 25.3)( 8 ) = − 203 mm/s Velocity and change in position of B after 8 s. v B = 203 mm/s W xB − ( x B ) 0 = ( v B ) 0 t + 1 1 2 aBt 2 = 0 + ( − 25.3)( 8 ) = − 811 mm 2 2 ΔxB = 811 mm W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 188. (a) Construction of the curves. Construct the a−t curve. 0 < t < 10 s: 10 s < t < 26 s: 26 s < t < 41 s: 41 s < t < 46 s: 46 s < t < 50 s: a = slope of v − t curve ∆t = 10 s, ∆t = 16 s, ∆t = 15 s, ∆t = 5 s, ∆t = 4 s, ∆v = 0 ∆v = − 80 m/s ∆v = 0 ∆v = 15 m/s ∆v = 0 a= a= a= a= a= ∆v =0 ∆t ∆v = − 5 m/s 2 ∆t ∆v =0 ∆t ∆v = 3 m/s 2 ∆t ∆v =0 ∆t Construct the x−t curve. x is maximum or minimum where v = 0. For 10 s ≤ t ≤ 26 s, v = 0 when ∆x = area of v−t curve. v = 60 − 5 ( t − 10 ) 60 − 5t + 50 = 0 or t = 22 s Also 0 to 10 s 10 s to 22 s 22 s to 26 s x0 = −540 m ∆x = (10 )( 60 ) = 600 m ∆x = ∆x = 1 (12 )( 60 ) = 360 m 2 1 ( 4 )( − 20 ) = − 40 m 2 x10 = − 540 + 600 = 60 m x22 = 60 + 360 = 420 m x26 = 420 − 40 = 380 m Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System 26 s to 41 s 41 s to 46 s ∆x = (15 )( − 20 ) = − 300 m x41 = 380 − 300 = 80 m  − 20 − 5  ∆x = ( 5 )   = − 62.5 m 2   ∆x = ( 4 )( − 5) = − 20 m x46 = 80 − 62.5 = 17.5 m x50 = 17.5 − 20 = − 2.5 m 46 s to 50 s (b) Total distance traveled. 0 ≤ t ≤ 22 s, 22 s ≤ t ≤ 50 s, d1 = x22 − x0 = 420 − ( − 540 ) = 960 m d 2 = x50 − x22 = − 2.5 − 420 = 422.5 d = 1383 m W Total: (c) Times when x = 0. For 0 ≤ t ≤ 10 s, At x = 0, For 46 s ≤ t ≤ 50, d = d1 + d 2 = 1382.5 m x = − 540 + 60t m − 540 + 60t = 0 x = 17.5 − 5 ( t − 46 ) m 17.5 − 5 ( t − 46 ) = 0 t − 46 = 3.5 t = 49.5 s W t = 9sW At x = 0, Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 189. ( vA )0 = 100 km/h = 27.778 m/s ( vB )0 = 25 km/h = 6.944 m/s Sketch acceleration curve for car B over 0 < t < 5 s. Using moment-area formula at t = 5 s. xB − ( xB )0 = ( vo ) t + ( aB )( 5 )( 2.5 ) 70 = ( 6.944 )( 5 ) + 12.5aB aB = 2.822 m/s 2 Determine when B reaches 100 km/h. ( vB ) f = ( vB )0 + A2 27.778 = 6.944 + 2.822t B t B = 7.38 s A2 = ( 2.822 )( 7.38 ) = 20.83 m/s Then, and Subtracting, xB = ( xB )0 + ( vB )0 t B + A2 tB 2 by moment-area formula x A = ( x A ) 0 + ( v A )0 t B tB ⎤ xB − x A = ( x B ) 0 − ( x A ) 0 + ⎡ ⎣( vB )0 − ( v A )0 ⎦ t B + A2 2 Then, ⎛ 7.38 ⎞ xB − x A = 120 + ( 6.944 − 27.778 )( 7.38 ) + ( 20.83) ⎜ ⎟ ⎝ 2 ⎠ Car B is ahead of car A. xB/ A = 43.1 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 190. (a) Vertical motion: y = y0 + v y y0 = 1.5 m, v y gt 2 ( )0 t − 1 2 or or ( )0 = 0 t= 2 ( y0 − y ) g At point B, y=h tB = 2 ( y0 − h ) g When h = 788 mm = 0.788 m, When h = 1068 mm = 1.068 m, Horizontal motion: tB = tB = x0 = 0, ( vx )0 = v0 , x = v0t or ( 2 )(1.5 − 0.788) 9.81 = 0.3810 s = 0.2968 s ( 2 )(1.5 − 1.068) 9.81 v0 = v0 = v0 = x xB = t tB With xB = 12.2 m, we get and 12.2 = 32.02 m/s 0.3810 12.2 = 41.11 m/s 0.2968 115.3 km/h ≤ v0 ≤ 148.0 km/h W 32.02 m/s ≤ v0 ≤ 41.11 m/s (b) Vertical motion: Horizontal motion: tan α = − or vy = vy vx = v0 vy dy gt B =− = B dx ( vx )B v0 ( )0 − gt = − gt ( ) For h = 0.788 m, For h = 1.068 m, tan α = tan α = ( 9.81)( 0.3810 ) = 0.11673, 32.02 α = 6.66° W α = 4.05° W ( 9.81)( 0.2968) = 0.07082, 41.11 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 191. The horizontal and vertical components of velocity are vx = v0 sin15° v y = v0 cos15° − gt At point B, vx v0 sin15° = = − tan12° v y v0 cos15° − gt or v0 sin15° + v0 cos15° tan12° = gt tan12° 0.46413v0 = gt tan12° t = 2.1836 v0 g Vertical motion: y − y0 = v0 cos15°t − 1 2 gt 2 2 v0 1 2⎛v ⎞ − g ( 2.1836 ) ⎜ 0 ⎟ 2 g ⎝g⎠ 2 = 2.1836 cos15° = − 0.27486 2 v0 g ⎛ 8 ⎞ 2 v0 = − 3.638g ( y − y0 ) = − ( 3.638 )( 32.2 ) ⎜ − − 0⎟ 12 ⎝ ⎠ = 78.10 ft 2 / s 2 v0 = 8.84 ft /s W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 192. First determine the velocity vC of the coal at the point where the coal impacts on the belt. Horizontal motion: ⎤ ( vC ) x = ⎡ ⎣( vC ) x ⎦ 0 = −1.8cos 50° = −1.1570 m/s Vertical motion: = ⎡( vC ) y ⎤ ( vC )2 y ⎣ ⎦ 2 0 − 2 g ( y − y0 ) 2 = (1.8sin 50° ) − ( 2 )( 9.81)( −1.5 ) = 31.331 m 2 / s 2 ( vC ) y tan β = = − 5.5974 m/s − 5.5974 = 4.8379, −1.1570 2 2 β = 78.32° 2 = ( vC ) x + ( vC ) y = 32.669 m 2 /s 2 vC vC = 5.7156 m/s, or vC = 5.7156 m/s 78.32° vC = ( −1.1570 m/s ) i + ( − 5.5974 m/s ) j v B = vB ( − cos10°i + sin10° j) Velocity of the belt: Relative velocity: (a) v C/B = v C − v B = v C + ( − v B ) v C/B is vertical. ( vC/B )x = 0 vB = 1.175 m/s ( vC/B )x = −1.1570 − vB ( − cos10°) = 0, (b) v B = 1.175 m/s vC/B is minimum. Sketch the vector addition as shown. 2 2 2 vB /C = vB + vC − 2vB vC cos 88.32° 10° W Set the derivative with respect to vB equal to zero. 2vB − 2vC cos88.32° = 0 vB = vC cos88.32° = 0.1676 m/s v B = 0.1676 m/s 10° W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 193. Given: Then, (a) t = 0, ( vA )0 = 0, ( a A )t = dv A = 0.8 in./s 2 dt v A = ( v A )0 + ( a A )t t = 0.8 t v A = 0, ( a A )n = 2 vA ρ =0 a A = ( a A )t (b) t = 2 s, a A = 0.800 in./s 2 W v A = 0 + ( 0.8 )( 2 ) = 1.6 in./s ( a A )n = 2 vA ρ = (1.6 )2 3.5 = 0.731 in./s 2 1/2 1/2 2 2⎤ ⎡( 0.8 )2 + ( 0.731)2 ⎤ aA = ⎡ ⎢( a A )t + ( a A )n ⎦ ⎥ =⎢ ⎥ ⎣ ⎣ ⎦ a A = 1.084 in./s 2 W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 194. (a) At point A. a A = g = 9.81 m/s 2 Sketch tangential and normal components of acceleration at A. ( aA )n ρA = (b) = gcos 50° ( 2) v A2 = ( a A )n 9.81cos 50° 2 ρ A = 0.634 m W At point B, 1 meter below point A. Horizontal motion: ( vB ) x = ( v A ) x = 2cos 50° = 1.286 m/s Vertical motion: 2 = ( v A ) y + 2a y ( y B − y A ) ( vB ) 2 y = ( 2 cos 40° ) + ( 2 )( − 9.81)( −1) = 21.97 m 2 /s 2 2 ( vB ) y = 4.687 m/s tan θ = ( vB ) y ( vB ) x = 4.687 , 1.286 or θ = 74.6° aB = gcos 74.6° ( v B ) x + ( vB ) y v 2 ρB = B = gcos 74.6° ( aB ) n = 2 2 (1.286 )2 + 21.97 9.81cos 74.6° ρ B = 9.07 m W Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. COSMOS: Complete Online Solutions Manual Organization System Chapter 11, Solution 195. Differentiate the expressions for r and θ with respect to time. ( ) θ = 2 ( 2t + 4e ) rad, r = 6 4 − 2e − t ft, − 2t & = 12e − t ft/s, r && r = −12e − t ft/s 2 θ& = 2 2 − 8e− 2t rad/s & = 12 ft/s, r && r = −12 ft/s 2 ( ) && = 32e− 2t rad/s 2 θ (a) At t = 0 s, r = 12 ft, θ = 8 rad, & = 12 ft/s, vr = r θ& = −12 rad/s, && = 32 rad/s 2 θ & = −144 ft/s vθ = rθ v = (12 ft/s ) er − (144 ft/s ) eθ W & 2 = −12 − (12 )(12 )2 = −1740 ft/s 2 ar = && r − rθ && + 2r & = (12 )( 32 ) + ( 2 )(12 )( −12 ) = 96 ft/s 2 &θ aθ = rθ a = − 1740 ft/s 2 er + 96 ft/s 2 eθ W (b) At t ∞, ( ) ( ) e− t r ≈ 24 ft, 0 and & ≈ 0, r e − 2t && r≈0 0 θ ≈ 4t rad, θ& ≈ 4 rad/s, && ≈ 0 θ & ≈ 96 rad/s & ≈ 0, vθ = rθ vr = r v = ( 96 ft/s ) eθ W & 2 = − ( 24 )( 4 )2 = − 384 ft/s 2 , ar ≈ && r − rθ aθ ≈ 0 a = − 384 ft/s 2 er W The particle is moving on a circular path of radius of 24 ft and with a speed of 96 ft/s. The acceleration is the normal acceleration v 2 /r = ( ) ( 96 )2 24 = 384 ft/s 2 directed toward the center of the circle. Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.