solucionario dinamica hibbeler

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  • 1.Table of Contents Chapter 121 Chapter 13 145 Chapter 14 242 Chapter 15 302 Chapter 16 396 Chapter 17 504 Chapter 18 591 Chapter 19 632 Chapter 20 666 Chapter 21 714 Chapter 22 786

2. Engineering Mechanics - Dynamics Chapter 12Problem 12-1A truck traveling along a straight road at speed v1, increases its speed to v2 in time t. If itsacceleration is constant, determine the distance traveled.Given: kmkmv1 = 20v2 = 120 t = 15 s hrhrSolution: v2 v1ma = a = 1.852 t2s1 2d = v1 t +atd = 291.67 m2 Problem 12-2 A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel.ft Given:v = 80d = 500 fts Solution: 2 2 v ftv = 2a d a = a = 6.4 2d 2 s vv = at t = t = 12.5 s a Problem 12-3 A baseball is thrown downward from a tower of height h with an initial speed v0. Determine the speed at which it hits the ground and the time of travel. Given: ftft h = 50 ftg = 32.2v0 = 18 2 s s Solution: 2ft v =v0 + 2g hv = 59.5s1 3. Engineering Mechanics - Dynamics Chapter 12 v v0 t = t = 1.29 s g *Problem 124 Starting from rest, a particle moving in a straight line has an acceleration of a = (bt + c). What is the particles velocity at t1 and what is its position at t2? mm Given:b = 2 c = 6t1 = 6 st2 = 11 s 32 ss Solution: tt a ( t) = b t + c v ( t ) = a ( t ) dtd ( t ) = v ( t ) dt 00v ( t1 ) = 0 d ( t2 ) = 80.7 m m s Problem 12-5 Traveling with an initial speed v0 a car accelerates at rate a along a straight road. How long will it take to reach a speed vf ? Also, through what distance does the car travel during this time? km kmkm Given:v0 = 70 a = 6000vf = 120 hr2hrhr Solution: vf v0 vf = v0 + a t t = t = 30 s a2222 vf v0 vf = v0 + 2a ss = s = 792 m2a Problem 12-6 A freight train travels at v = v0 1 e (bt ) where t is the elapsed time. Determine the distance traveled in time t1, and the acceleration at this time. 2 4. Engineering Mechanics - Dynamics Chapter 12Given:ftv0 = 60s 1b = st1 = 3 s Solution:()tbt dv ( t) = v0 1 e a ( t) = v ( t)d ( t ) = v ( t ) dtdt 0d ( t1 ) = 123.0 fta ( t1 ) = 2.99ft 2s Problem 12-7 The position of a particle along a straight line is given by sp = at3 + bt2 + ct. Determine its maximum acceleration and maximum velocity during the time interval t0 t tf.ft ft ft Given:a = 1 b = 9 c = 15t0 = 0 s tf = 10 s32sss Solution: 32 sp = a t + b t + c t d 2 vp = sp = 3a t + 2b t + c dt 2 d d ap = vp = s = 6a t + 2b dt 2 p dtSince the acceleration is linear in time then the maximum will occur at the start or at the end.We check both possibilities. amax = max ( 6a t0 + b , 6a tf + 2b)ft amax = 422s The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero. We will check all three locations.b tcr = tcr = 3 s3a 3 5. Engineering Mechanics - DynamicsChapter 12(2 2vmax = max 3a t0 + 2b t0 + c , 3a tf + 2b tf + c , 3a tcr + 2b tcr + c 2) vmax = 135 ft s *Problem 12-8 From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed vf when it hits the ground? Each floor is a distance h higher than the one below it. (Note: You may want to remember this when traveling at speed vf ) ft Given:vf = 55 mph h = 12 ftg = 32.2 2 s Solution: 2 2vf ac = gvf = 0 + 2ac s H = H = 101.124 ft2acNumber of floors NHeight of one floorh = 12 ftH N =N = 8.427N = ceil ( N)hThe car must be dropped from floor number N = 9 Problem 129 A particle moves along a straight line such that its position is defined by sp = at3 + bt2 + c. Determine the average velocity, the average speed, and the acceleration of the particle at time t1. m m Given:a = 1 b = 3 c = 2m t0 = 0 s t1 = 4 s 3 2 s sSolution: 3 2 d d sp ( t) = a t + b t + c vp ( t) =sp ( t)ap ( t ) = vp ( t) dtdt Find the critical velocity where vp = 0.4 6. Engineering Mechanics - Dynamics Chapter 12 t2 = 1.5 sGiven vp ( t2 ) = 0 t2 = Find ( t2 )t2 = 2 s sp ( t1 ) sp ( t0 )m vave = vave = 4t1 ssp ( t2 ) sp ( t0 ) + sp ( t1 ) sp ( t2 )m vavespeed =vavespeed = 6t1 s a1 = ap ( t1 ) ma1 = 182 s Problem 1210 A particle is moving along a straight line such that its acceleration is defined as a = kv. If v = v0 when d = 0 and t = 0, determine the particles velocity as a function of position and the distance the particle moves before it stops.2m Given:k = v0 = 20ssv d Solution: ap ( v) = k vvv = k v 1 dv = k sp dsv 0 Velocity as a function of position v = v0 k sp Distance it travels before it stops0 = v0 k sp v0sp =sp = 10 m k Problem 12-11 The acceleration of a particle as it moves along a straight line is given by a = bt + c. If s = s0 and v = v0 when t = 0, determine the particles velocity and position when t = t1. Also, determine the total distance the particle travels during this time period. mm m Given: b = 2c = 1 s0 = 1 m v0 = 2 t1 = 6 s 3 2s ss 5 7. Engineering Mechanics - Dynamics Chapter 12Solution: v t bt2 1 dv = ( b t + c) dtv = v0 + + ct v 020t s2 btb 3 c 2 1 ds = v0 ++ c t dt s = s0 + v0 t +t + t0 s 262 0 2b t1 mWhen t = t1 v1 = v0 ++ c t1v1 = 322sb 3 c 2s1 = s0 + v0 t1 + t1 + t1s1 = 67 m6 2The total distance traveled depends on whether the particle turned around or not. To tell wewill plot the velocity and see if it is zero at any point in the interval 2bt t = 0 , 0.01t1 .. t1 v ( t) = v0 ++ ctIf v never goes to zero 2 then d = s1 s0d = 66 m 40 v( t ) 2000 24 6 t *Problem 1212 A particle, initially at the origin, moves along a straight line through a fluid medium such that its velocity is defined as v = b(1 ect). Determine the displacement of the particle during the time 0 < t < t1.m0.3 Given: b = 1.8 c =t1 = 3 ss s6 8. Engineering Mechanics - Dynamics Chapter 12Solution:= b( 1 e c t) t v ( t)sp ( t) = v ( t) dt sp ( t1 ) = 1.839 m 0 Problem 1213 The velocity of a particle traveling in a straight line is given v = bt + ct2. If s = 0 when t = 0, determine the particles deceleration and position when t = t1. How far has the particle traveled during the time t1, and what is its average speed?mm Given: b = 6c = 3 t0 = 0 st1 = 3 s 23ss 2 dt Solution:v ( t) = b t + c ta ( t) =v ( t)sp ( t) = v ( t) dt dt 0a1 = a ( t 1 ) mDeceleration a1 = 12 2 sFind the turning time t2t2 = 1.5 s Givenv ( t2 ) = 0 t2 = Find ( t2 )t2 = 2 sTotal distance traveled d = sp ( t1 ) sp ( t2 ) + sp ( t2 ) sp ( t0 )d=8md mAverage speed vavespeed =vavespeed = 2.667 t1 t0 s Problem 1214 A particle moves along a straight line such that its position is defined by s = bt2 + ct + d. Determine the average velocity, the average speed, and the acceleration of the particle when t = t1. mm Given: b = 1 c = 6d = 5mt0 = 0 st1 = 6 s 2 s s Solution:2 d d sp ( t) = b t + c t + d v ( t) =sp ( t) a ( t) =v ( t)dtdt Find the critical timet2 = 2s Given v ( t2 ) = 0t2 = Find ( t2 ) t2 = 3 ssp ( t1 ) sp ( t0 ) m vavevel =vavevel = 0 t1 s 7 9. Engineering Mechanics - DynamicsChapter 12sp ( t1 ) sp ( t2 ) + sp ( t2 ) sp ( t0 ) mvavespeed =vavespeed = 3t1sa1 = a ( t 1 )m a1 = 22s Problem 1215A particle is moving along a straight line such that when it is at the origin it has a velocity v0. If it begins to decelerate at the rate a = bv1/2 determine the particles position and velocity when t = t1. Given: m m v0 = 4b = 1.5 t1 = 2 sa ( v) = b v s 3 s Solution: vda ( v) = b v = vdt 1dv = 2 ( v )v0 = b tvv 02 v + 1 b tv ( t1 ) = 0.25 mv ( t) = 0 2 s tsp ( t) = v ( t) dtsp ( t1 ) = 3.5 m0 *Problem 12-16 A particle travels to the right along a straight line with a velocity vp = a / (b + sp). Determine its deceleration when sp = sp1.2 m Given:a = 5b = 4m sp1 = 2 ms2advp a a a Solution: vp =ap = vp ==b + sp b + sp dsp(b + sp) (b + sp)32 2 a m ap1 = ap1 = 0.116(b + sp1)32s 8 10. Engineering Mechanics - Dynamics Chapter 12 Problem 1217 Two particles A and B start from rest at the origin s = 0 and move along a straight line such that aA = (at b) and aB = (ct2 d), where t is in seconds. Determine the distance between them at t and the total distance each has traveled in time t. Given:ft ftftft a = 6 b = 3 c = 12d = 8t = 4s 3 2 3 2ss s s Solution:dvA a t2= at b vA = b t dt 2 a t3 b t2 sA = 6 2 dvB c t3 c t4 d t2 2= ct d vB = d t sB = dt3 s 12 s 2 Distance between A and B 3 2 4 2 at bt ct dt dAB =+ dAB = 46.33 m62 12 s2 Total distance A and B has travelled. 3 2 4 2 at bt ct dt D =+ D = 70.714 m62 12 s2 Problem 1218A car is to be hoisted by elevator to the fourth floor of a parking garage, which is at a height habove the ground. If the elevator can accelerate at a1, decelerate at a2, and reach a maximum speed v, determine the shortest time to make the lift, starting from rest and ending at rest. ft ftft Given:h = 48 ft a1 = 0.6a2 = 0.3 v = 8 22 s ss Solution: Assume that the elevator never reaches its maximum speed.ft Guesses t1 = 1 s t2 = 2 s vmax = 1 h1 = 1 fts Given vmax = a1 t1 9 11. Engineering Mechanics - DynamicsChapter 121 2 h1 = a1 t12 0 = vmax a2 ( t2 t1 ) h = h1 + vmax( t2 t1 ) a2 ( t 2 t 1 ) 12 2 t1 t2 = Find ( t , t , v , h ) t2 = 21.909 s vmax 1 2 max 1 h1 ftftSince vmax = 4.382< v =8then our original assumption is correct. s s Problem 12-19 A stone A is dropped from rest down a well, and at time t1 another stone B is dropped from rest. Determine the distance between the stones at a later time t2. ft Given:d = 80 ftt1 = 1 s t2 = 2 s g = 32.2 2 s Solution: g 2aA = gvA = g t sA =t 2vB = g( t t1 ) sB = ( t t1 ) g 2aB = g 2 At time t2g 2sA2 = t2sA2 = 64.4 ft2( t2 t1 ) 2gsB2 = sB2 = 16.1 ft2d = sA2 sB2 d = 48.3 ft*Problem 12-20A stone A is dropped from rest down a well, and at time t1 another stone B is dropped from rest.Determine the time interval between the instant A strikes the water and the instant B strikes thewater. Also, at what speed do they strike the water? 10 12. Engineering Mechanics - Dynamics Chapter 12 ftGiven: d = 80 ft t1 = 1 s g = 32.2 2sSolution: g 2 aA = gvA = g tsA =t 2 vB = g( t t1 )sB = ( t t1 ) g 2 aB = g 2 Time to hit for each particle 2d tA = tA = 2.229 sg 2d