mecanica de materiales 6 ed riley & sturges

MECHANICS OF MATERIALS, 6th Edition

RILEY, STURGES AND MORRIS

Chapter 11-1* From a free-body diagram of the forearm, the equilibrium equations give

Fy = 0 :M F = 0 :

T F W 20 = 0

1.5T 5.5W 11.5 ( 20 ) = 0

T = 3.667W + 153.33 lb ............................... Ans. F = 2.667W + 133.33 lb ............................... Ans.

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MECHANICS OF MATERIALS, 6th Edition 1-2* From a free-body diagram of the ring, the equations of equilibrium


Fx = 0 : Fy = 0 :are solved to get

T2 cos10 T1 sin10 = 0 T1 cos10 T2 sin10 175 ( 9.81) = 0

T1 = 5.67128T2 T1 = 1799 N .............................................................................. Ans. T2 = 317 N ................................................................................ Ans. T3 = 175 ( 9.81) = 1717 N ....................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-3 The equations of equilibrium


Fx = 0 : Fy = 0 :are solved to get

N A sin 30 N B sin 30 = 0 N A cos 30 + N B cos 30 800 = 0

N A = NB

N A = 462 lb N B = 462 lb

60 ................................................................ Ans. 60 ................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-4* The equations of equilibrium


Fx = 0 : Fy = 0 :

Ax N B sin 45o = 0Ay + N B cos 45o 300 = 0

M A = 0 :are solved to get

1.5 ( N B cos 45o ) 1.5 ( 300 ) = 045............................. Ans.

N B = 424.264 N 424 N Ax = 300 NA = 300 N

Ay = 0 N........................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-5 The equations of equilibrium


Fx = 0 : Fy = 0 :

Ax = 0 Ay 250 = 0M A 3 ( 250 ) = 0


Ax = 0 lb

Ay = 250 lb.....................................................Ans. .................................................Ans.

A = 250 lb M A = 750 lb ft


MECHANICS OF MATERIALS, 6th Edition 1-6 From an overall free-body diagram, the equations of equilibrium


Fx = 0 : Fy = 0 :

Ax = 0

Ay 10 15 + N F = 09 N F 3 (10 ) 6 (15 ) = 0


Ax = 0 kNAy = 11.6667 kN

N F = 13.3333 kN Then, from a free-body diagram of the right hand section of the truss, the equations of equilibrium

M C = 0 : M D = 0 : Fy = 0 :are solved to get

3 (13.3333) 3TDE = 0 3TBC 3 (15 ) + 6 (13.3333) = 0

13.3333 15 TCD cos 45o = 0

TDE = 13.33 kN (T) ............................................................................................................ Ans. TBC = 11.67 kN = 11.67 kN (C) .................................................................................. Ans. TCD = 2.36 kN = 2.36 kN (C) ...................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-7*


M A = 0 : Z Fx = 0 :^ Fy = 0 :

27 N F 13 ( 33cos15 ) + 4 ( 35sin15 ) = 075 ............. Ans. 15 ................ Ans. P cos 30 35sin15 = 0

N F = 14.93561 lb 14.94 lbP = 10.46005 lb 10.46 lb

N R + N F P sin 30 35cos15 = 075 .......................................... Ans.

N R = 24.1 lb


MECHANICS OF MATERIALS, 6th Edition 1-8


Fy = 0 : M A = 0 :

P + 4060 5210 = 0 C 16 ( 4060 ) 37 ( 5210 ) = 0

P = 1150 N .......................................................................................Ans.

C = 257, 730 N mm 258 N m ...............................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-9* From an overall free-body diagram, the equations of equilibrium


Fx = 0 : Fy = 0 :

Ax = 0 Ay 10 20 + N E = 045 N E 15 (10 ) 30 ( 20 ) = 0


Ax = 0 kipAy = 13.3333 kip

N E = 16.6667 kip Then, from a free-body diagram of the right hand section of the truss, the equations of equilibrium

M F = 0 : M E = 0 : M C = 0 :are solved to get

15 N E + 15 (TDE sin 30 ) = 0 15 ( 20 ) 15 (TCF sin 60 ) = 0 22.5 N E 7.5 ( 20 ) (15cos 30 ) TFG = 0

TCD = 33.3333 kip 33.3 kip (C) ............................................................................... Ans. TCF = +23.094 kip 23.1 kip (T) .................................................................................. Ans. TFG = +17.32 kip = 17.32 kip (T) .................................................................................. Ans.




W = 2000 ( 9.81) = 19, 620 N P W sin 30 = 0 2 (W sin 30 ) 2 (W cos 30 ) + 3 N F 1P = 0

Z Fx = 0 :M A = 0 : ^ Fy = 0 :

N R + N F W cos 30 = 030 ............................... Ans. 60 .................... Ans. 60 ................... Ans.

P = 9810 N = 9.81 kN

N R = 8933.806 N 8.93 kN N F = 8057.612 N 8.06 kN


MECHANICS OF MATERIALS, 6th Edition 1-11 From a free-body diagram of the brake pedal, the equilibrium equations are solved to get the forces


M A = 0 :

5.5Q ( 30 cos 30 )(11) ( 30sin 30 )( 4 ) = 0

Q = 62.871 lb Fx = 0 : Fy = 0 :

Ax Q + 30 cos 30 = 0 Ay 30sin 30 = 0

Ax = 36.890 lb Ay = 15.00 lbA = 39.8 lb 22.13 ........................................................... Ans. Q = 62.9 lb ........................................................................ Ans.




1-12* From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment

Fx = 0 : Fy = 0 :

Ax = 0 Ay 2 = 0M A 2 ( 4) 3 = 0

Ay = 2 kNM A = 0 : M A = 11 kN m

A = 2 kN ................................................................................................................................ Ans. M A = 11 kN m ..................................................................................................................... Ans.




1-13 From a free-body diagram of the beam, the equilibrium equations are solved to get the forces

Fx = 0 :M A = 0 :

Ax = 015 B 3 ( 500 ) 6 ( 800 ) 9 ( 700 ) 12 ( 400 ) = 0 Ay + B 500 800 700 400 = 0

B = 1160 lb Fy = 0 :

Ay = 1240 lb

A = 1240 lb ........................................................................................................................... Ans. B = 1160 lb ............................................................................................................................ Ans.




W = 450 ( 9.81) = 4414.50 N

From a free-body diagram of the lower pulley, vertical equilibrium equation gives the tension

Fy = 0 :

2T1 4414.50 = 0

T1 = 2207.25 NThen, from a free-body diagram of the upper pulley, moment equilibrium equation gives the force F

M axle = 0 :

100T1 90T1 100 F = 0

F = 221 N ............................................................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-15* From a free-body diagram of the bracket, the equilibrium equations


M A = 0 :

18 B (12 )(10 ) 2 (12 3) = 0 Ax + B = 0 Ay (12 )(10 ) 2 = 0

Fx = 0 : Fy = 0 :

Are solved to get the forces

B = 13.333 lb Ax = 13.333 lb

Ay = 60.0 lb

A = 61.464 lb 61.5 lb 77.47 ..................................... Ans. B = 13.33 lb ....................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-16 (a) From a free-body diagram of the plane, the force equilibrium equations are solved to get the forces


Fx = 0 : Fy = 0 :

V 40 cos 70 70 cos16 = 0 N 40sin 70 70sin16 = 0

V = 81.0 N .........................................Ans. N = 56.882 N 56.9 N .................Ans.(b) Then the moment equilibrium equation gives the required location of the normal force

M A = 0 :

( 70 cos16 )( 75) ( 70sin16 )( 280 ) + Nd + ( 40 cos 70 )( 60 ) ( 40sin 70 )( 60 ) = 0

d = 31.5 mm 31.5 mm (from the left end of plane) ....................................................................... Ans.




M A = 0 :

1.25 N D 9 ( 25 ) = 0

N D = 180.0 lb ....................................Ans. Fx = 0 : Fy = 0 :

Ax + N D sin 38 = 0 Ay 25 N D cos 38 = 0 Ay = 166.8 lb56.4 ................................................................... Ans.

Ax = 110.8 lbA = 200.3 lb


MECHANICS OF MATERIALS, 6th Edition 1-18* From a free-body diagram of the upper handle, moment equilibrium gives


Fx = 0 : Fy = 0 :

FA cos + Bx = 0 FA sin 100 + By = 093 (100 ) 28 ( FA sin ) + 5 ( FA cos ) = 0

M B = 0 :

= tan 1

30 = 30.964 50 FA = 919.116 NBy = 372.881 N

Bx = 788.136 N

Then a free body diagram of the upper jaw gives

M C = 0 :Therefore

Fd + 12 By + 35 Bx = 0

d = 352 + 152 = 38.0789 mm F = 842 N F = 842 N 66.8 on the jaw 66.8 on the block.......................................................................................... Ans.




1-19 Use W = 3750 lb (the weight carried by one truss). Then by symmetry (or from equilibrium of a free-body diagram of the entire truss) each support carries half of the total weight

E y = Ay = W 2 = 1875 lbAlso by symmetry (or equilibrium of a free-body diagram of the truck), the trucks weight is divided equally between its front and rear wheels.

N F = N R = W 2 = 1875 lbThen equilibrium of the floor panel between pins G and H gives

M H = 0 : M G = 0 :

10G1 4 (1875 ) = 0 6 (1875 ) 10 H1 = 0

G1 = 750 lb H1 = 1125 lbNext, from a free-body diagram of a section of the left side of the truss

= tan 1M B = 0 : M G = 0 : Fy = 0 :

5 = 26.565 108TGH 5 (1875 ) = 0

= tan 1

8 = 38.660 10

TGH = 1171.875 lb15 (1875 ) 8 (TBC cos ) 10 (TBC sin ) + 10 (1125 ) = 0

TBC = 1451.295 lb 1451 lb (C) ............................................. Ans.

1875 + TBC sin TBG sin 1125 = 0

TBG = +161.6 lb = 161.6 lb (T) ................................................... Ans.Finally, from a free-body diagram of pin C

Fx = 0 : Fy = 0 :

TCD cos TBC cos = 0

TBC sin TCD sin TCG = 0

TCD = 1451.295 lb TCG = +1298 lb = 1298 lb (T) ...................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-20* Cut a section through CD, DG, and FG, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give


M G = 0 :M D = 0 :

( 5cos 30 )( 9 ) ( 5sin 30 )( 4 ) + TCD ( 3) = 0 ( 5cos 30 )( 6 ) TFG ( 3) = 0

TCD = 9.66 kN = 9.66 kN (C) ................... Ans. TFG = 8.66 kN (T) .......................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-21 A free-body diagram of cylinder A gives


Fx = 0 : Fy = 0 :

N AB cos 60 N AC cos 60 = 0 N AB sin 60 + N AC sin 60 100 = 0

N AB = N AC = 57.73503 lbThen, from a free-body diagram of cylinder B

Fx = 0 : Fy = 0 :

B sin N AB cos 60 N BC = 0 B cos N AB sin 60 200 = 0

The minimum force occurs when N BC = 0 , therefore

N AB cos 60 B sin = tan = B cos N AB sin 60 + 200

= 6.59 ........................................................................................... Ans.




W = 250 ( 9.81) = 2452.50 N

From a free-body diagram of the pulley

Fx = 0 : Fy = 0 :

T cos T cos = 0 T sin + T sin 2452.50 = 0

=From the geometry of the cable

a + b = 42 m

( a + b ) cos = 40 m = cos 140 = 17.7528 42

Also from the geometry of the cable

h = a sin = 6 + b sin Therefore

( a b ) = 6 sin = 19.67789 mwhich together with a + b = 42 m gives

a = 30.83895 m b = 11.16105 m x = a cos = 29.4 m ...................................................................... Ans. T = 4020 N ...................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-23* Max pull occurs when either the rear wheels begin to slip B = 0.8 By


ground Ay = 0 . The force which makes the front wheels lift off of the ground is

(

)

(

) or when the front wheels start to lift off the

M B = 0 :

8 (15, 000 ) 5 ( P cos 30 ) 10 ( P sin 30 ) = 0

P = 12,862 lb 12.86 kip ................................................................................................... Ans.Checking the amount of friction required and the amount of friction available for this pulling force

Fx = 0 : Fy = 0 :

P sin 30 Bx = 0 Ay + By 15, 000 P cos 30 = 0

Ay = 0 Bx = 6431 lb (friction required) By = 26,138 lb0.8 ( 26138 ) = 20,911 lb (friction available)Since the friction required is much less than the friction available, we made the correct guess.


MECHANICS OF MATERIALS, 6th Edition 1-24 Divide the weight by 2 since there are two frames


W = 200 ( 9.81) 2 = 981 NThen from a free-body diagram of the drum

Fx = 0 : Fy = 0 :

N1 cos 45 N 2 cos 45 = 0 0.4 N1 sin 45 + N 2 sin 45 981 = 0

N1 = N 2 = 693.672 N 694 N ................................................Ans.Finally from a free-body diagram of one leg

M C = 0 : Fx = 0 : Fy = 0 :

1T 1A 0.8 N 2 = 0 T + Cx + N 2 sin 45 = 0 A + C y N 2 cos 45 = 0

where by symmetry (or from overall equilibrium)

A = 981 2 = 490.5 N .............................................................. Ans.and then

T = 1045.4376 N 1045 N .................................................. Ans.

Cx = 1535.94 N 1536 N ............................................ Ans. C y = 0 N .................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-25* The components of the three tension forces are


TA = TA

20i + 30 j 50k

202 + 302 + 502 = 0.32444TAi + 0.48666TA j 0.81111TAk 16i 25 j 50k 162 + 252 + 502 = 0.27517TB i 0.42995TB j 0.85990TB k25i 15 j 50k 252 + 152 + 502 = 0.43193TC i 0.25916TC j 0.86387TC k

TB = TB

TC = TC

Then the x-, y-, and z-components of the force equilibrium equation give

x:

0.32444TA + 0.27517TB 0.43193TC = 0

y:z:

0.48666TA 0.42995TB 0.25916TC = 00.81111TA 0.85990TB 0.86387TC + 900 = 0

TA = 418.214 lb 418 lb ..................................................................................................... Ans. TB = 205.219 lb 205 lb ..................................................................................................... Ans. TC = 444.876 lb 445 lb .................................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-26* The components of the forces are


W = 100 ( 9.81) k = 981k N

TA = TA

4i 8 j + 5k

4 2 + 82 + 5 2 = 0.39036TAi 0.78072TA j + 0.48795TAk6i 8 j + 5k

TB = TB

6 2 + 8 2 + 52 = 0.53666TB i 0.71554TB j + 0.44721TB k 8 j + 5k 82 + 5 2= 0.84800TC j + 0.53000TC k

TC = TCx:


0.39036TA 0.53666TB = 00.78072TA 0.71554TB + 0.84800TC = 0

y:z:

0.48795TA + 0.44721TB + 0.53000TC 981 = 0 TA = 603.139 N 603 N ..................................................................................................... Ans. TB = 438.716 N 439 N ..................................................................................................... Ans. TC = 925.473 N 925 N ..................................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-27 From a free-body diagram of pin A


Fx = 0 : Fy = 0 :

TAF cos 45 TAB cos 45 = 0 20 TAB sin 45 TAF sin 45 = 0

TAB = TAF = 14.14214 lb 14.14 lb .................................... Ans.Finally from a free-body diagram of BCD

Fx = 0 : Fy = 0 :

TAB cos 45 + Cx + Dx = 0 TAB sin 45 + C y Dy = 02 Dx 1.5 (10 ) 2 (TAB cos 45 ) 1(TAB sin 45 ) = 0

M C = 0 :

where by symmetry (or from overall equilibrium)

Dy = 20 2 = 10 lb and then

Cx = 32.500 lb 32.5 lb Dx = 22.500 lb 22.5 lb

C y = 0 lb

C = 32.5 lb ................................................................................. Ans. D = 24.6 lb 24.0 ..................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-28 From a free body diagram of the wheel and arm BC


Fx = 0 : Fy = 0 :

TCD + Bx = 0 2700 By = 0150 ( 2700 ) 325TCD = 0

M B = 0 :

Bx = 1246 NB = 2970 N

By = 2700 N65.2 (on AB) ....................................................... Ans.

TCD = 1246.154 N 1246 N (C) ............................................. Ans.Then from a free-body diagram of the arm AB (and assuming that the spring pushes perpendicularly against the arm)

M A = 0 : Fx = 0 : Fy = 0 :

100 (1246.154 ) 500 ( 2700 ) + bFS = 0

Ax 1246.154 + FS sin = 0 2700 FS cos + Ay = 0 50 = 11.310 250

= tan 1

b = 502 + 2502 = 254.951 mm

FS = 5783.92 N 5780 N (C) .................................................... Ans. Ax = 111.827 NA = 2970 N

Ay = 2971.60 N87.8 .................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-29* First equilibrium of an overall free-body diagram gives


Fx = 0 : Fy = 0 : M A = 0 :

Ax 100 = 0 200 Ay = 012 N F 24 (100 ) = 0

Ax = 100 lb N F = 200 lbA = 224 lb

Ay = 200 lb63.4 ...............................Ans.

Then from a free-body diagram of the bar ABCD

Fx = 0 : Fy = 0 :

Ax + Cx FBE cos 45 100 = 0 C y + FE sin 45 Ay = 018 Ax + 6 ( FBE cos 45 ) 6 (100 ) = 0

M C = 0 :

Cx = 400 lbC = 447 lb

C y = 200 lb45 .................................. Ans. 26.6 ............................................................ Ans.

FBE = 565.685 lb 566 lb


MECHANICS OF MATERIALS, 6th Edition 1-30 The total weight of water carried by each truss is


WT = 1000 ( 9.81)( 0.2 )( 3.6 )( 2 ) = 14,126.40 N

By symmetry (or overall equilibrium) the support reactions at A and F are

Ay = F = WT 2 = 7063.20 NThe weight of water carried by each roof panel is

WP = 1000 ( 9.81)( 0.2 )(1.2 )( 2 ) = 4708.80 Nand by symmetry (or equilibrium of the panel) half of this load is carried by the pins at each end of the panel

B = E = WP 2 = 2354.40 Nsin = 0.8000 cos = 0.6000

C = D = 2 (WP 2 ) = 4708.80 N

Then, equilibrium of Pin B gives

Fx = 0 : Fy = 0 :

TBC = 0 TAB 2354.40 = 0

TAB = 2354.40 NPin E:

Fx = 0 : Fy = 0 :

TDE = 0 TEF 2354.40 = 0

TEF = 2354.40 NPin A:

Fx = 0 : Fy = 0 :Pin H:

TAH + TAC cos = 07063.20 + ( 2354.40 ) + TAC sin = 0

TAH = 3531.60 N TAC = 5886.00 N TGH = 3531.60 N

Fx = 0 : Fy = 0 :Pin C:

TGH ( 3531.60 ) = 0

TCH = 0TCD + TCG cos ( 5886 ) cos ( 0 ) = 0

Fx = 0 :

TCD = 3531.60 N TCG = 0 N TFG = 3531.60 N TDG = 0 N TDF = 5886.00 N

Fy = 0 : 4708.80 ( 5886 ) sin ( 0 ) TCG sin = 0Pin G:

Fx = 0 : Fy = 0 :Pin D:

TFG ( 3531.60 ) ( 0 ) cos = 0

( 0 ) sin + TDG = 04708.80 ( 0 ) TDF sin = 0

Fy = 0 :


MECHANICS OF MATERIALS, 6th Edition 1-30 (cont.) The member forces are AB: 2350 N (C) ...............................DE: AH: 3530 N (T) ................................DG: BC:


0 N ............................................................ Ans. 0 N ............................................................ Ans.

AC: 5890 N (C) ................................DF: 5890 N (C) .................................................... Ans.

0 N .......................................EF: 2350 N (C) .................................................... Ans. 0 N .......................................GH: 3530 N (T) .................................................... Ans. 0 N ............................................................................................................................ Ans.

CD: 3530 N (C) ................................FG: 3530 N (T) .................................................... Ans. CG: CH:


MECHANICS OF MATERIALS, 6th Edition 1-31 The equations of equilibrium for the two blocks are


Fx = 0 : Fy = 0 :

T sin N 2 cos = 0 T cos + N1 sin = 0 T cos + N 2 sin 150 = 0 T sin + N1 cos 200 = 0

Adding the second and third equation together gives

N1 sin + N 2 sin = 150while subtracting the first equation from the last equation gives

N1 cos + N 2 cos = 200Dividing these two equations gives

( N1 + N 2 ) sin ( N1 + N 2 ) cos = 36.87

= tan =

150 200

N1 + N 2 = 250 lbNow the first two equations can be rewritten

T sin 2 N 2 sin cos = 0 T cos 2 + N1 sin cos = 0and subtracting the second equation from the first gives

T ( sin 2 + cos 2 ) = ( N1 + N 2 ) sin cos T (1) = ( 250 )( 0.6000 )( 0.8000 )T = 120 lb

(a) (b) (c)

N1 = 160 lb ......................................... N 2 = 90 lb ............................................................... Ans.T = 120 lb ................................................................................................................................ Ans. = 36.87 ............................................................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-32* ABC is an equilateral triangle with a height of h = 950sin 60 = 822.724 mm , and


W = 60 ( 9.81) k = 588.60k N

FAB = FAB

475 j + 822.724k 950 = 0.50000 FAB j + 0.86603FAB k

Moment equilibrium about C

M C = 0 :

( 0.95i ) ( By j + Bz k )( 0.95 j) ( 0.50000 FAB j + 0.86603FABk ) + ( 0.475i + 0.23750 j + 0.411362k ) ( 588.6k ) = 0

has components

x:

0.82272 FAB 139.7925 = 0

y:z:

0.95Bz 279.585 = 0 0 0.95By = 0 FAB = 169.915 N 169.9 N (C) ........................................................................................ Ans. By = 0 N .................................................................................................................................. Ans. Bz = 294.300 N 294 N .................................................................................................... Ans.


x:

Cx = 0C y + ( 0 ) 0.5 (169.915 ) = 0 294.300 + Cz 588.6 + 0.86603 (169.915 ) = 0

y:z:

Cx = 0 N .................................................................................................................................. Ans. C y = 84.958 N 85.0 N ...................................................................................................... Ans. Cz = 147.15 N 147.2 N .................................................................................................... Ans.




P = Pj

Moment equilibrium about C

M C = 0 :

( 50k ) ( Dx i + Dy j)+ ( 20k + 2.5i ) ( 40 j) + (11k 17.32051i + 10 j) ( Pj) = 0has components

x:

+50 Dy 800 + 11P = 0 50 Dx = 0100 + 17.32051P = 0 P = 5.77350 lb 5.77 lb ..................................................................................................... Ans.

y:z:

Dx = 0 lb .................................................................................................................................. Ans. Dy = 14.7298 lb 14.73 lb ................................................................................................. Ans.Then the x-, y-, and z-components of the force equilibrium equation give

x:

Cx + 0 = 0 C y + 14.7298 40 5.77350 = 0 Cz = 0

Cx = 0 lb ............................................ Ans. C y = 31.044 lb 31.0 lb ............... Ans. Cz = 0 lb ............................................ Ans.

y:z:


MECHANICS OF MATERIALS, 6th Edition 1-34* From a free-body diagram of the platform the equilibrium equations give


Fx = 0 : Fy = 0 :

FAC cos FBD cos = 0 FAC sin + FBD sin P = 0

FBD = FAC P = 2 FAC sin Then from a free-body diagram of the screw-block A, the equilibrium equations give

Fx = 0 : Fy = 0 :

800 FAC cos FAE cos = 0FAE sin FAC sin = 0400 cos P = 214 N .................... Ans. P = 462 N .................... Ans. P = 800 N .................... Ans.

FAE = FAC =Therefore

= 15 = 30 = 45

FAC = 414.110 N FAC = 461.880 N FAC = 565.685 N


MECHANICS OF MATERIALS, 6th Edition 1-35* From an overall free-body diagram, the equations of equilibrium give


Fx = 0 : Fy = 0 :M A = 0 :

Ax E = 0 Ay 25 = 03 ( 25 ) 21E = 0

Ax = E = 3.57143 lb Ay = 25 lbAssume that the weight of the seat back is very small compared to the weight of the seat. Then the center of gravity of the seat and the center of gravity of the entire chair are the same point. A free-body diagram of the seat gives

Fx = 0 : Fy = 0 :

TBD cos Cx = 0 C y + TBD sin 25 = 03 ( 25 ) 12 (TBD sin ) + 1(TBD cos ) = 0

M C = 0 :in which

TBD

10 = 39.806 12 = 10.8476 lb Cx = 4.76188 lb

= tan 1

C y = 18.0555 lb

A = 25.3 lb

81.87 .......................................................................................................... Ans. 39.81 .................................................................................................... Ans. 75.23 ........................................................................................................ Ans.

TBD = 10.85 lbC = 18.67 lb


MECHANICS OF MATERIALS, 6th Edition 1-36 From an overall free-body diagram, the equations of equilibrium give


Fx = 0 : Fy = 0 : M A = 0 : C = 2587.50 N Ax = 2250 N

Ax + ( 500 3) + 750 = 0

Ay 900 + C = 04C ( 3 500 )( 3) 4 ( 900 ) 3 ( 750 ) = 0

Ay = 1687.50 N

Next, draw a free-body diagram of the bar ABC. Note that the forces Ax and Ay are the forces exerted on bar ABC by the support (the same forces that are shown on the overall free-body diagram) and that the forces FAx and FAy are the forces exerted on bar ABC by the bar ADE. The equations of equilibrium give

Fx = 0 : Fy = 0 : M A = 0 :

Ax + Bx FAx = 0 Ay FAy By + C = 04 ( 2587.50 ) 2 By = 0

By = 5175.00 N Fx = 0 : Fy = 0 :

FAy = 4275.00 N FAx + Dx + 750 = 0 FAy + Dy 900 = 0

Finally, from a free-body diagram of the bar ADE

M D = 0 :

1.5 FAx 2 ( 4275 ) 2 ( 900 ) 1.5 ( 750 ) = 0

FAx = 3750 N Dx = 3000 NA = 5690 N D = 5980 N E = 1170 N

Dy = 5175 N48.7 ........................................................................................................... Ans. 59.9 ........................................................................................................... Ans. 50.2 ............................................................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-37 First draw a free-body diagram of the lower jaw, and write the equations of equilibrium


Fx = 0 : Fy = 0 :

C x = 0 D Cy E = 0 3C y 2 E = 0

M D = 0 : Cx = 0 E = 1.5C y

Next, from a free-body diagram of the lower handle, the equations of equilibrium give

Fx = 0 : Fy = 0 :

0 Bx = 0 50 + C y By = 01C y 20 ( 50 ) = 0

M B = 0 :

C y = 1000 lb Bx = 0 lb By = 1050 lbE = 1500 lb ............................................................................................................................. Ans.




W = 100 ( 9.81) = 981 N

cos =

2500 1875 1200 300 150sin tan = 1875 150 cos

= 58.612

= 5.466

There are two rollers (one on each side of the door) at B and D, hence the 2 N B and 2 N C on the free-body diagram

Fx = 0 : Fy = 0 :

T cos 2 N B = 0 T sin + 2 N C 981 = 0

M D = 0 :

(1050 cos )( 981) (150 cos )( 2 NC ) (1350sin )( 2 N B ) = 0

Substituting the first two equations into the third gives

(1050 cos )( 981) (150 cos )( 981 T sin ) (1350sin )(T cos ) = 0T = 403.455 N 403 N ...................................................................................................... Ans.

N B = 200.812 N 201 N ................................................................................................... Ans. N C = 471.283 N 471 N ................................................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-39* (a) First draw an overall free-body diagram. The force of the ground on the track of the crane is equivalent to a single concentrated force N acting at some location on the treads. As the load increases, the distance d gets smaller and smaller. The maximum load that the crane can lift corresponds to d = 0 . Then, summing moments about the point where the normal force acts gives


9 (12, 000 ) (12 cos 30 1)( 600 ) ( 24 cos 30 1 + 1)(W ) = 0 W = 4930 lb ..................................................... Ans.

(b) Next, from a free-body diagram of the pulley at the end of the boom the equations of equilibrium give

Fx = 0 : Fy = 0 :

Bx 3600 cos10 = 0 By 3600 3600sin10 = 0

Bx = 3545.3079 lb By = 4225.1334 lbFinally, from a free-body diagram of the boom the equations of equilibrium give

Fx = 0 : Fy = 0 :

Ax ( 3545.3079 ) T cos10 = 0 Ay 600 T sin10 ( 4225.1334 ) = 0

M A = 0 :24 (T sin 20 ) + ( 24sin 30 )( 3545.3079 ) (12 cos 30 )( 600 ) ( 24 cos 30 )( 4225.1334 ) = 0 T = 6275.1466 lb 6280 lb ............................................................................................... Ans.(c)

Ax = 9725.1209 lb

Ay = 5914.8012 lb

A = 11,380 lb

31.3 ........................................................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-40 From an overall free-body diagram, the equations of equilibrium give


M A = 0 :

(1.8cos 30 ) P ( 0.9sin 30 )( 2 ) ( 2.7 sin 30 + 0.6sin 30 )(1) ( 2.7 sin 30 + 1.8sin 30 + 0.3)(10 ) = 0P = 17.99 kN .......................................................... Ans.Next, from a free-body diagram of the bucket, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Gx TEI = 0 G y 10 = 0

M G = 0 :

(1.2 cos 30 ) TEI 0.3 (10 ) = 0G y = 10.00 kN

TEI = 2.88675 kN 2.89 kN (T) ........................................ Ans. Gx = 2.88675 kNFinally, from a free-body diagram of the arm DEFG, the equations of equilibrium give

Fx = 0 : Fy = 0 :

M D = 0 :

( 2.88675) Dx ( 2.88675) + FBF cos = 0 Dy (10 ) 1 + FBF sin = 0 ( 0.6 cos 30 )( 2.88675) ( 0.6sin 30 )(1) + (1.2 cos 30 )( FBF cos ) + (1.2sin 30 )( FBF sin ) (1.8cos 30 )( 2.88675 ) (1.8sin 30 )(10 ) = 0

in which

tan = FBF

1.8cos 30 1.2 cos 30 = 19.107 1.8sin 30 + 1.2sin 30 = 10.43807 kN 10.44 kN (C) .....................................Ans. Dy = 7.58627 kN37.6 ........................................................................................................ Ans. 73.9 ........................................................................................................ Ans.

Dx = 9.86303 kND = 12.44 kN G = 10.41 kN


MECHANICS OF MATERIALS, 6th Edition 1-41 First draw a free-body diagram of the wheel. It is stated that the pin B is at or near the surface of the wheel. Then, the equations of equilibrium give


M axle = 0 : P=B

2P 2B = 0

Next, from a free-body diagram of the platform, the equations of equilibrium give

Fx = 0 : Fy = 0 :

FDE cos Cx = 0 C y + FDE sin 80 = 03 ( FDE cos ) 2 ( 80 ) = 0

M C = 0 :

= sin 1FDE

2 = 30 4 = 61.5840 lb C y = 49.2080 lb

Cx = 53.3333 lb

Finally, from a free-body diagram of the arm ABC, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Ax + ( 53.3333) B sin = 0 = 0 Ay + B cos ( 49.2080 ) = 0 2 B ( 4 cos )( 49.2080 ) ( 4sin )( 53.3333) = 0

M A = 0 :

B = 138.5641 lb

Ax = 15.9487 lbA = 72.6 lb B = 138.6 lb C = 72.6 lb

Ay = 70.7920 lb

77.3 ............................................................................................................ Ans. 60 .............................................................................................................. Ans. 42.7 ............................................................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-42* The weight of the bar AB is


W = 25 ( 9.81) = 245.25 NFrom a free-body diagram of the bar AB, the equations of equilibrium give

Fx = 0 : Fy = 0 :

A sin 30 B sin 45 = 0 A cos 30 + B cos 45 W = 0

M B = 0 :

( 0.5cos )W (1cos )( A cos 30 ) + (1sin )( A sin 30 ) = 0

Since sin 45 = cos 45 adding the first two equations together gives

A ( sin 30 + cos 30 ) = WSubstituting this result into the third equation gives

A ( cos 30 cos sin 30 sin ) = 0.5cos A ( sin 30 + cos 30 ) Dividing by A cos gives

2 ( cos 30 sin 30 tan ) = ( sin 30 + cos 30 )

2 cos 30 ( sin 30 + cos 30 ) 2sin 30 = 20.10 ............................................................................................................................... Ans. tan =(Actually, this result is independent of both the length and weight of the bar.)




W = 25 ( 9.81) = 245.25 N

FCD = FCD

0.65 j + 0.95k

0.652 + 0.952 = 0.56468 FCD j + 0.82531FCD k

Moment equilibrium about B

M B = 0 :

(1.2i ) ( Ay j + Az k ) + ( 0.6i + 0.5 j) ( 245.25k )+ (1.6i + 0.65 j) ( 0.56468FCD j + 0.82531FCD k ) = 0has components

x:

122.625 + 0.53645FCD = 0 1.2 Az + 147.150 1.32050 FCD = 01.2 Ay + 0.90349 FCD = 0 FCD = 228.586 N 229 N .................................................................................................. Ans. Ay = 172.104 N 172.1 N ............................................................................................ Ans. Az = 129.915 N 129.9 N ............................................................................................... Ans.

y:z:


x:

Bx = 0

y:z:

( 172.104 ) + By + 0.56468 ( 228.586 ) = 0 ( 129.915) + Bz + 0.82531( 228.586 ) 245.25 = 0Bx = 0 N .................................................................................................................................. Ans. By = 43.026 N 43.0 N ..................................................................................................... Ans. Bz = 185.511 N 185.5 N .................................................................................................. Ans.


MECHANICS OF MATERIALS, 6th Edition 1-44 From a free-body diagram of pipe A, the equations of equilibrium are


Fx = 0 : Fy = 0 :where

N A FAB sin = 0

FAB cos 5 ( 9.81) = 0

= sin 1Therefore

b 150 150

FAB =

49.05 N cos N A = FAB sin NN A < 50 N for b < 260 mm ..................Ans.FAB < 100 N for b < 280 mm ...............Ans. FAB < 200 N for b < 295 mm ...............Ans.

(a) (b) (c)


MECHANICS OF MATERIALS, 6th Edition 1-45 From a free-body diagram of the ring, the equations of equilibrium are


Fx = 0 : Fy = 0 :where

TAB sin TBD cos = 0 TAB cos TBD sin 250 = 0

= sin 1 = sin 1Therefore

b 5 5 (1 cos ) 3

TAB =TBD =

TBD cos sin 250sin lb cos cos sin sin 250 cos lb cos cos sin sin

TAB =(a)

bmax occurs when TBD goes negative(after it goes to infinity);

bmax 3.90 ft ................................................... Ans.(At this point, the rope will straight from D to B to A.) (c) To pull further to the side, the worker needs a longer rope to pull on or he needs to attach his rope lower - closer to the crate. .......................... Ans.




W = 50 ( 9.81) = 490.50 N

The cable is continuous, therefore the tension in the cable is continuous (equal to the force P ); and from a free-body diagram of the pulley, the equations of equilibrium give

Fx = 0 : Fy = 0 :

P cos P cos = 0 P sin + P sin W = 0 d 1.5

= = tan 1P= W 2sin

(a) (b) (c)

P < 2W P < 4W P < 8W

d > 387 mm ............................. Ans. d > 189 mm .............................. Ans. d > 94 mm ................................ Ans.




AB = tan 1

d 20

BC = tan 1

d 10

From a free-body diagram of the stop light, the equations of equilibrium give

Fx = 0 : Fy = 0 :Solving yields

TBC cos BC TAB cos AB = 0 TAB sin AB + TBC sin BC 75 = 0

TAB = TBC =

75cos BC sin BC cos AB + cos BC sin AB 75cos AB sin BC cos AB + cos BC sin AB




W = 6800 ( 9.81) = 66, 708 N

From a free-body diagram of the truck, the equilibrium equations are

Fx = 0 : Fy = 0 : M B = 0 :

P sin Bx = 0

Ay + By 66, 708 P cos = 0

( 66, 708 )( 2.4 ) 4.4 Ay ( P cos )(1.5) ( P sin )( 3) = 0

The fourth equation needed to solve for the four unknowns is either Ay = 0 (the front wheels are on the verge of lifting off the ground) or Bx = 0.8 By (the rear wheels are on the verge of slipping). Guessing that the front wheels are on the verge of lifting off the ground gives the solution

Ay = 0 N 66, 708 ( 2.4 ) N 1.5cos + 3sin Bx = P sin N P= By = 66, 708 + P cos NThe forces Bx and 0.8 By are plotted on the same graph as the force P . Since Bx is always less than

0.8By , the guess that the front wheels are on the vergeof lifting off the ground was the correct guess, and the solution is valid for all values of .

E = 1500 lb ............................................................................................................................. Ans.

TBD = 10.85 lbC = 18.67 lb

39.81 .................................................................................................... Ans. 75.23 ........................................................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-49 From a free-body diagram of the I-beam, the equations of equilibrium give


M D = 0 : M F = 0 : D= F=

400 ( 3) + 1000 ( 7 d ) + 6 F = 0 6 D 1000 ( d 1) 400 ( 3) = 0

1000 ( d 1) + 1200 6 1000 ( 7 d ) + 1200 6

Next, from a free-body diagram of the joint D, the equations of equilibrium give

Fx = 0 : Fy = 0 :

TCD cos TDE = 0 TCD sin D = 0 D sin 1 3 TDE = D tan

TCD =in which

= tan 1 = 18.435By inspection, members CE and CF are both zero-force members. Therefore the tension force in member BC will be the same as the tension force in member CD and the compression force in member EF will be the same as the compression force in member DE,

TBC = TEF =

D sin D tan

TCF = 0


MECHANICS OF MATERIALS, 6th Edition 1-50 From an overall free-body diagram of the light pole,


= tan 1M A = 0 : TGH =

1.75 b

= tan 1

2.75 b 5

the moment equation of equilibrium gives

2 ( 7500 ) 2.75 (TGH cos ) = 0

2 ( 7500 ) 2.75cos

It will be assumed that the cable DG supports the end of the arm BG and that the connection of the horizontal arm BG to the vertical pole ABCD exerts negligible moment on the arm. (If the connection could provide a sufficient moment, then the cable between D and G would not be necessary.) Then, from a free-body diagram of the pin G, the equations of equilibrium give

Fx = 0 : Fy = 0 :

FBG + TGH sin TDG cos = 0 TDG sin TGH cos = 0 TGH cos sin cos cos sin sin = TGH sin

TDG = FBG


MECHANICS OF MATERIALS, 6th Edition 1-51 From an overall free-body diagram of the crane, the equations of equilibrium give


Fy = 0 :

N 12, 000 600 W = 0

M C = 0 :

( 9 )(12, 000 ) (12 cos 1)( 600 ) ( 24 cos 1 + 1) W Nd = 0 N = (12, 600 + W ) lb 108, 600 ( 7200 + 24W ) cos d= ft12, 600 + W

(a) For W = 3600 lb

d=

108, 600 93, 600 cos ft 16, 200

(b) From a free-body diagram of the pulley at B,

tan = Fx = 0 : Fy = 0 :

24sin 6 24 cos + 9 Bx 3600 cos = 0 By 3600 3600sin = 0By = 3600 (1 + sin )

and the equations of equilibrium give

Bx = 3600 cos

From a free-body diagram of the boom, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Ax Bx T cos = 0 Ay By T sin 600 = 0 24sin ( ) T (12 cos )( 600 ) + ( 24sin ) Bx ( 24 cos ) By = 0

M A = 0 :

T=

( 7200 + 24B ) cos 24B sin 24sin ( )y x

Ax = Bx + T cos Ay = By + T sin + 6002 A = Ax2 + Ay

Note that the tension force becomes negative for an angle of about 66 . Since negative forces in the cable are not possible, the boom would topple over onto the top of the cab of the crane if the operator tried to lift higher than 66 .


MECHANICS OF MATERIALS, 6th Edition 1-51 (cont.) (c) For d = 1 ft


d=

108, 600 ( 7200 + 24W ) cos = 1 ft 12, 600 + W

Wmax =

96, 000 7200 cos lb 1 + 24 cos




W = 250 ( 9.81) = 2452.50 N 6 (TBC sin 60 ) ( 3cos ) W = 0

(a) From a free-body diagram of the post AB, moment equilibrium gives

M B = 0 : TBC =

( 2452.50 )( 3cos )6sin 60

Since the pin at A is frictionless and the weight of the brace AC is neglected, the brace AC is a two-force member and from a free-body diagram of the pin C, the equations of equilibrium give

Fx = 0 : Fy = 0 :

FAC cos ( 60 + ) + TBC cos ( 60 ) TCD cos = 0 FAC sin ( 60 + ) TBC sin ( 60 ) TCD sin = 0

FAC =TCD =in which

cos ( 60 ) sin + sin ( 60 ) cos TBC sin ( 60 + ) cos cos ( 60 + ) sin FAC cos ( 60 + ) + TBC cos ( 60 ) =P cos

tan =

6sin ( 60 + ) 6sin = 7 + 6 cos 7 + 6 cos ( 60 + )

For this arrangement, the force P necessary to start weight of the post ( 2450 N ) .

raising the post ( 2300 N ) is almost as large as the


MECHANICS OF MATERIALS, 6th Edition 1-52 (cont.) (b) From a free-body diagram of the post AB, moment equilibrium now gives


M B = 0 :

6 (TBC sin 67.5 ) ( 3cos ) W = 0

TBC =

( 2452.50 )( 3cos )6sin 67.5

Again, since the pin at A is frictionless and the weight of the brace AC is neglected, the brace AC is a two-force member and from a free-body diagram of the pin C, the equations of equilibrium give

Fx = 0 :FAC cos ( 45 + ) + TBC cos ( 67.5 ) TCD cos ( 22.5 ) = 0 Fy = 0 : FAC sin ( 45 + ) TBC sin ( 67.5 ) + TCD sin ( 22.5 ) = 0

FAC = TCD =

sin ( 67.5 ) cos ( 22.5 ) cos ( 67.5 ) sin ( 22.5 ) TBC sin ( 45 + ) cos ( 22.5 ) + cos ( 45 + ) sin ( 22.5 )

FAC cos ( 45 + ) + TBC cos ( 67.5 ) cos ( 22.5 )

Finally, since the weight of the brace AD is also neglected, the brace AD is also a two-force member and from a free-body diagram of the pin D, the equations of equilibrium give

Fx = 0 : FAD cos ( 90 ) + TCD cos ( 22.5 ) TDE cos = 0 Fy = 0 : FAD sin ( 90 ) TCD sin ( 22.5 ) TDE sin = 0

FAD =TDE =

cos ( 22.5 ) sin + sin ( 22.5 ) cos TCD cos ( 90 ) sin + sin ( 90 ) cos TCD cos ( 22.5 ) FAD cos ( 90 ) =P cos

Note that the force in the brace AD goes to zero at about = 75 . For 75 90 , the solution becomes similar to that of part a (with the angle between the post and the brace AC 45 rather than 60 ). For this arrangement, the force P necessary to start the force required using a single brace (part a).

raising the post (1700 N ) is about 25% less than




a = 12.5cos 2 b = 12.5sin + 0.5 b = tan 1 a

From a free-body diagram of the truck box, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Ax C cos = 0Ay + C sin 22, 000 = 08.5 ( 22, 000 cos ) 2.5 ( 22, 000sin )

M A = 0 :

12.5 C sin ( ) = 0 C= 187, 000 cos 55, 000sin 12.5sin ( )

Ax = C cos Ay = 22, 000 C sin 2 A = Ax2 + Ay




Fx = 0 : Fx = 0 : Fx = 0 :

TAB 75 + 100 50 = 0 TBC + 100 50 = 0 TCD 50 = 0

TAB = 25 kN = 25 kN (C) ............................................... Ans. TBC = +50 kN = 50 kN (T) ............................................... Ans. TCD = 50 kN = 50 kN (C) ............................................... Ans.




Fx = 0 : M cut = 0 :

V = 0 90 120 P = 0 M 90 ( 5 8 ) 120 (1.5 ) = 0

Fy = 0 :

V = 0 lb .................................................................................................Ans. P = 30 lb = 30 lb (C) ......................................................................Ans. M = 236 lb in. .................................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-56 Next, from a free-body diagram of the man standing on the beam, the equations of equilibrium give


Fx = 0 : Fy = 0 :

Cx = 0T 75 ( 9.81) + T 40 ( 9.81) + C y = 0

M C = 0 :

3 75 ( 9.81) + 1.5 40 ( 9.81) 3T 1.5T = 0 C y = 114.450 N

T = 621.300 N

Next, draw a free-body diagram of the portion of the beam between section aa and the right end of the beam. Note that since one-fourth of the beam has been cut away, only threefourths of the total beam weight is included on the free-body diagram. The equations of equilibrium give

Fx = 0 : Fy = 0 :

Pa + 0 = 0Va + ( 621.3) 30 ( 9.81) 114.450 = 0

M a = 0 :

M a + 0.75 ( 621.3) 1.125 30 ( 9.81) 2.25 (114.450 ) = 0

Pa = 0 N ........................ Va = 213 N ................................................................................. Ans. M a = 122.6 N m ............................................................................................................... Ans.Finally, draw a free-body diagram of the portion of the beam between section bb and the right end of the beam. This time three-fourths of the beam has been cut away and only one-fourth of the total beam weight is included on the free-body diagram. The equations of equilibrium give

Fx = 0 : Fy = 0 :

Pb + 0 = 0Vb 10 ( 9.81) 114.450 = 0

M b = 0 :

M b 0.375 10 ( 9.81) 0.75 (114.450 ) = 0

Pb = 0 N ........................ Vb = 213 N .................................................................................... Ans. M b = 122.6 N m ................................................................................................................ Ans.


MECHANICS OF MATERIALS, 6th Edition 1-57* From an overall free-body diagram of the bracket, the equations of equilibrium give


Fx = 0 : Fy = 0 :

Ax + B cos 45 = 0 Ay + B sin 45 500 = 05 ( 500 ) 15 ( B cos 45 ) 15 ( B sin 45 ) = 0

M A = 0 :

B = 117.85113 lb

Ax = 83.3333 lb

Ay = 416.6667 lb

Next, draw a free-body diagram of the portion of the bracket between section aa and pin A. The equations of equilibrium give

Fx = 0 : Fy = 0 :

V 83.3333 = 0 P + 416.6667 = 0 M 8 ( 83.3333) = 0

M a = 0 :

P = 417 lb = 417 lb (C) .................................................. Ans.V = 83.3 lb ............................................................................ Ans. M = 667 lb in. .................................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-58 From a free-body diagram of the lower half of the clamp, the equations of equilibrium give


Fx = 0 : Fy = 0 :

V =0 P 2000 = 0 M + 0.075 ( 2000 ) = 0

M cut = 0 :

P = 2000 N (T) .................................................................... Ans.V = 0 N .................................................................................. Ans. M = 150.0 N m ............................................................... Ans.


MECHANICS OF MATERIALS, 6th Edition 1-59* Draw free-body diagrams of the sections of the tower between the points of interest and the top. For each section, the vertical component of the equation of equilibrium gives


Fy = 0 :

PA 2000 3 (1500 cos 30 ) W = 0

W = 40 ( 50 ) = 2000 lb

PA = 7900 lb .................................................... Ans. Fy = 0 : PB 2000 3 (1500 cos 30 ) 3 (1250 cos 40 ) W = 0

W = 40 (150 ) = 6000 lb

PB = 14, 770 lb .....................................................................................Ans. Fy = 0 : PC 2000 3 (1500 cos 30 ) W 3 (1250 cos 40 ) 3 (1000 cos 50 ) = 0 W = 40 ( 250 ) = 10, 000 lb

PC = 20, 700 lb ............................................ Ans. Fy = 0 : PD 2000 3 (1500 cos 30 ) W 3 (1250 cos 40 ) 3 (1000 cos 50 ) 3 ( 750 cos 60 ) = 0 W = 40 ( 350 ) = 14, 000 lb

PD = 25,800 lb ....................................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-60* The weights of the two cylinders are the same


W = 50 ( 9.81) = 490.5 NFirst draw a free-body diagram of the upper cylinder, and write the equations of equilibrium

Z Fx = 0 :^ Fy = 0 :

N 2 W sin 27 = 0 N1 W cos 27 = 0

N1 = 437.0387 N N 2 = 222.6823 NNext, from a free-body diagram of the lower cylinder, the equations of equilibrium give

Z Fx = 0 :^ Fy = 0 :

N 4 N 2 W sin 27 = 0 N 3 W cos 27 = 0 N 4 = 445.3647 N

N 3 = 437.0387 N Fx = 0 : Fy = 0 :

Next, from a free-body diagram of the rack, the equations of equilibrium give

Ax N 4 cos 27 + ( N1 + N 3 ) sin 27 B sin 27 = 0 Ay N 4 sin 27 ( N1 + N 3 ) cos 27 + B cos 27 = 0 120 ( 445.3647 ) 320 ( 437.0387 ) 120 ( 437.0387 ) + 580 B = 0

M A = 0 :

B = 239.4022 N Ax = 902.3320 N

Ay = 767.6911 N

Finally, from a free-body diagram of the upper portion of the rack (between section a and the roller support B), the equations of equilibrium give

Z Fx = 0 :^ Fy = 0 :

P = 0

B V N1 = 0 0.360 B M 0.100 N1 = 0

M cut = 0 :

P = 0 N ...............................................................................Ans. V = 197.6 N ....................................................................Ans. M = 42.5 N m .................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-61 From a free-body diagram of the lower portion of the crutch (between section a and the bottom B), the equations of equilibrium give


Z Fx = 0 :^ Fy = 0 :

V 35sin 25 = 0 P + 35cos 25 = 0 M 2 ( 35sin 25 ) = 0

M cut = 0 :

P = 31.7 lb = 31.7 lb (C) ............................................Ans.V = 14.79 lb .......................................................................Ans. M = 29.6 lb ft ..................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-62 From a free-body diagram of the wheel, the equations of equilibrium give


M B = 0 :

325 FCD 150 ( 2700 ) = 0

FCD = 1246 NSince CD is a two-force member, the axial force on every cross section is the same

P = 1246 N (C) ........................................................................ Ans.and the shear force and the bending moment are both zero

V = 0 N ...................................................................................... Ans. M = 0 N m .............................................................................. Ans.


MECHANICS OF MATERIALS, 6th Edition 1-63 From a free-body diagram of the handle, the equations of equilibrium give


M F = 0 :

30 (1000 ) 8 FDE = 0

FDE = 3750 lbSince DE is a two-force member, the axial force on every cross section is the same

P = 3750 lb (C) ................................................................... Ans.and the shear force and the bending moment are both zero

V = 0 lb .................................................................................................Ans. M = 0 lb ft ..........................................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-64* First draw an overall free-body diagram of the shaft and sum moments about the bearing B


M C = 0 :

( 800 j + 250i ) ( 5k ) + ( 800 j 250i ) ( 30k ) + ( 2000 j + 250k ) ( 30i ) + ( 2000 j 250k ) ( 5i ) + ( 2800 j) ( Bx i + Bz k ) = 0The x-, y-, and z-components of this equation give

x:

4000 + 24, 000 2800 Bz = 01250 7500 + 7500 1250 = 0

y:z:

60, 000 + 10, 000 + 2800 Bx = 0 Bx = 25.00 kN Bz = 10.00 kN

Next draw a free-body diagram of the portion of the shaft between the bearing B and the section at A and write the equations of equilibrium

Fx = 0 : Fy = 0 : Fz = 0 :

Vx + 30 + 5 25 = 0 Py = 0 Vz + 10 = 0

Vx = 10 kN .......................................................................... Ans. Py = 0 kN .............................................................................. Ans. Vz = 10 kN .......................................................................... Ans.M cut = 0 :

( M i + T j + M k ) + ( 0.4 j + 0.25k ) ( 30i )x y z

+ ( 0.4 j 0.25k ) ( 5i ) + ( 1.2 j) ( 25i + 10k ) = 0The x-, y-, and z-components of this equation give

x:

M x 12 = 0 Ty + 7.5 1.25 = 0 M z 30 + 12 + 2 = 0

M x = 12 kN m ................................. Ans. Ty = 6.25 kN m ............................ Ans. M z = 16 kN m ................................. Ans.

y:z:


MECHANICS OF MATERIALS, 6th Edition 1-65* First draw a free-body diagram of the blocks, and write the equations of equilibrium


Fx = 0 : Fy = 0 :

NC N A = 0 FC + FA 30 = 08 ( 30 ) 16 ( FC ) = 0

M A = 0 :

NC = N A

FC = FA = 15 lb

Next, from a free-body diagram of the left handle, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Bx N C = 030 By (15 ) = 0 4 ( 30 ) + 4 (15 ) 4 N C = 0

M B = 0 : N C = 45 lb Bx = 45 lb

By = 15 lb

Next, from a free-body diagram of the lower section of the left handle, the equations of equilibrium give

Fx = 0 : Fy = 0 :

P 45 = 0 V 15 = 0 M 4 ( 45 ) = 0

M cut = 0 :

P = 45 lb ........................................................................................Ans. V = 15 lb ......................................................................................................Ans. M = 180 lb in. ...........................................................................................Ans.




W = 360 ( 9.81) = 3531.60 N

From a free-body diagram of the bar ABC, the equations of equilibrium give

Fx = 0 : Fy = 0 :

Ax + FBD cos = 0 Ay + FBD sin W = 01430 ( FBD sin ) ( 2700 cos16 ) W = 0

M A = 0 :tan =

h 900 + 1430sin16 = b 1430 cos16 890 = 69.471 = 16 = 53.471 FBD = 7976.730 N Ax = 2797.2911 N Ay = 3938.5663 NThen, from a free-body diagram of the left section of the bar, the equations of equilibrium give

Z Fx = 0 :^ Fy = 0 :

P Ax cos16 Ay sin16 = 0 V + Ax sin16 Ay cos16 = 0M ( 0.530sin16 ) Ax + ( 0.530 cos16 ) Ay = 0

M cut = 0 :

P = 3770 N .................................................................... Ans. V = 3010 N .................................................................................................Ans. M = 1598 N m .......................................................................................Ans.


MECHANICS OF MATERIALS, 6th Edition 1-67 From a free-body diagram of the pipe the moment equilibrium equation


M cut = 0 :

( M i + T j + M k ) + ( 7i + 18j + 10k ) ( 50k ) = 0x y z

has x-, y-, and z-components

x:

M x 900 = 0 Ty 350 = 0 Mz = 0 Vx = 0 Py = 0 Vz 50 = 0

M x = 900 lb in. .............. Ans. Ty = 350 lb in. ................ Ans. M z = 0 lb in. .................. Ans. Vx = 0 lb ................................................................................. Ans. Py = 0 lb ................................................................................. Ans. Vz = 50 lb .............................................................................. Ans.

y:z:

and the force equilibrium equation has components

Fx = 0 : Fy = 0 : Fz = 0 :


MECHANICS OF MATERIALS, 6th Edition 1-68* From a free-body diagram of the bar ABC, the equations of equilibrium give


Fx = 0 : Fy = 0 :

Ax = 0

Ay + FBD 3 = 0200 FBD 400 ( 3) = 0

M A = 0 :

FBD = 6 N Ax = 0 N Ay = 3 NP ( 3) sin = 0 V ( 3) cos = 0Then, from a free-body diagram of the left section of the bar, the equations of equilibrium give

Z Fx = 0 :^ Fy = 0 :

M cut = 0 :in which

( 0.1)( 3) M = 0120 = 30.964 200

= tan 1Then

P = 1.543 kN ................................................................. Ans. V = 2.57 kN ................................................................................................Ans. M = 0.3 kN m ..........................................................................................Ans.

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mecanica de materiales 6 ed riley & sturges

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