ch03

STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS

40

3-4 A 750-kg body is supported by the flexible cable system shown in Fig. P3-4. Determine the tensions in cables AC, BC, and CD.

SOLUTION From a free-body diagram for the body

0 :yF 750 9.81 0CDT

7357.50 7360 NCDT .......................... Ans.

From a free-body diagram of the ring, the equations of equilibrium

0 :xF cos60 cos30 0BC ACT T

0 :yF sin 60 sin 30 7357.50 0BC ACT T

are solved to get

1.73205BC ACT T

7360 NACT ..................................................................................................................................Ans.

12,740 NBCT ..............................................................................................................................Ans.

3-5 An 800-lb homogeneous cylinder is supported by two rollers as shown in Fig. P3-5. Determine the forces exerted by the rollers on the cylinder. All surfaces are smooth (frictionless).

SOLUTION The equations of equilibrium

0 :xF sin 30 sin 30 0A BN N

0 :yF cos30 cos30 800 0A BN N

are solved to get

A BN N

462 lbAN ..............................................................................Ans.

462 lbBN ..............................................................................Ans.

3-6 A worker is using a hoist and cable to lift a 175-kg engine from a car as shown in Fig. P3-6. Determine the forces in the three cables attached to the ring.

SOLUTION From a free-body diagram of the ring, the equations of equilibrium

0 :xF 2 1cos10 sin10 0T T

0 :yF 1 2cos10 sin10 175 9.81 0T T

are solved to get

1 25.67128T T

1 1799 NT ...................................................................................... Ans.

2 317 NT ....................................................................................... Ans.

3-7 The lightweight collar A shown in Fig. P3-7 is free to slide on the smooth rod BC. Determine the forces exerted on the collar by the cable and by the rod when the 900 lb downward force F is applied to the collar.

SOLUTION From a free-body diagram of the collar, the equations of equilibrium


41

0 :xF cos 20 sin 30 0T N

0 :yF sin 20 cos30 900 0T N

are solved to get 1.87939N T

700 lbT ............................................................................................... Ans.

1316 lbN ............................................................................................ Ans.

3-8 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P3-8. When a 500-N sideways force P is applied to the cable, the cable is pulled 5 to the side as shown. For this position, determine the x- and y-components of the cable force being applied to the automobile.


0 :xF 2 1cos5 cos5 0T T

0 :yF 1 2500 sin 5 sin 5 0T T

are solved to get

1 2 2868.43 NT T

The x- and y-components of the cable force being applied to the automobile are then

1 1 cos5 2860 NxT T ..................................................Ans.

1 1 sin 5 250 NyT T .....................................................Ans.

3-9 Two flower pots are supported with cables as shown in Fig. P3-9. If pot A weighs 10 lb and pot B weighs 8 lb, determine the tension in each of the cables and the slope of cable BC.

SOLUTION From a free-body diagram of the upper ring

0 :xF cos 45 cos 0CD BCT T (a)

0 :yF sin 45 sin 8 0CD BCT T (b)

From a free-body diagram for the lower ring

0 :xF cos cos 45 0BC ABT T (c)

0 :yF sin 45 sin 10 0AB BCT T (d)

Adding equations (a) and (c) gives

cos 45 cos 45 0CD ABT T

CD ABT T

Then adding Eqs. (b) and (d) gives

sin 45 sin 45 18 lbCD ABT T

12.7279 12.73 lbCD ABT T ............... Ans.

Now Eqs. (b) and (a) can be written

sin 1.0000BCT (e)

cos 9.0000BCT (f)

Dividing Eq. (e) by Eq. (f) gives


42

tan 1 9

6.340 .................................................... 9.06 lbBCT ........................................................Ans.

3-10 Three smooth homogeneous cylinders A, B, and C are stacked in a box as shown in Fig. P3-10. Each cylinder has a diameter of 250 mm and a mass of 245 kg. Determine

(a) The force exerted by cylinder B on cylinder A. (b) The forces exerted on cylinder B by the vertical and horizontal surfaces at D and E. SOLUTION

245 9.81 2403.45 NW

(a) From a free-body diagram of cylinder A, the equations of equilibrium

0 :xF cos 40 cos 40 0AB ACN N

0 :yF sin 40 sin 40 2403.45 0AB ACN N

are solved to get

1869.55 1870 NAB ACN N ............. Ans.

(b) From a free-body diagram of cylinder B, the equations of equilibrium

0 :xF cos 40 0D ABN N

0 :yF sin 40 2403.45 0E ABN N

are solved to get

1432 NDN .....................................................................Ans.

3610 NEN .....................................................................Ans.

3-11 Three smooth homogeneous cylinders A, B, and C are stacked in a V-shaped trough as shown in Fig. P3-11. Each cylinder weighs 100 lb and has a diameter of 5 in. Determine the minimum angle for equilibrium.

SOLUTION From a free-body diagram of cylinder A, the equations of equilibrium

0 :xF sin 30 sin 30 0AB ACN N

0 :yF cos30 cos30 100 0AB ACN N

are solved to get

57.735 lbAB ACN N

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

0 :xF sin 57.735sin 30 0B BCF N

0 :yF cos 100 57.735cos30 0BF

But the contact force between cylinders B and C cannot be negative. Therefore, the minimum angle corresponds to 0BCN and

sin 28.868 lbBF

cos 150.000 lbBF

sin 28.868tan 0.19245cos 150.000

B

B

FF

10.89 ........................................................................................................................................Ans.


43

3-12 A 250-kg body is supported by the flexible cable system shown in Fig. P3-12. Determine the tensions in cables A, B, C, and D.

SOLUTION From a free-body diagram of the lower ring, the equations of equilibrium

0 :xF cos60 0C DT T

0 :yF sin 60 250 9.81 0CT

are solved to get

2831.90 2830 NCT ..................................................Ans.

1415.95 1416 NDT ...................................................Ans.

From a free-body diagram of the upper ring, the equations of equilibrium

0 :xF cos 40 cos30 2831.90cos 60 0A BT T

0 :yF sin 40 sin 30 2831.90sin 60 0A BT T

are solved to get

1507 NAT .......................................................................Ans.

2970 NBT .......................................................................Ans.

3-13 A 500-lb lawn roller is to be pulled over a curb as shown in Fig. P3-13. Determine the minimum pulling force that must be applied by the man to just start the 3-ft diameter roller over the curb. Also determine the angle which gives the minimum pulling force. Assume that the pulling force is along the handle, which makes an angle of with the horizontal.

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

0 :xF cos cos 0CT N (a)

0 :yF sin sin 500 0F CN N T (b)

where

1sin 12 18 41.810

The roller just begins to roll over the curb when 0FN . Therefore, Eqs. (a) and (b) become

sin 500 0.66667 CT N (c)

cos 0.74536 CT N (d)

Dividing Eq. (c) by Eq. (d) gives

500 0.66667tan

0.74536C

C

NN

.......................................(e)

Solving Eqs. (d) and (e) by trial and error for the angle that makes the tension a minimum NC T 40 386.968 376.519 45 354.101 373.257 48.19 333.332 372.679 50 321.553 372.865 55 288.825 375.327

min 373 lb @ 48.19T ........................................................................................................... Ans.


44

3-14 In order to hold a 130-kg crate in a stationary position, a worker exerts a force P at an angle on a rope as shown in Fig. P3-14. Determine the force exerted by the worker when = 20 .


0 :xF sin 8 cos 20 0ABT P

0 :yF cos8 sin 20 130 9.81 0ABT P

are solved to get

6.75197ABT P

201.0 NP ..............................................................................Ans.

1357 NABT

3-15 A farmer is extracting a post from the ground using the structure shown in Fig. P3-15. What force must the farmer apply to the cable system if the force required to remove the post is 2000 lb?

SOLUTION From a free-body diagram of the right ring, the equations of equilibrium

0 :xF sin15 0CE BCT T

0 :yF cos15 2000 0CET

are solved to get

535.898 lbBCT

2070.552 lbCET

From a free-body diagram of the left ring, the equations of equilibrium

0 :xF 535.898 cos15 0ABT

0 :yF sin15 0ABT P

are solved to get

554.803 lbABT 143.6 lbP ....................................................................................................................................Ans.

3-16 A continuous cable is used to support two blocks as shown in Fig. P3-16. Block A is supported by a small wheel that is free to roll on the cable. Determine the displacement y of block A for equilibrium if the masses of blocks A and B are 22 kg and 34 kg, respectively.

SOLUTION From a free-body diagram of the hanging weight, the equations of equilibrium

0 :yF 34 9.81 0T

are solved to get

333.54 NT Since the pulleys are free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 22 kg weight, the equations of equilibrium

0 :xF 2 1cos cos 0T T


46

Adding Eqs. (a) and (c) together gives

sin 60 sin 30 0 NB AN N (e)

while adding Eqs. (b) and (d) together gives

cos60 cos30 1000 NB AN N (f)

Solving Eqs. (e) and (f) gives

866.03 N 866 NAN ........................... Ans.

500.00 N 500 NBN ........................... Ans.

Then Eqs. (b) and (a) can be written

sin 50 NT (g)

cos 433.013 NT (h) Dividing Eq. (g) by Eq. (h) gives

50tan

433.013

6.587 .......................................................................................................................................Ans.

436 NT .......................................................................................................................................Ans.

3-19 Concrete is to be moved from a mixer to the second floor of a building under construction using a container as shown in Fig. P3-19. The container and its contents weigh 3000 lb, and it is supported by three cables equally spaced around the top of the 4-ft diameter container. Determine the force in each cable.

SOLUTION The coordinates of points A, B, C, and D are:

A: 2cos30 1.73205 ftx

2sin 30 1.0000 fty

B: 0 ftx

2.0000 fty

C: 2cos30 1.73205 ftx

2sin 30 1.0000 fty

D: 0 ftx y

4.5sinz h where

1cos 2 4.5 63.612

Therefore

4.0311 fth

2 2 2

1.73205 4.0311 0.38490 0.22222 0.895811.73205 1 4.0311

A A A A AT T T Ti j kT i j k

0.38490 0.22222 0.89581C C C CT T TT i j k

2 2

2 4.0311 0.44445 0.895812 4.0311

B B B BT T Tj kT j k

Then, the equations of equilibrium

0 :xF 0.38490 0.38490 0A CT T


47

0 :yF 0.22222 0.44445 0.22222 0A B CT T T

0 :zF 0.89581 0.89581 0.89581 3000 0A B CT T T

are solved simultaneously to get

1116 lbA B CT T T .................................................................................................................Ans.

3-20 A mass m is to be supported by two cables (A and B) as shown Fig. P3-20. If the maximum force that the cables can withstand is 15 kN, determine the maximum mass m that can be supported.

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

0 :xF cos 25 cos 40 0B AT T

0 :yF sin 40 sin 25 0A BT T mg

If max 15 kNA AT T , then

12.6785 kNBT

which is okay since 12.6785 kN 15 kNBT . Then

sin 40 sin 25 15 kNA Bmg T T

1529 kgm ....................................................................................................................................Ans.

3-21 The hot-air balloon shown in Fig. P3-21 is tethered with three mooring cables. If the net lift of the balloon is 900 lb, determine the force exerted on the balloon by each of the three cables.

SOLUTION

2 2 2

20 30 5020 30 50

0.32444 0.48666 0.81111

A A

A A A

T

T T T

i j kT

i j k

2 2 2

16 25 5016 25 50

0.27517 0.42995 0.85990

B B

B B B

T

T T T

i j kT

i j k

2 2 2

25 15 5025 15 50

0.43193 0.25916 0.86387

C C

C C C

T

T T T

i j kT

i j k


0 :xF 0.32444 0.27517 0.43193 0A B CT T T

0 :yF 0.48666 0.42995 0.25916 0A B CT T T

0 :zF 0.81111 0.85990 0.86387 900 0A B CT T T


418 lbAT ...........................................................................................................................Ans.

205 lbBT ...........................................................................................................................Ans.

445 lbCT ...........................................................................................................................Ans.


48

3-22 A 100-kg traffic light is supported by a system of cables as shown in Fig. P3-22. Determine the tensions in each of the three cables.

SOLUTION

2 2 2

4 8 5 0.39036 0.78072 0.487954 8 5


2 2 2

6 8 5 0.53666 0.71554 0.447216 8 5

B B B B BT T T Ti j kT i j k

2 2

8 5 0.84800 0.530008 5

C C C CT T Tj kT j k


0 :xF 0.39036 0.53666 0A BT T

0 :yF 0.78072 0.71554 0.84800 0A B CT T T

0 :zF 0.48795 0.44721 0.53000 100 9.81 0A B CT T T


603 NAT ...........................................................................................................................Ans.

439 NBT ...........................................................................................................................Ans.

925 NCT ...........................................................................................................................Ans.

3-23 A 250-lb force is applied to the joint at the top of the structure shown in Fig. P3-23. The joint is held in position by the slender members AD, BD, and CD, which can only exert forces that act along the members. If the 250-lb force is in the xy-plane, determine the forces in members AD, BD, and CD.

SOLUTION

250sin 30 250cos30 125 216.506 lbF i j i j

2 2

12 18 0.55470 0.8320512 18

AD AD AD ADT T Ti kT i k

BD BDTT k

2 2

9 18 0.44721 0.894439 18

CD CD CD CDT T Tj kT j k


0 :xF 0.55470 125 0ADT

0 :yF 0.44721 216.506 0CDT

0 :zF 0.83205 0.89443 0AD BD CDT T T


225 lbADT .......................................................................................................................Ans.

621 lbBDT .........................................................................................................................Ans.

484 lbCDT .......................................................................................................................Ans.


49

3-24 Three cables are used to support a 250-kg homogeneous plate as shown in Fig. P3-24. Determine the force in each of the three cables.

SOLUTION

2 2 2

0.5 0.6 1.5 0.29566 0.35479 0.886970.5 0.6 1.5


2 2 2

0.5 0.6 1.5 0.29566 0.35479 0.886970.5 0.6 1.5

B B B B BT T T Ti j kT i j k

2 2

0.5 1.5 0.31623 0.948680.5 1.5

C C C CT T Ti kT i k


0 :xF 0.29566 0.29566 0.31623 0A B CT T T

0 :yF 0.35479 0.35479 0A BT T

0 :zF 0.88697 0.88697 0.94868 250 9.81 0A B CT T T


691 NAT ............................................................................................................................Ans.

691 NBT ............................................................................................................................Ans.

1293 NCT .........................................................................................................................Ans.

3-25 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-25. Determine the magnitude and the coordinate direction angles of the unknown force F4.

SOLUTION

1 4 5 80 3 5 80 64 48 lbF i j i j

2 150 lbF k

3 200cos30 200sin 30 173.205 100.000 lbF j k j k


0 :xF 464 0xF

0 :yF 448 173.205 0yF

0 :zF 4150 100 0zF


4 64.000 lbxF 4 125.205 lbyF 4 50.000 lbzF

4 149.239 lb 149.2 lbF ...............................................................................................Ans.

1 64cos 115.39149.239x ..........................................................................................Ans.

1 125.205cos 147.03149.239y .......................................................................................Ans.

1 50cos 70.43149.239z ............................................................................................Ans.


50

3-26 A pair of steel pipes is stacked in a box as shown in Fig. P3-26. The masses and diameters of the smooth pipes are mA = 5 kg, mB = 20 kg, dA = 100 mm, and dB = 200 mm. Plot the two forces exerted on pipe A (by pipe B and by the side wall) as a function of the distance b between the walls of the box (200 mm b 300 mm). Determine the range of b for which

(a) The force at the side wall is less than WA, the weight of pipe A. (b) Neither of the two forces exceed 2WA. (c) Neither of the two forces exceed 4WA. SOLUTION From a free-body diagram of pipe A, the equations of equilibrium are

0 :xF sin 0A ABN F

0 :yF cos 5 9.81 0ABF

where

1 150sin150b

Therefore

49.05 NcosABF ............................................ Ans.

sin NA ABN F ........................................ Ans.

(a) 50 NAN for 260 mmb ..................... Ans.

(b) 100 NABF for 280 mmb .................. Ans.

(c) 200 NABF for 295 mmb .................. Ans.

3-27 A 75-lb stoplight is suspended between two poles as shown in Fig. P3-27. Neglect the weight of the flexible cables and plot the tension in both cables as a function of the sag distance d (0 d 8 ft). Determine the minimum sag d for which both tensions are less than

(a) 100 lb (b) 250 lb (c) 500 lb SOLUTION From a free-body diagram of the ring, the equations of equilibrium are

0 :xF cos cos 0B B A AT T

0 :yF sin sin 75 0A A B BT T

where

1tan20Ad

1tan10Bd

Therefore

cos

cosA A

BB

TT

75cos 75cos lb

sin cos sin cos sinB B

AA B B A A B

T ........................................................Ans.


51

75cos lb

sinA

BA B

T ................................................................................................................Ans.

(a) 100 lbBT for 5.6 ftd ......................... Ans.

(b) 250 lbBT for 2.0 ftd ......................... Ans.

(c) 500 lbBT for 1.0 ftd ......................... Ans.

3-28 A 50-kg load is suspended from a pulley as shown in Fig. P3-28. The tension in the flexible cable does not change as it passes around the small frictionless pulleys, and the weight of the cable may be neglected. Plot the force P required for equilibrium as a function of the sag distance d (0 d 1 m). Determine the minimum sag d for which P is less than

(a) Twice the weight of the load. (b) Four times the weight of the load. (c) Eight times the weight of the load. SOLUTION From a free-body diagram of pulley supporting the load, the equations of equilibrium

0 :xF cos cos 0C AT T

0 :yF sin sin 50 9.81 0A CT T

give

A C

49.05 N

2sin A

P T .................................... Ans.

Since the two angles are equal, the pulley is midway between the supports, and

1tan1.5Ad

(a) 1000 NP for 380 mmd ................... Ans.

(b) 2000 NP for 185 mmd .................... Ans.

(c) 4000 NP for 92 mmd ...................... Ans.


52

3-29 A worker positions a 250-lb crate by pulling on the rope BD as shown in Fig. P3-29. The 3-ft long rope BD is horizontal ( = 0) when the 5-ft long rope AB is vertical ( = 0).

(a) What is the maximum distance bmax that the crate can be pulled to the side using this arrangement? (b) Calculate and plot the forces in ropes AB and BD as a function of the distance b for 0 b bmax. (c) How could the worker pull the crate to the side more than the bmax calculated in part a? SOLUTION From a free-body diagram of the ring, the equations of equilibrium are

0 :xF sin cos 0AB BDT T

0 :yF cos sin 250 0AB BDT T

where

1sin5b

1 5 1 cossin

3

Therefore

cos

sinBD

ABTT

250sin lb

cos cos sin sinBDT ............Ans.

250cos lb

cos cos sin sinABT ............Ans.

(a) maxb occurs when BDT goes negative (after it goes to infinity);

max 3.90 ftb ..................................... Ans.

(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.

3-30 Two small wheels are connected by a light-weight rigid rod as shown in Fig. P3-30. Plot the angle (between the rod and the horizontal) as a function of the weight W1 (0.25W2 W1 10W2). Determine the weight W1 for which

(a) = 50 (b) = 10 (c) = 25 SOLUTION From a free-body diagram of body 1, the equations of equilibrium are

0 :xF 1cos sin 30 0T N (a)

0 :yF 1 1cos30 sin 0N T W (b)

From a free-body diagram of body 2, the equations of equilibrium are

0 :xF 2 sin 60 cos 0N T (c)

0 :yF 2 cos60 sin 5 9.81 0N T (d)

Adding Eqs. (a) and (c) together gives

2 1sin 60 sin 30 0N N (e)

while adding Eqs. (b) and (d) together gives


53

2 1 1cos60 cos30 49.05N N W (f)

Solving Eqs. (e) and (f) gives

1 21.73205N N

12

49.052

WN

Then Eqs. (b) and (a) can be written

2sin cos 60 49.05T N (g)

2cos sin 60T N (h)

Dividing Eq. (g) by Eq. (h) gives

2

2

cos60 49.05tansin 60

NN

(a) 50 for 1 14.98 NW ....................... Ans.

(b) 10 for 1 233 NW .............................. Ans.

(c) 25 for 1 971 NW .............................. Ans.

3-31 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P3-31. If the side force P has a magnitude of 150 lb,

(a) Calculate and plot FC, the force applied to the car as a function of the angle (0 45 ). (b) What is the maximum angle for which this method is effective (that is, for which P FC)? SOLUTION From a free-body diagram of the knot, the equations of equilibrium are

0 :xF 1 1 2 2cos cos 0T T

0 :yF 1 1 2 2sin sin 0P T T

(a) If 1 2 , then 1 2 CT T F and

150 lb

2sin 2sinCPF .......................... Ans.

(b) CP F when

30 .......................................................... Ans.


54

3-32 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-32. Determine the magnitude and the direction angle of the unknown force F4.

SOLUTION

1 500cos63 226.995 NxF 1 500sin 63 445.503 NyF

2 750cos30 649.519 NxF 2 750sin 30 375.000 NyF

3 1000cos60 500.000 NxF 3 1000sin 60 866.025 NyF


0 :xF 4226.995 649.519 500 0xF

0 :yF 4445.503 375 866.025 0yF

give

4 1376.514 NxF 4 45.522 NyF

4 1377 NF 1.89 ........................................................................................................Ans.

3-33 Two 10-in.-diameter pipes and a 6-in.-diameter pipe are supported in a pipe rack as shown in Fig. P3-33. The 10-in.-diameter pipes each weigh 300 lb and the 6-in.-diameter pipe weighs 175 lb. Determine the forces exerted on the pipes by the supports at the contact surfaces A, B, and C. Assume all surfaces to be smooth.

SOLUTION

1 6cos 41.4108AB BC

From a free-body diagram of cylinder B, the equations of equilibrium

0 :xF cos 0B BCN F

0 :yF sin 300 0BCF

are solved to get

453.557 lbBCF

340 lbBN ................................................. Ans.

From a free-body diagram of cylinder C, the equations of equilibrium

0 :xF cos cos 0BC AB CF F N

0 :yF sin sin 175 0AB BCF F

are solved to get

718.132 lbABF

879 lbCN ................................................. Ans.

Finally, from a free-body diagram of cylinder A, the equations of equilibrium

0 :xF cos 0A ABN F

0 :yF 300 sin 0D ABN F

are solved to get

539 lbAN ................................................. Ans.


55

3-34 The 250-kg block A of Fig. P3-34 is supported by a small wheel that is free to roll on the continuous cable between supports B and C. If the length of the cable is 42 m, determine the distance x and the tension T in the cable when the system is in equilibrium.

SOLUTION Since the pulley is free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 250 kg weight, the equations of equilibrium

0 :xF 2 1cos cos 0T T

0 :yF 1 2sin sin 250 9.81 0T T

are solved to get

1 2

1226.25 N

sinT (a)

From the geometry of the cable

1 2 42 m (b)

1 2cos cos 40 m (c)

1 2sin sin 6 m (d)

Therefore, from Eqs. (b) and (c) 42cos 40

17.753

4020 NT .................................................. Ans. Finally, Eq. (d) can be written

1 2 19.6779 m (e)

and adding Eqs. (b) and (e) gives

1 30.839 m

1 cos 29.4 mx ................................. Ans.

3-35 A joint in a bridge truss is subjected to the forces shown in Fig. P3-35. Determine the forces C and T required for equilibrium.


0 :xF 9 cos 45 9cos 60 12 0T

0 :yF sin 45 9sin 60 0T C

are solved to give

10.61 kipT .......................................................................................................................Ans.

15.29 kipC .......................................................................................................................Ans.


56

3-36 Four forces act on a small airplane in flight, as shown in Fig. P3-36; the weight W (25 kN), the thrust provided by the engine FT (10 kN), the lift provided by the wings FL, and the drag resulting from motion through the air FD. If the airplane is flying with a constant velocity in the x’ direction, determine the magnitudes of the lift and drag forces.


0 :xF cos10 sin10 10cos10 0D LF F

0 :yF cos10 10sin10 sin10 25 0L DF F

are solved to give

24.6 kNLF .......................................................................................................................Ans.

5.66 kNDF .......................................................................................................................Ans.

3-37 A 500-lb block is supported by three cables as shown in Fig. P3-37. Determine the tensions in the cables AB, AC, and AD.

SOLUTION

2 2

6 12 0.44721 0.894436 12

AB AB AB ABT T Tj kT j k

2 2 2

4 3 12 0.30769 0.23077 0.923084 3 12

AC AC AC AC ACT T T Ti j kT i j k

2 2 2

4 8 12 0.26726 0.53452 0.801784 8 12

AD AD AD AD ADT T T Ti j kT i j k


0 :xF 0.30769 0.26726 0AC ADT T

0 :yF 0.44721 0.23077 0.53452 0AB AC ADT T T

0 :zF 0.89443 0.92308 0.80178 500 0AB AC ADT T T


267 lbABT .........................................................................................................................Ans.

141.3 lbACT ......................................................................................................................Ans.

162.7 lbADT ......................................................................................................................Ans.

3-38 A 1250-N force F is supported by a cable AD and by struts AB and AC, as shown in Fig. P3-38. If the struts can transmit only axial tensile or compressive forces, determine the forces in the struts and the tension in the cable.

SOLUTION

2 2

6 3 0.89443 0.447216 3

AD AD AD ADT T Tj kT j k

2 2 2

2 6 2 0.30151 0.90453 0.301512 6 2

AB AB AB AB ABF F F Fi j kF i j k

2 2 2

3 6 2 0.42857 0.85714 0.285713 6 2

AC AC AC AC ACF F F Fi j kF i j k



57

0 :xF 0.30151 0.42857 0AB ACF F

0 :yF 0.89443 0.90453 0.85714 0AD AB ACT F F

0 :zF 0.44721 0.30151 0.28571 1250 0AD AB ACT F F


1677 NADT .......................................................................................................................Ans.

995 NABF .........................................................................................................................Ans.

700 NACF .........................................................................................................................Ans.

3-39 A 75-lb force F is supported by a tripod as shown in Fig. P3-39. If the legs can transmit only axial tensile or compressive forces, determine the forces in the legs AB, AC, and AD.

SOLUTION

2 2 2

25 20 54 0.39823 0.31859 0.8601825 20 54

AB AB AB AB ABF F F Fi j kF i j k

2 2 2

25 40 54 0.34867 0.55787 0.7531325 40 54

AC AC AC AC ACF F F Fi j kF i j k

2 2 2

25 6 54 0.41800 0.10032 0.9028925 6 54

AD AD AD AD ADF F F Fi j kF i j k


0 :xF 0.39823 0.34867 0.41800 0AB AC ADF F F

0 :yF 0.31859 0.55787 0.10032 0AB AC ADF F F

0 :zF 0.86018 0.75313 0.90289 75 0AB AC ADF F F


33.4 lbABF ........................................................................................................................Ans.

11.62 lbACF .....................................................................................................................Ans.

41.5 lbADF .......................................................................................................................Ans.

3-40 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-40. Determine the magnitude of the unknown forces F1, F2, and F3.

SOLUTION

1 1 1 1 12 2 2

3 2.5 3.5 0.57208 0.47673 0.667423 2.5 3.5

F F F Fi j kF i j k

2 2 2 2 22 2 2

3 4.5 3.5 0.46569 0.69854 0.543313 4.5 3.5

F F F Fi j kF i j k

3 3 3 3 32 2 2

3 5 4 0.42426 0.70711 0.565693 5 4

F F F Fi j kF i j k


0 :xF 1 2 30.57208 0.46569 0.42426 0F F F

0 :yF 1 2 30.47673 0.69854 0.70711 0F F F


58

0 :zF 1 2 30.66742 0.54331 0.56569 50 0F F F


1 2.34 kNF ........................................................................................................................Ans.

2 43.1 kNF ........................................................................................................................Ans.

3 44.2 kNF ........................................................................................................................Ans.

ch03

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