Solution manual for engineering mechanics statics and dynamics 14th edition by hibbeler isbn 0133915 by thomas.konkel646

Solution Manual for Engineering Mechanics Statics and Dynamics 14th Edition by Hibbeler ISBN 0133915425

9780133915426

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3–1.

The members of a truss are pin connected at joint O Determine the magnitudes of F1 and F2 for equilibrium. Set

161 1

u = 60°. SOLUTION 5 kN 5 4 30 3 y 70 F2 x O u : + ©Fx = 0; F2 sin 70° + F1 cos 60° - 5 cos 30°4 5 (7) = 0 7 kN F1 0.9397F2 + 0.5F1 = 9.930 3 + c ©Fy = 0; F2 cos 70° + 5 sin 30° - F1 sin 60°5 (7) = 0 0.3420F2 - 0.8660F1 = 1.7 Solving: F2 = 9.60 kN F1 = 1.83 kN Ans Ans.

161 2 Ans: F2 = 9.60 kN F1 = 1.83 kN

The members of a truss are pin connected at joint O. y Determine the magnitude of F1 equilibrium.Set F2 = 6 kN.

SOLUTION

and its angle

161 3

–2.

u 4 for 5 kN 5 4 30 3 70 F2 x O u : + ©Fx = 0; 6 sin 70° + F1 cos u - 5 cos 30°5 F1 cos u = 4.2920 3 (7) = 0 7 kN F1 + c ©Fy = 0; 6 cos 70° + 5 sin 30° - F1 sin u - 5(7) = 0 F1 sin u = 0.3521 Solving: u = 4.69° F1 = 4.31 kN Ans. Ans.

161 4

Ans: u = 4.69° F1 = 4.31 kN

Determine the magnitude and direction u of F so that the y particle is in equilibrium.

Solution

Equations of Equilibrium. Referring to the FBD shown in Fig a, S + ΣFx = 0; F sin u + 5 - 4 cos 60° - 8 cos 30° = 0 F sin u = 3.9282 (1)

cΣFy = 0; 8 sin 30° - 4 sin 60° - F cos u = 0 F cos u = 0.5359

Divide Eq (1) by (2), sin u cos u = 7.3301 sin u

Realizing that tan u = cos u , then tan u = 7.3301

u = 82.23° = 82.2°

Substitute this result into Eq. (1),

F sin 82.23° = 3.9282

F = 3.9646 kN = 3.96 kN

163163

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3–3.

8 kN 30 60 x 5 kN

u 4 kN F

(2)

Ans.

Ans. Ans:

u = 82.2°

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The bearing consists of rollers, symmetrically confined within the housing The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium.

165165 N *3–4.

40° C + c ©Fy = 0; 125 - NC cos 40° = 0 NC = 163.176 = 163 N : + ©Fx = 0; NB - 163.176 sin 40° = 0 NB = 105 N Ans. Ans. B C NB A 125 N

SOLUTION

The members of a truss are connected to the gusset plate If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium.Take u = 90°

167167 3–5.

SOLUTION f

1 a 3

= 53.13° 4 y 9 kN A O u F 5 3 B 4 x : + ©Fx = 0; T cos 53.13° - F a 4 5 b = 0 C 3 + c ©Fy = 0; 9 - T sin 53.13° - F a 5 b = 0 T Solving, T = 7.20 kN F = 5.40 kN Ans. Ans.

= 90° - tan -

Ans: T = 7.20 kN F = 5.40 kN

The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle u for equilibrium. The forces are concurrent at point O. Take

169169 3–6.

F = 8 kN

y 9 kN A F 5 3 B 4 : + ©Fx = 0; T cos f - 8 a 4 5 b = 0 (1) O u x 3 + c ©Fy = 0; 9 - 8 a 5 b - T sin f = 0 (2) C Rearrange then divide Eq. (1) into Eq. (2): T tan f = 0.656, f = 33.27° T = 7.66 kN u = f + tan - 1 a 3 b = 70.1° 4 Ans. Ans.

SOLUTION

Ans:

T = 7.66 kN

u = 70.1°

170170

C θ

20°

+ R©Fx¿ = 0; 60 cos 10° - T - T cos u = 0

+ Q©Fy¿ = 0; T sin u - 60 sin 10° = 0

Thus,

T(1 + cos u) = 60 cos 10°

T(2cos2 u ) = 60 cos 10° 2 (1)

u 2T sin 2 u cos 2 = 60 sin 10° (2)

Divide Eq.(2) by Eq.(1)

u tan 2 = tan 10°

u = 20°

Ans

Ans.

–

The man attempts to pull down the tree using the cable and small pulley arrangement shown. If the tension in AB is 60 lb, determine the tension in cable CAD and the angle u which the cable makes at the pulley. A B

30°

SOLUTION

T 30.5 lb

Ans:

*3–8.

The cords ABC and BD can each support a maximum load of 100 lb. Determine the maximum weight of the crate, and

the angle u for equilibrium.

Solution

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of rope ABC is constant.Assuming that the tension in cable BD reaches the limit first. Then, TBD = 100 lb. Referring to the FBD shown in Fig a,

Substitute this result into Eq. (1), 5 100 cos 78.69° = 13 W

W = 50.99 lb = 51.0 lb 6 100 lb (O.K) Ans.

Ans: u = 78.7°

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u B

5 S + ΣFx = 0; W a 13 b - 100 cos u = 0 5W 12 13 A 5 C 100 cos u = (1) 13 12 +cΣFy = 0; 100 sin u - W - W a 13 b = 0 25 100 sin u = 13 W (2)

cos

sin

Divide Eq. (2) by (1), sin u

u = 5

u Realizing that tan u = cos u , tan u = 5

u = 78.69° = 78.7° Ans.

= 51.0 lb

Determine the maximum force F that can be supported in the position shown if each chain can support a maximum

tension of 600 lb before it fails

Solution

Equations of Equilibrium. Referring to the FBD shown in Fig a

Since chain AB is subjected to a higher tension, its tension will reach the limit first. Thus,

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the

3 4 5 30 A C F

4 +cΣFy = 0; TAB a 5 b - F sin 30° = 0 TAB = 0.625 F 3 S + ΣFx = 0; TAC + 0.625 F a 5 b - F cos 30° = 0 TAC = 0.4910 F

TAB = 600; 0.625 F = 600 F = 960 lb Ans.

176176 Ans: F = 960 lb

3–10.

The block has a weight of 20 lb and is being hoisted at uniform velocity Determine the angle u for equilibrium and B the force in cord AB

Solution

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of cord CAD is constant. Thus, F = 20 lb. Referring to the FBD shown in Fig. a,

S + ΣFx = 0; 20 sin u - TAB sin 20° = 0 20 sin u

TAB = (1) sin 20° +cΣFy = 0; TAB cos 20° - 20 cos u - 20 = 0 (2)

Substitute Eq (1) into (2), 20 sin u sin 20° cos 20° - 20 cos u = 20

sin u cos 20° - cos u sin 20° = sin 20°

Realizing that sin (u - 20°) = sin u cos 20° - cos u sin 20°, then sin (u - 20°) = sin 20°

- 20° = 20°

170170

20 A u D C F

Ans.

= 40°

Substitute this result into Eq (1)

sin 40°

AB = sin 20° = 37.59 lb = 37.6 lb

Ans:

u = 40°

TAB = 37.6 lb

171171

Determine the maximum weight W of the block that can be suspended in the position shown if cords AB and CAD can each support a maximum tension of 80 lb. Also, what is the angle u for equilibrium?

Solution

Equations of Equilibrium. Assume that for equilibrium, the tension along the length of cord CAD is constant. Thus, F = W Assuming that the tension in cord AB reaches the limit first, then TAB = 80 lb Referring to the FBD shown in Fig a,

S + ΣFx = 0; W sin u - 80 sin 20° = 0

80 sin 20° W = (1) sin u

+cΣFy = 0; 80 cos 20° - W - W cos u = 0

80 cos 20°

W = 1 + cos u (2)

Equating Eqs (1) and (2), 80 sin 20° 80 cos 20°

sin u = 1 + cos u

sin u cos 20° - cos u sin 20° = sin 20°

Realizing then sin (u - 20°) = sin u cos 20° - cos u sin 20°, then sin (u - 20°) = sin 20°

u - 20° = 20°

u = 40° Ans.

Substitute this result into Eq (1)

80 sin 20°

W = sin 40° = 42.56 lb = 42.6 lb 6 80 lb (O.K) Ans.

Ans:

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–11.

B 20 A u D C F

*3–12.

The lift sling is used to hoist a container having a mass of F 500 kg Determine the force in each of the cables AB and AC as a function of u If the maximum tension allowed in

each cable is 5 kN, determine the shortest lengths of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G

SOLUTION

Free-Body Diagram: By observation, the force F1 has to support the entire weight

If the maximum allowable tension in the cable is 5 kN, then

174174

B θ θ C Equations of Equilibrium: 1.5 m 1.5 m : + ©Fx = 0; + c ©Fy = 0; FAC cos u - FAB cos u =

FAC = FAB = F 4905 - 2F sin u = 0 F = 52452.5 cos u6 N G

FAC = FAB = F = 52.45 cos u6 kN Ans.

Thus,

u = 29.37° 1.5 From the geometry, l = cos u and u = 29.37°. Therefore 1.5 l = cos 29.37° = 1.72 m Ans.

2452.5 cos u = 5000

175175 Ans: FAC = {2.45 cos u} kN l = 1.72 m

3–13.

A nuclear-reactor vessel has a weight of 500(103) lb Determine the horizontal compressive force that the spreader bar AB exerts on point A and the force that each cable segment CA and AD exert on this point while the

vessel is hoisted upward at constant velocity. A

Solution

At point C : S + ΣFx = 0; FCB cos 30° - FCA cos 30° = 0

FCB = FCA +cΣFy = 0; 500(103) - FCA sin 30° - FCB sin 30° = 0 500(103) - 2FCA sin 30° = 0

FCA = 500(103) lb

At point A :

+ ΣFx = 0; 500(103) cos 30° - FAB = 0

FAB = 433(103) lb

+cΣFy = 0; 500(103) sin 30° - FAD = 0

FAD = 500(103) sin 30°

FAD = 250(103) lb

Ans: FCA = 500(103) lb

176176

30 C 30

B D E

Ans.

Determine the stretch in each spring for equlibrium of the 2-kg block The springs are shown in the equilibrium position.

178178 4 3–14.

3 m 4 m C B SOLUTION FAD = 2(9.81) = xAD(40) xAD = 0.4905 m 3 m Ans. kAC 20 N/m A kAB 30 N/m : + ©Fx = 0; FAB a 5 1 b - FAC a b = 0 22 kAD 40 N/m 1 3 + c ©Fy = 0; FAC a b + FAB a 22 5 D b - 2(9.81) = 0 FAC = 15.86 N 15.86 xAC = 20 = 0.793 m Ans. FAB = 14.01 N 14.01 xAB = 30 = 0.467 m Ans.

179179

Ans: xAD = 0.4905 m xAC = 0.793 m xAB = 0.467 m

The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D

SOLUTION F = kx = 30(5 - 3) = 60 N

180180 4 3 3–

15.

3 m 4 m C B

: + ©F

a 5 b = 0 3 m kAC 20 N/m A kAB 30 N/m T = 67.88 N + c ©Fy = 0; - W + 67.88 sin 45° + 60 a 5 W = 84 N 84 m = 9.81 = 8.56 kg b = 0 D Ans.

0; Tcos 45° - 60

Ans: m = 8.56 kg

181181

Determine the mass of each of the two cylinders if they cause a sag of s = 0.5 m when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed.

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16.

2 m 1 m 2 m C D 1.5 m s k 100 N/m k 100 N/m

2.5)

32.84

SOLUTION T

= 100

(2.828 -

Ans:

m = 2.37 kg

183183

17.

Determine the stiffness kT of the single spring such that the force F will stretch it by the same amount s as the force F stretches the two springs Express kT in terms of stiffness k1 and k2 of the two springs

184184 s s k k k k k k

–

Unstretched position kT F k1 k2 F Solution F = ks s = s1 + s2 F F F s = = + T 1 2 1 1 1 = + T k1 k2 Ans. Ans: 1 = T 1 1 + 1 k2

3–18.

If the spring DB has an unstretched length of 2 m, determine the stiffness of the spring to hold the 40-kg crate in the position shown.

Solution

Equations of Equilibrium. Referring to the FBD shown in Fig a,

185185

2 m 3 m C B 2 m k D

3 1 S + ΣFx = 0; TBD a 113 b - TCDa 12 b =

(1) 2 1 +cΣFy = 0; TBD a 113 b + TCDa 12 b - 40(9.81) = 0 (2) Solving Eqs (1) and (2) TBD = 282.96 N TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m

Fsp = kx

282.96 = k

113 - 2) k = 176.24 N>m = 176 N>m Ans. Ans: k = 176 N>m

Then, x = l - l0 = (113 - 2) m. Thus,

;

(

3–19.

Determine the unstretched length of DB to hold the 40-kg crate in the position shown. Take k = 180 N>m.

Solution

Equations of Equilibrium. Referring to the FBD shown in Fig. a,

186186

2 m 3 m C B 2 m k D

3 1 S + ΣFx = 0; TBD a 113 b - TCDa 12 b

(1) 2 1 +cΣFy = 0; TBD a 113 b + TCDa 12 b - 40(9.81) = 0 (2) Solving Eqs (1) and (2) TBD = 282.96 N TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m Then,

Fsp = kx; 282.96 = 180(113 - l0) l0 = 2.034 m = 2.03 m Ans. Ans:

= 0

x = l - l0 = 113 - l0. Thus

A vertical force P = 10 lb is applied to the ends of the 2-ft cord AB and spring AC If the spring has an unstretched length of 2 ft, determine the angle u for equilibrium. Take k = 15 lb>ft.

180180

*3–20.

2 ft B u 2 ft C k

Fs cos

cos u

+ c ©Fy = 0; T sin u + Fs sin f - 10 = 0 s = 2(4)2 + (2)2 - 2(4)(2) cos u - 2 = 225 - 4 cos u - 2 Fs = ks = 2k(25 - 4 cos u - 1) A (1) (2) P From Eq. (1): T = Fs a cos f b cos u T = 2kA25 - 4 cos u - 1B¢ 2 - cos u ≤ a 1 b 25 - 4 cos u cos u From Eq. (2): 2ka 25 - 4 cos - 1b(2 - cos u) 25 - 4 cos u 2ka 25 - 4 cos u - 1b2 sin u tanu + 225 - 4 cos u = 10 a 25 - 4 cos u - 1 b 10 (2tanu - sin u + sin u) = 25 - 4 cos u 2k tan u a 25 - 4 cos u - 1 b 25 - 4 cos u Set k = 15 lb>ft 10 = 4k Solving for u by trial and error, u = 35.0° Ans.

SOLUTION

Fx = 0;

f - T

= 0

181181 Ans: u = 35.0°

3–21.

Determine the unstretched length of spring AC if a force P = 80 lb causes the angle u = 60° for equilibrium. Cord AB is 2 ft long Take k = 50 lb>ft

SOLUTION

l = 242 + 22 - 2(2)(4) cos 60°

l = 212

212 2 sin 60° = sin f

f = sin- 1 ¢ 2sin 60° ≤ = 30° 212

182182

B u 2 ft C k A

50(212 - l¿) l = 21240 = 2.66 ft 50 Ans.

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The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m Determine the horizontal force F applied to the cord which is attached to the small ring B so that the displacement of the ring from the wall is d = 1.5 m.

184184 x 3–22.

A k 500 N/m SOLUTION B 6 m F k 500 N/m : + ©F = 0; 1.5 (T)(2) - F = 0 C 211.25 d T = ks = 500(232 + (1.5)2 - 3) = 177.05 N F = 158 N Ans.

3–23.

The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m Determine the displacement d of the cord from the wall when a force F = 175 N is applied to the cord.

SOLUTION

T sin u = 87.5

T C 232 2 S = 87.5 + d

T = ks = 500(232 + d2 - 3)

By trial and error:

d a13 = 0.175 29 + d2

d = 1.56 m

d A k 500 N/m B 6 m F k 500 N/m C d

Ans.

Ans: d = 1.56

*3–24.

Determine the distances x and y for equilibrium if F1 = 800 N and F2 = 1000 N

Solution

Equations of Equilibrium. The tension throughout rope ABCD is constant, that is A

F1 = 800 N. Referring to the FBD shown in Fig. a, x +cΣFy = 0; 800 sin f - 800 sin u = 0 f = 0

S + ΣFx = 0; 1000 - 2[800 cos u] = 0 u = 51.32°

Referring to the geometry shown in Fig b, y = 2 m Ans. and 2 x = tan 51.32°; x = 1.601 m = 1.60 m

188188 1

F C D y F2 B 2 m

Ans:

Ans.

y = 2 m