5–15.

Determine the horizontal and vertical components ofreaction at the pin at A and the reaction of the roller at B onthe lever.

AB

F � 50 lb

20 in. 18 in.

14 in. 30�

SOLUTION

Equations of Equilibrium: From the free-body diagram, and can be obtainedby writing the moment equation of equilibrium about point and the forceequation of equilibrium along the axis, respectively.

Ans.

Ans.

Using the result and writing the force equation of equilibrium alongthe axis, we have

Ans.Ay = 110.86 lb = 111 lb

Ay - 50 cos 30° - 67.56 = 0+ c ©Fy = 0;

yFB = 67.56 lb

Ax = 25 lb

Ax - 50 sin 30° = 0©Fx = 0;:+FB = 67.56 lb = 67.6 lb

50 cos 30°(20) + 50 sin 30°(14) - FB(18) = 0a + ©MA = 0;

xA

AxFB

*5–20.

The pad footing is used to support the load of 12 000 lb.Determine the intensities and of the distributedloading acting on the base of the footing for theequilibrium.

w2w1

w1

w2

5 in.9 in.9 in.

35 in.

12 000 lb

SOLUTION

a

Ans.

Ans.w1 = 6.58 kip>ft

121w1 - 1.6462a35

12b + 1.646a35

12b - 12 = 0+ c ©Fy = 0;

w2 = 1.646 kip>ft = 1.65 kip>ftw2 a35

12b117.5 - 11.672 - 12114 - 11.672 = 0+ ©MA = 0;

Equations of Equilibrium: The load intensity w can be determined directly by summing moments about point A.

2

5–23.

The smooth disks D and E have a weight of 200 lb and 100 lb,respectively. Determine the largest horizontal force P thatcan be applied to the center of disk E without causing thedisk D to move up the incline. P

1.5 ft

AB

DE

C

3

541 ft

SOLUTIONFor disk E:

For disk D:

Require for Pmax. Solving,

Ans.

NC = 143 lb

NA = 262 lb

Pmax = 210 lb

N¿ = 214 lb

NB = 0

+ c ©Fy = 0; NA a 35b + NB - 200 + N¿ a 1

5b = 0

:+ ©Fx = 0; NA a45b - N¿ ¢224

5≤ = 0

+ c ©Fy = 0; NC - 100 - N¿ a 15b = 0

:+ ©Fx = 0; - P + N¿ ¢2245≤ = 0

5–33.

The woman exercises on the rowing machine. If she exerts aholding force of on handle , determine thehorizontal and vertical components of reaction at pin andthe force developed along the hydraulic cylinder on thehandle.

SOLUTION

Equations of Equilibrium: Since the hydraulic cylinder is pinned at both ends, it canbe considered as a two-force member and therefore exerts a force directedalong its axis on the handle, as shown on the free-body diagram in Fig. a. From thefree-body diagram, can be obtained by writing the moment equation ofequilibrium about point .

Ans.

Using the above result and writing the force equations of equilibrium along the andaxes, we have

Ans.

Ans.Cy = 68.19 N = 68.2 N

200 sin 30° - 628.42 sin 15.52° + Cy = 0+ c ©Fy = 0;

Cx = 432.29 N = 432 N

Cx + 200 cos 30° - 628.42 cos 15.52° = 0+: ©Fx = 0;

yx

FBD = 628.42 N = 628 N

-200 sin 30°(750 + 150) = 0

FBD cos 15.52°(250) + FBD sin 15.52°(150) - 200 cos 30°(250 + 250)a+ ©MC = 0;

CFBD

FBD

BDC

ABCF = 200 N

F � 200 N

C

AB

D

0.25 m

0.75 m0.15 m0.15 m

0.25 m

30�

*5–56.

SOLUTION

Since

Solving,

Ans.u = 27.1° and u = 50.2°

880 sin u = 400 tan u + 20(9.81)

2.2 cos u = 1.0 +20(9.81)

400 tan u

cos u =1.0 + F

400

2.2

tan u =20(9.81)

F

:+ ©Fx = 0; R cos u - F = 0

+ c ©Fy = 0; R sin u - 20(9.81) = 0

The disk B has a mass of 20 kg and is supported on thesmooth cylindrical surface by a spring having a stiffness of

and unstretched length of The springremains in the horizontal position since its end A is attachedto the small roller guide which has negligible weight.Determine the angle for equilibrium of the roller.u

l0 = 1 m.k = 400 N>m0.2 m

r 2 m

BA k

u

*5–72.

SOLUTIONForce Vector and Position Vectors:

Equations of Equilibrium: Force equilibrium requires

Equating i, j and k components, we have

(1)

(2)

(3)Az - 608.11 - 153.91 -23

FED -8

17 FBC = 0©Fz = 0;

Ay + 211.43 -13

FED -9

17FBC = 0©Fy = 0;

Ax + 608.11 - 366.21 -23

FED +1217

FBC = 0©Fx = 0;

+ aAz - 608.11 - 153.91 -23

FED -8

17FBCbk 0

+ aAy + 211.43 -13

FED -9

17FBCb j

aAx + 608.11 - 366.21 -23

FED +1217

FBCb i

FA + F1 + F2 + FED + FBC = 0©F = 0;

r1 = 54k6m r2 = 58k6 m r3 = 56k6m

=1217

FBC i -9

17FBC j -

817

FBC k

FBC = FBCB 16 - 02i + 1-4.5 - 02j + 10 - 42k216 - 022 + 1-4.5 - 022 + 10 - 422 R

= - 23

FED i -13

FED j -23

FED k

FED = FEDB 1-6 - 02i + 1-3 - 02j + 10 - 62k21-6 - 022 + 1-3 - 022 + 10 - 622 R

= 5-366.21i + 211.43j - 153.91k6 N

F2 = 4505-cos 20° cos 30°i + cos 20° sin 30°k - sin 20°k6 N

F1 = 8605cos 45°i - sin 45°k6 N = 5608.11i - 608.11k6 N

FA = Ax i + Ay j + Az k

The pole is subjected to the two forces shown. Determinethe components of reaction of A assuming it to be a ball-and-socket joint. Also, compute the tension in each of theguy wires, BC and ED.

x

B

E

4 m

4.5 m 6 m

6 m

y

6 m

3 m

2 m

30°

20°

45°F2 = 450 N

F1 = 860 N

z

D

AC

*5–72. (continued)

Moment equilibrium requires

Equating i, j and k components, we have

(4)

(5)

Solving Eqs. (4) and (5) yields

Ans.

Substituting the results into Eqs. (1), (2) and (3) yields

Ans.Ax = 32.4 N Ay = 107 N Az = 1277.58 N = 1.28 kN

FBC = 205.09 N = 205 N FED = 628.57 N = 629 N

4817

FBC - 4FED + 1935.22 = 0©My = 0;

3617

FBC + 2FED - 1691.45 = 0©Mx = 0;

+ 6k * a -23

FEDi -13

FEDj -23

FEDkb = 0

+ 8k * 1241.90i + 211.43j - 762.02k24k * a12

17 FBCi -

917

FBCj -8

17FBCkb

r1 * FBC + r2 * 1F1 + F22 + r3 * FED = 0©MA = 0;

5–75.

If the pulleys are fixed to the shaft, determine the magnitudeof tension and the x, y, z components of reaction at thesmooth thrust bearing A and smooth journal bearing B.

SOLUTIONEquations of Equilibrium: From the free-body diagram of the shaft, Fig. a, , T,and can be obtained by writing the force equation of equilibrium along the y axisand the moment equations of equilibrium about the y and z axes, respectively.

Ans.

Ans.

Ans.

Using the above results and writing the moment equation of equilibrium about the x axis and the force equation of equilibrium along the x axis, we have

Ans.

Ans.

Finally, writing the force equation of equilibrium along the z axis, yields

Ans.Az = 711.11 N = 711 N

©Fz = 0; Az - 1233.33 - 900 + 1422.22 = 0

Ax = 866.67 N = 867 N©Fx = 0; 400 + 900 - 433.33 - Ax = 0

Bz = 1422.22 N = 1.42 kN

©Mx = 0; Bz(3) - 900(2) - 1233.33(2) = 0

Bx = -433.33 N = -433N

©Mz = 0; -Bx(3) - 400(1) - 900(1) = 0

T = 1233.33 N = 1.23 kN

©My = 0; 400(0.2) - 900(0.2) - 900(0.3) + T(0.3) = 0

©Fy = 0; Ay = 0

Bx

Ay

T

y

B

A

x

z

1 m1 m

1 m

0.3 m0.2 m

{400 i} N

{900 i} N{�900 k} N

T

*5–84.

Both pulleys are fixed to the shaft and as the shaft turnswith constant angular velocity, the power of pulley A istransmitted to pulley B. Determine the horizontal tension Tin the belt on pulley B and the x, y, z components ofreaction at the journal bearing C and thrust bearing D if

The bearings are in proper alignment and exertonly force reactions on the shaft.u = 45°.

300 mm

250 mm

150 mm

80 mm

200 mm

θ

T

50 N

z

y

A

BC

D

x

80 N65 N

SOLUTIONEquations of Equilibrium:

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.Dz = 32.1 N

Dz + 77.57 + 50 sin 45° - 80 - 65 = 0©Fz = 0;

Dy = 68.5 N

Dy + 24.89 - 50 cos 45° - 58.0 = 0©Fy = 0;

Dx = 0©Fx = 0;

Cy = 24.89 N = 24.9 N

58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0©Mz = 0;

Cz = 77.57 N = 77.6 N

165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0©My = 0;

T = 58.0 N

6510.082 - 8010.082 + T10.152 - 5010.152 = 0©Mx = 0;