8–3.If the coefficient of static friction at Ais m s 0.4 B x A · Blocks A and B have a mass of 100 kg and 150 kg, respectively. If the coefficient of static friction between A

8–3. If the coefficient of static friction at A is and the collar at B is smooth so it only exerts a horizontalforce on the pipe, determine the minimum distance sothat the bracket can support the cylinder of any masswithout slipping. Neglect the mass of the bracket.

x

m s= 0.4

200 mm

x100 mm

B

A

C

8–7.

SOLUTIONTo hold lever:

a

Require

Lever,

a

a) Ans.

b) Ans.P = 70 N 7 39.8 N Yes

P = 30 N 6 39.8 N No

PReqd. = 39.8 N

+ ©MA = 0; PReqd. (0.6) - 111.1(0.2) - 33.333(0.05) = 0

NB =33.333 N

0.3= 111.1 N

+ ©MO = 0; FB (0.15) - 5 = 0; FB = 33.333 N

The block brake consists of a pin-connected lever andfriction block at B. The coefficient of static friction betweenthe wheel and the lever is and a torque of is applied to the wheel. Determine if the brake can holdthe wheel stationary when the force applied to the lever is (a) (b) P = 70 N.P = 30 N,

5 N # mms = 0.3,

200 mm 400 mm

P150 mm O

B

A

5 N m

50 mm

.

8–56.

The uniform 6-kg slender rod rests on the top center of the3-kg block. If the coefficients of static friction at the pointsof contact are and determinethe largest couple moment M which can be applied to therod without causing motion of the rod.

mC = 0.3,mB = 0.6,mA = 0.4,

SOLUTIONEquations of Equilibrium: From FBD (a),

(1)

(2)

(3)

From FBD (b),

(4)

(5)

(6)

Ans.

Since the block does not slip.ton seod gnippils neht,oslA

occur at point B. Therefore, the above assumption is correct.1FB2max = ms B NB = 0.6150.832 = 30.50 N 7 FB ,1FA2max = ms A NA = 0.4180.262 = 32.11 N 7 FA ,

NB = 50.83 N NA = 80.26 N FA = FB = NC = 26.75 N

M = 8.561 N # m = 8.56 N # m

FB 10.32 - NB 1x2 - 29.431x2 = 0a+ ©MO = 0;

FA - FB = 0:+ ©Fx = 0;

NA - NB - 29.43 = 0+ c ©Fy = 0;

FC10.62 + NC10.82 - M - 58.8610.32 = 0©MB = 0;

NB + FC - 58.86 = 0+

a+

c ©Fy = 0;

FB - NC = 0:+ ©Fx = 0;

M

C

B

600 mm

100 mm 100 mm

800 mm

300 mmA

Friction: Assume slipping occurs at point C and the block tips, then FC = msCNC = 0.3NC and x = 0.1 m. Substituting these values into Eqs. (1), (2), (3), (4), (5), and (6) and solving, we have

8–78.

SOLUTION

Frictional Forces on Screw: Here,

owt ta noitcirf ecniS dnascrews must be overcome, then, Applying Eq. 8–3, we have

Ans.

Note: Since the screw is self-locking. It will not unscrew even if moment Mis removed.

Equations of Equilibrium and Friction: Since the shaft is on the verge to rotateabout point O, then, and From FBD (a),

a

From FBD (b),

a Ans.230.511759.222410.22 - M = 0 M = 352 N # m+ ©MO = 0;

879.61 10.62 - NB 10.32 = 0 NB = 1759.22 N+ ©MD = 0;

FB = ms¿NB = 0.5NB .FA = ms¿NA = 0.5NA

fs 7 u,

P = 879.61 N = 880 N

5 = 2P10.0062 tan16.057° + 19.290°2M = Wr tan1u + f2

W = 2P.fs = tan-1ms = tan-1 10.352 = 19.290°.M = 5 N # m

u = tan-1a l

2prb = tan-1 c 4

2p162 d = 6.057°,

The braking mechanism consists of two pinned arms and asquare-threaded screw with left and righthand threads.Thuswhen turned, the screw draws the two arms together. If thelead of the screw is 4 mm, the mean diameter 12 mm, andthe coefficient of static friction is determine thetension in the screw when a torque of is applied totighten the screw. If the coefficient of static friction betweenthe brake pads A and B and the circular shaft is determine the maximum torque M the brake can resist.

mœs = 0.5,

5 N # mms = 0.35, 300 mm

300 mm

5 N · m

200 mm

C

A B

M

D

8–80.

Determine the horizontal force P that must be appliedperpendicular to the handle of the lever at A in order todevelop a compressive force of 12 kN on the material. Eachsingle square-threaded screw has a mean diameter of 25 mmand a lead of 7.5 mm. The coefficient of static friction at allcontacting surfaces of the wedges is and thecoefficient of static friction at the screw is

SOLUTION

Referring to the free-body diagram of wedge C shown in Fig. a, we have

Using the result of N and referring to the free-body diagram of wedge B shown inFig. b, we have

Since the screw is being tightened, Eq. 8–3 should be used. Here,

and W = T = 4166.68N. SinceM must overcome the friction of two screws,

Ans.P = 104 NP(0.25) = 234166.68(0.0125) tan (8.531° + 5.455°)4M = 23Wr tan (fs + u)4

fs = tan-1ms = tan-1(0.15) = 8.531°; M = P(0.25);

u = tan- 1 c L

2prd = tan-1 c 7.5

2p(12.5)d = 5.455°;

T = 4166.68 N

T - 6563.39 sin 15° - 0.2(6563.39) cos 15° - 0.2(6000) = 0©Fx = 0;:+N¿ = 6000 N

N¿ - 6563.39 cos 15° + 0.2(6563.39) sin 15° = 0+ c ©Fy = 0;

N = 6563.39 N

2N cos 15° - 230.2N sin 15°4 - 12000 = 0+ c ©Fy = 0;

mœs = 0.15.

ms = 0.2,

A

B

250 mm15� 15�C

8 9

A cable is attached to the plate B of mass MB, passes over a fixed peg at C, and is attached tothe block at A. Using the coefficients of static friction shown, determine the smallest mass ofblock A so that it will prevent sliding motion of B down the plane.

Given:

MB 20 kg� �A 0.2�

� 30 deg� �B 0.3�

�C 0.3�g 9.81m

s2�

Solution:

Iniitial guesses: T1 1 N� T2 1 N� NA 1 N� NB 1 N� MA 1 kg�

Given

Block A:

�Fx = 0; T1 �A NA� MA g sin �� � 0�

�Fy = 0; NA MA g cos �� � 0�

Plate B:

�Fx = 0; T2 MB g sin �� � �B NB �A NA 0�

NB NA� MB g cos �� � 0��Fy = 0;

Peg C: T2 T1 e�C��

T1

T2

NA

NB

MA

��������

��������

Find T1 T2� NA� NB� MA�� �

MA 2.22 kg�

1.

Ans.

–

8–102.

SOLUTIONEquations of Equilibrium: From FBD (a),

a (1)

From FBD (b),

a (2)

Frictional Force on Flat Belt: Here, Applying Eq. 8–6,we have

(3)

Solving Eqs. (1), (2), and (3) yields

Ans.

T1 = 39.22 N T2 = 117.8 N

M = 3.93 N m#

T2 = T1 e0.3p = 3.003T1

T2 = T1 emb,b = 180° = p rad.

M + T1 10.052 - T2 10.052 = 0+ ©MO = 0;

T2 11002 + T1 12002 - 196.211002 = 0+ ©MC = 0;

The 20-kg motor has a center of gravity at G and is pin-connected at C to maintain a tension in the drive belt.Determine the smallest counterclockwise twist or torque Mthat must be supplied by the motor to turn the disk B ifwheel A locks and causes the belt to slip over the disk. Noslipping occurs at A. The coefficient of static frictionbetween the belt and the disk is ms = 0.35.

50 mm

M

50 mm

150 mm

100 mm

B

C

A G

8–103.

Blocks A and B have a mass of 100 kg and 150 kg,respectively. If the coefficient of static friction between Aand B and between B and C is and between theropes and the pegs D and E determine thesmallest force F needed to cause motion of block B if P 30 N.

SOLUTION

Assume no slipping between A and B.

Ans.

Note: Since B moves to the right,

Hence, no slipping occurs between A and B as originally assumed.

Pmax = 112 N 7 30 N

245.25 = Pmax e0.5(p2 )

(FAB)max = 0.25 (981) = 245.25 N

T2 = T1emb; F = 768.1e0.5(3p

4 ) = 2.49 kN

Peg E :

NBC = 1909.4 N

FBE = 768.1 N

NBC - 981 + FBE sin 45° - 150 (9.81) = 0+ c ©Fy = 0;

- 65.80 - 0.25 NBC + FBE cos 45° = 0©Fx = 0;:+Block B :

T2 = T1 emb; FAD = 30 e0.5(p2 ) = 65.80 N

Peg D :

=

m¿s = 0.5ms = 0.25

P

F

DA

B

E

C

45�

4000 N. If µs = 0.35,

50 mm 18.75 mm

25 mm

P = 4000 N

3 3

2 2

2 50 25= (0.35)(4000)3 50 25

= 54 444 N mm

= 54.4 N m Ans.M

8–116.

SOLUTIONEquations of Equilibrium and Friction: The resultant normal force on the post is

Since the post is on the verge of rotating,

a

Ans.M = 17.0 N # m

M - 54.0p10.12 = 0+ ©MO = 0;

F = ms N = 0.31180p2 = 54.0p N.

N =121600 + 021321p210.22 = 180p N.

A 200-mm diameter post is driven 3 m into sand for whichIf the normal pressure acting completely around the

post varies linearly with depth as shown, determine thefrictional torque M that must be overcome to rotate the post.

ms = 0.3.M

200 mm

3 m

600 Pa

8–131.

The cylinder is subjected to a load that has a weight W.If the coefficients of rolling resistance for the cylinder’stop and bottom surfaces are and , respectively,show that a horizontal force having a magnitude of

is required to move the load andthereby roll the cylinder forward. Neglect the weight of thecylinder.

P = [W(aA + aB)]>2r

aBaA

SOLUTION

a (1)

Since and are very small, . Hence, from Eq. (1)

(QED)P =W(aA + aB)

2r

cos fA - cos fB = 1fBfA

+ ©MB = 0; P(r cos fA + r cos fB) - W(aA + aB) = 0

+ c ©Fy = 0; (RA)y - W = 0 (RA)y = W

:+ ©Fx = 0; (RA)x - P = 0 (RA)x = P

W

P

r

A

B