806

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8–58.

Determine the largest angle that will cause the wedge tobe self-locking regardless of the magnitude of horizontalforce P applied to the blocks. The coefficient of staticfriction between the wedge and the blocks is .Neglect the weight of the wedge.

SOLUTION

Free-Body Diagram: For the wedge to be self-locking, the frictional force Findicated on the free-body diagram of the wedge shown in Fig. a must act downwardand its magnitude must be

Equations of Equilibrium: Referring to Fig. a, we have

Using the requirement we obtain

Ans.u = 33.4°

N tan u>2 … 0.3N

F … 0.3N,

F = N tan u>2

2N sin u>2 - 2F cos u>2 = 0+ c ©Fy = 0;

F … msN = 0.3N .

ms = 0.3

P Pu

Ans:u = 33.4°

811

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8–63.

SOLUTION

From FBD (b),

Ans.P = 2.39 kN

- 2.841 sin 10° = 0

P - 0.3512.6252 - 0.3512.8412 cos 10°:+ ©Fx = 0;

NA = 2.841 kN

NA cos 10° - 0.35NA sin 10° - 2.625 = 0+ c ©Fy = 0;

NB - 2.625 = 0 NB = 2.625 kN+ c ©Fy = 0;

Determine the minimum applied force P required to movewedge A to the right.The spring is compressed a distance of175 mm. Neglect the weight of A and B. The coefficient ofstatic friction for all contacting surfaces is Neglect friction at the rollers.

ms = 0.35. k = 15 kN/m

AP

B

10°Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx = 1510.1752 = 2.625 kN. If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = msNA = 0.35NA and FB = msNB = 0.35NB. From FBD (a),

Ans:P = 2.39 kN

813

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8–65.

Determine the smallest force P needed to lift the 3000-lbload. The coefficient of static friction between A and C andbetween B and D is , and between A and B .Neglect the weight of each wedge.

ms¿ = 0.4ms = 0.3

SOLUTION

From FBD (a):

(1)

(2)

Solving Eqs. (1) and (2) yields:

From FBD (b):

Ans. P = 4054 lb = 4.05 kip

:+ ©Fx = 0; P - 0.3(3868.2) - 4485.4 sin 15° - 1794.1 cos 15° = 0

+ c ©Fy = 0; NC + 0.4 (4485.4) sin 15° - 4485.4 cos 15° = 0 NC = 3868.2 lb

N = 4485.4 lb ND = 2893.9 lb

+ c ©Fy = 0; N cos 15° - 0.4N sin 15° - 0.3ND - 3000 = 0

:+ ©Fx = 0; 0.4N cos 15° + N sin 15° - ND = 0

3000 lb

15°PA

BD

C

Ans:P = 4.05 kip

831

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8–83.

A cylinder having a mass of 250 kg is to be supported by thecord which wraps over the pipe. Determine the smallestvertical force F needed to support the load if the cordpasses (a) once over the pipe, , and (b) two timesover the pipe, . Take .ms = 0.2b = 540°

b = 180°

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

a) If

Ans.

b) If

Ans.F = 372.38 N = 372 N

2452.5 = Fe 0.2(3p)

T2 = T1 e mb

b = 540° = 3 p rad

F = 1308.38 N = 1.31 kN

2452.5 = Fe 0.2p

T2 = T1 e mb

b = 180° = p rad

T2 = 250(9.81) = 2452.5 NT1 = F F

Ans:F = 1.31 kNF = 372 N

834

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8–86.

SOLUTION

Yes, just barely. Ans.

Fmax = 0.8 (185) = 148 lb 7 136.9 lb

F = 136.9 lb

:+ ©Fx = 0; 136.9 - F = 0

N = 185 lb

+ c©Fy = 0; N - 185 = 0

T2 = T1 emb = 100 e 0.2 p2 = 136.9 lb

b =p

2

The 100-lb boy at A is suspended from the cable that passesover the quarter circular cliff rock. Determine if it ispossible for the 185-lb woman to hoist him up; and if this ispossible, what smallest force must she exert on thehorizontal cable? The coefficient of static friction betweenthe cable and the rock is , and between the shoes ofthe woman and the ground .ms

œ = 0.8ms = 0.2

A

Ans:Yes, it is possible.F = 137 lb

840

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*8–92.

Determine the force P that must be applied to the handle of the lever so that B the wheel is on the verge of turning if M = 300 N # m. The coefficient of static friction between the belt and the wheel is ms = 0.3.

Ans:19.6 N

Solution

Frictional Force on Flat Belt. Here b = 270° =3p 2

rad.

TD = TAems b

TD = TAe0.3 (3p2 )

TD = 4.1112 TA (1)

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

a+ΣMB = 0; 300 + TA (0.3) - TD (0.3) = 0 (2)

Solving Eqs. (1) and (2),

TA = 321.42 N TD = 1321.42 N

Subsequently, from the FBD of the lever, Fig. b

a+ΣMC = 0; 1321.42(0.025) - 321.42(0.06) - P(0.7) = 0

P = 19.64 N = 19.6 N Ans.

700 mm60 mm

25 mm

C

300 mm

B

M

A

D

P