SOLUTION 8 7. To hold lever: a+ M O = 0; F B (0.15) - 5 = 0; F B = N. Require = N N B = N 0.3. Lever,

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1 8 3. If the coefficient of static friction at is m s = 0.4 and the collar at is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support the cylinder of any mass without slipping. Neglect the mass of the bracket. 100 mm x C 200 mm

8 7. The block brake consists of a pin-connected lever and friction block at. The coefficient of static friction between the wheel and the lever is ms = 0.

2 8 7. The block brake consists of a pin-connected lever and friction block at. The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5N# m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N. 50 mm 150 mm O 5N m P 200 mm 400 mm To hold lever: a+ M O = 0; F (0.15) - 5 = 0; F = N Require N = N 0.3 = N Lever, a+ M = 0; P Reqd. (0.6) (0.2) (0.05) = 0 P Reqd. = 39.8 N a) P = 30 N N No b) P = 70 N N Yes

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4 8 56. The uniform 6-kg slender rod rests on the top center of the 3-kg block. If the coefficients of static friction at the points of contact are m = 0.4, m = 0.6, and m C = 0.3, determine the largest couple moment M which can be applied to the rod without causing motion of the rod. C 800 mm M Equations of Equilibrium: From FD (a), : + F x = 0; F - N C = 0 (1) 300 mm 600 mm + c F y = 0; a+ M = 0; N + F C = 0 F C N C M = 0 (2) (3) 100 mm 100 mm From FD (b), + c F y = 0; N - N = 0 (4) : + F x = 0; F - F = 0 (5) a+ M O = 0; F N 1x x2 = 0 (6) Friction: ssume slipping occurs at point C and the block tips, then F C = ms C N C = 0.3NC and x = 0.1 m. Substituting these values into Eqs. (1), (2), (3), (4), (5), and (6) and solving, we have M = N # m = 8.56 N # m N = N N = N F = F = N C = N Since 1F 2 max = m s N = = N 7 F, the block does not slip. lso, 1F 2 max = m s N = = N 7 F, then slipping does not occur at point. Therefore, the above assumption is correct.

5 8 78. The braking mechanism consists of two pinned arms and a square-threaded screw with left and righthand threads. Thus when turned, the screw draws the two arms together. If the lead of the screw is 4 mm, the mean diameter 12 mm, and the coefficient of static friction is m s = 0.35, determine the tension in the screw when a torque of 5N# m is applied to tighten the screw. If the coefficient of static friction between the brake pads and and the circular shaft is ms œ = 0.5, determine the maximum torque M the brake can resist. 200 mm M 5N m 300 mm 300 mm C D Frictional Forces on Screw: Here, u = tan -1 a l 2pr b = 4 tan-1 c 2p162 d = 6.057, M = 5N# m and f s = tan -1 m s = tan = Since friction at two screws must be overcome, then, W = 2P. pplying Eq. 8 3, we have M = Wr tan1u + f2 5 = 2P tan P = N = 880 N Note: Since f s 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Equations of Equilibrium and Friction: Since the shaft is on the verge to rotate about point O, then, F = m s N = 0.5N and F = m s N = 0.5N. From FD (a), a+ M D = 0; N = 0 N = N From FD (b), a + MO = 0; M = 0 M = 352 N # m

6 8 80. Determine the horizontal force P that must be applied perpendicular to the handle of the lever at in order to develop a compressive force of 12 kn on the material. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is m s = 0.2, and the coefficient of static friction at the screw is ms œ = C mm Referring to the free-body diagram of wedge C shown in Fig. a, we have + c F y = 0; 2N cos N sin = 0 N = N Using the result of N and referring to the free-body diagram of wedge shown in Fig. b, we have + c F y = 0; N cos ( ) sin 15 = 0 N =6000 N : + F x = 0; T sin ( ) cos (6000) = 0 T = N Since the screw is being tightened, Eq. 8 3 should be used. Here, u = tan - 1 L c 2pr d = 7.5 tan-1 c 2p(12.5) d = ; f s = tan -1 m s = tan -1 (0.15) = ; M = P(0.25); and W = T = N. Since M must overcome the friction of two screws, M = 23Wr tan (f s + u)4 P(0.25) = (0.0125) tan ( )4 P = 104 N

7 8 91. cable is attached to the plate of mass M, passes over a fixed peg at C, and is attached to the block at. Using the coefficients of static friction shown, determine the smallest mass of block so that it will prevent sliding motion of down the plane. Given: M 20 kg deg 0.3 g 9.81 m s 2 C 0.3 Solution: Iniitial guesses: T 1 1N T 2 1N N 1N N 1N M 1kg Given lock : Plate : F x = 0; T 1 N M g sin 0 F y = 0; N M g cos 0 F x = 0; T 2 M g sin N N 0 F y = 0; N N M g cos 0 Peg C: T 2 T 1 e C T 1 T 2 N N M Find T 1 T 2 N N M M 2.22 kg

8 The 20-kg motor has a center of gravity at G and is pinconnected at C to maintain a tension in the drive belt. Determine the smallest counterclockwise twist or torque M that must be supplied by the motor to turn the disk if wheel locks and causes the belt to slip over the disk. No slipping occurs at. The coefficient of static friction between the belt and the disk is m s = mm C G M 50 mm 150 mm 100 mm Equations of Equilibrium: From FD (a), a + M C = 0; T T = 0 (1) From FD (b), a + M O = 0; M + T T = 0 (2) Frictional Force on Flat elt: Here, b = 180 = p rad. pplying Eq. 8 6, T 2 = T 1 e mb, we have T 2 = T 1 e 0.3p = 3.003T 1 Solving Eqs. (1), (2), and (3) yields M = 3.93 N m # (3) T 1 = N T 2 = N

8 103. locks and have a mass of 100 kg and 150 kg, respectively. If the coefficient of static friction between and and between and C is m s = 0.25 and between the ropes and the pegs D and E m s = 0.

9 locks and have a mass of 100 kg and 150 kg, respectively. If the coefficient of static friction between and and between and C is m s = 0.25 and between the ropes and the pegs D and E m s = 0.5 determine the smallest force F needed to cause motion of block if P = 30 N. D 45 E P C F ssume no slipping between and. Peg D : T 2 = T 1 e mb ; F D = 30 e 0.5(p 2 ) = N lock : : + F x = 0; + c F y = 0; N C + F E cos 45 = 0 N C F E sin (9.81) = 0 F E = N N C = N Peg E : T 2 = T 1 e mb ; F = 768.1e 0.5(3p 4 ) = 2.49 kn Note: Since moves to the right, (F ) max = 0.25 (981) = N = P max e 0.5(p 2 ) P max = 112 N 7 30 N Hence, no slipping occurs between and as originally assumed.

10 4000 N. If µ s = 0.35, mm 50 mm P = 4000 N = 3 (0.35)(4000) mm = Nmm M = 54.4 N m

11 mm diameter post is driven 3 m into sand for which m s = 0.3. If the normal pressure acting completely around the post varies linearly with depth as shown, determine the frictional torque M that must be overcome to rotate the post. M 200 mm 3m Equations of Equilibrium and Friction: The resultant normal force on the post is N = p = 180p N. 2 F = m s N = p2 = 54.0p N. Since the post is on the verge of rotating, 600 Pa a+ M O = 0; M p10.12 = 0 M = 17.0 N # m

12 The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder s top and bottom surfaces are a and a, respectively, show that a horizontal force having a magnitude of P = [W(a + a )]>2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder. : + F x = 0; (R ) x - P = 0 (R ) x = P + c F y = 0; (R ) y - W = 0 (R ) y = W a + M = 0; P(r cos f + r cos f ) - W(a + a ) = 0 (1) f f Since and are very small, cos f - cos f = 1. Hence, from Eq. (1) P W r P = W(a + a ) 2r (QED)

SOLUTION 8 1. a+ M B = 0; N A = 0. N A = kn = 16.5 kn. Ans. + c F y = 0; N B = 0

SOLUTION 8 1. a+ M B = 0; N A = 0. N A = kn = 16.5 kn. Ans. + c F y = 0; N B = 0 8 1. The mine car and its contents have a total mass of 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and the tracks is m s = 0.4 when the wheels are locked,