E PROBLEM 8.6 C θ D T B A The 20-lb block A hangs from a cable as shown. Pulley C is connected by a short link to block E, which rests on a horizontal rail. Knowing that the coefficient of static friction between block E and the rail is ms = 0.35 and neglecting the weight of block E and the friction in the pulleys, determine the maximum allowable value of θ if the system is to remain in equilibrium. 20 lb SOLUTION Free-body diagrams: ms = 0.35 fs = tan-1 0.35 = 19.29 Force triangle for pulley C: Law of sines: sin a sin fs = T 2T sin a = 2sin fs = 2 sin19.29o a = 41.35o q = fs + a = 19.29 + 41.35 q = 60.6o ◀ Note: Answer is independent of WA . Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1241 PROBLEM 8.11 A B The 50-lb block A and the 25-lb block B are supported by an incline that is held in the position shown. Knowing that the coefficient of static friction is 0.15 between the two blocks and zero between block B and the incline, determine the value of q for which motion is impending. θ SOLUTION Since motion impends, F = ms N between A + B Free body: Block A Impending motion: SFy = 0: N1 - 50 cos q = 0 N1 = 50 cos q SFx = 0: T - 50 sin q + m1 N1 = 0 T = 50 sin q - m1 (50)cos q (1) Free body: Block B Impending motion: SFy = 0: N 2 - N1 - 25 cos q = 0 N 2 = 75cos q SFx : T - m1N1 - m2 N 2 - 25sin q T = m1 (50) cos q + m2 (75) cos q + 25sin q Eq. (1) - Eq. (2): (2) 0 = 25sin q - m1 (100) cos q - m2 (75) cos q = 0 Substituting in for m1 = 0.15, m2 = 0, we have: 15cos q = 25sin q : tan q = 15 ; 25 q = 31.0 ◀ Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1249 500 N 45 mm 90 mm PROBLEM 8.32 45 mm 90 mm B A C A 500-N concrete block is to be lifted by the pair of tongs shown. Determine the smallest allowable value of the coefficient of static friction between the block and the tongs at F and G. 75 mm D 105 mm E 360 mm 315 mm F 500 N G SOLUTION Free body: Members CA, AB, BD 1 C y = Dy = (500) = 250 N 2 By symmetry: Since CA is a two-force member, Cy Cx 250 N = = 90 mm 75 mm 75 mm C x = 300 N SFx = 0: Dx = C x Dx = 300 N Free body: Tong DEF SM E = 0: (300 N)(105 mm) + (250 N)(135 mm) + (250 N)(157.5 mm) - Fx (360 mm) = 0 Fx = + 290.625 N Minimum value of ms : ms = Fy Fx = 250 N 290.625 N ms = 0.860 ◀ Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1275 y PROBLEM 5.8 Locate the centroid of the plane area shown. r = 38 in. 16 in. x 20 in. SOLUTION Then A, in 2 x , in. y , in. xA, in3 yA, in3 1 p (38)2 = 2268.2 2 0 16.1277 0 36,581 2 - 20 ´ 16 = 320 3200 -2560 S 1948.23 3200 34,021 X= Y= S xA SA S yA SA -10 8 = 3200 in 3 1948.23 in 2 X = 1.643 in. ◀ = 34,021 in3 1948.23 in 2 Y = 17.46 in. ◀ Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 566