Explanation Given information : The weight W A of block A is 600 lb. The weight W B of block B is 150 lb. The weight W C of block C is 500 lb. The coefficient of static friction μ s between the blocks is 0.30. The coefficient of static friction μ s ′ between the floor and each block is 0.50. Assumption : Consider that block B slips up and block A does not move. Show the free body diagram of the whole blocks as in Figure (1). Use Figure (1). Along the horizontal direction : Apply the Equation of equilibrium in the x axis. ∑ F x = 0 − P + F = 0 − P + μ s ′ N = 0 P = μ s ′ N (I) Along the vertical direction : Apply the Equation of equilibrium in the y axis. ∑ F y = 0 N − ( W A + W B + W C ) = 0 N = W A + W B + W C (II) Consider block A : Show the free body diagram of block A as in Figure (2). Use Figure (2). Along the horizontal direction : Apply the Equation of equilibrium in the x axis. ∑ F x = 0 F A − N 1 = 0 F A = N 1 (III) Along the vertical direction : Apply the Equation of equilibrium in the y axis. ∑ F y = 0 N A − W A + F 1 = 0 N A + μ s N 1 = W A (IV) Consider block B : Show the free body diagram of block B as in Figure (3). Use Figure (3). Since there is no contact, the value of N B and F B are Equals to zero. Use Figure (3). Along the horizontal direction : Apply the Equation of equilibrium in the x axis. ∑ F x = 0 N 1 − N 2 cos 45 ° − F 2 sin 45 ° = 0 N 1 − N 2 cos 45 ° − μ s N 2 sin 45 ° = 0 (V) Along the vertical direction : Apply the Equation of equilibrium in the y axis. ∑ F y = 0 − F 1 − F 2 cos 45 ° + N 2 sin 45 ° − W B = 0 − μ s N 1 − μ s N 2 cos 45 ° + N 2 sin 45 ° = W B (VI) Consider block C : Show the free body diagram of block C as in Figure (4). Use Figure (4). Apply the Equation of equilibrium in the x axis. ∑ F x = 0 F C + N 2 cos 45 ° + F 2 cos 45 ° − P = 0 μ s ′ N C + N 2 cos 45 ° + μ s N 2 cos 45 ° − P = 0 (VII) Along the vertical direction : Apply the Equation of equilibrium in the y axis. ∑ F y = 0 N C + F 2 sin 45 ° − N 2 sin 45 ° − W C = 0 N C + μ s N 2 sin 45 ° − N 2 sin 45 ° = W C (VIII) Conclusion : Substitute 0.50 for μ s ′ in Equation (I). P = 0.50 N (IX) Substitute 600 lb for W A , 150 lb for W B , and 500 lb for W C in Equation (II). N = 600 + 150 + 500 = 1 , 250 lb Substitute 1,250 lb for N in Equation (IX)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 8.8, Problem 5RP

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The smallest horizontal force P that must be applied to block C in order to move this block.

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Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ... title('Velocity Triangle'), ... xlabel('x'),ylabel('y'), grid

Chapter 8 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The smallest horizontal force P that must be applied to block C in order to move this block.

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