Question

Two blocks of masses m A 1.2 kg and mp 4.5 kg can slide one relative to one another, depending on the magnitude of the applied force P. The coefficient of friction between blocks A and B as well as between block B and the horizontal surface is 0.25. Case a: assume that the system is just on the verge of motion due to force P and calculate: • The frictional force FA/B = between A and B, N N • The friction force FB = between B and the horizontal surface, • The tension in the cable T = N and it [N] Case b: a new force of magnitude (P = 1) 22 ( NV is applied with and the coefficient of friction is 0.18. Assess the system in the new condition and calculate, • The frictional force FA/B = N between A and B N The friction force FB = between B and the horizontal surface,

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Answer 1

Case-a) A B verge of Since the system is Just on the ver motion. .. FA/B = 0.25% Maxg şg=9.8 m/s² 125 x 1•2 x gos ]N = 2.94N FA/B = 20 FB = 0.25%(MA+mg) xg = 0.25X5•7x978 = 13.965N Tension in the cable=13-975NON Tension in the table = 5 2094 A A/B Tension in the cable = ON 2.94 N.

Now case b) FA/B = (0.18 x 1•2x 9:8)N = 2.1168N FB = (0.18% 5-7x908) N = 10.0548 N acceleration system :9=122- 10.0548 of the 2 5:7 S és a = 2.0956 m/s? Tenstion in the cable = T = 2.1168+ j .Rx a .. 1= 463 N speed of block B after 0.06 sec ath = V = 0 + ax 0.06 = 200956 x0.06 = 0.125%