Question

Please solve questions one and two with clear explanation and clear hand writing

Thank you

1. For the two blocks below, draw free body diagrams for each (assume the larger box is on the floor). What force (F) must be applied to have the smaller not slide down; the coefficient of static friction is 0.6 0.5 Kg 10 Kg 2. Draw a free body diagram for each block in the system. Find the acceleration of the system and the tensions in the rope. The coefficient of kinetic friction is 0.3 and exists between the two blocks. There is no friction between the larger block and the table 2 Kg 15 N

Answer 1

acceleration of both blocks = a = F/10.5 so f = N_2 = 0.5a = 0.5 times F/10.5 Therefore frictional force on 0.5 kg block in upward direction = f_r = mu N_2 = 0.6 times 0.5 F/10.5 the smaller block will not slide down if f_r = mg 0.3 F/10.5 = 0.5 times 9.8 OR F = 0.5 times 9.8 times 10.5/0.3 = 171.5 N so F = 171.5 N. Free Body diagram is shown in figure Normal force on 2 kg block N = 2 times 9.8 = 19.6 so friction = f_1 = 0.3 times 19.6 = 5.88 N so 15 - T - f_1 = 3a equation (1) and T - f_1 = 2a equation (2) By adding (1) and (2) rightarrow 15 - 2f_1 = 5a OR 15 - 2 times 5.88 = 5a implies a = 0.648 m/s^2 so Tension = T = 5.88 + 2 times 0.648 = 7.176 N