Question

Friction coefficient of friction support force, sliding or static coefficient Three horizontally stacked blocks was pushed with 100 N. The first block was 10 kg, the second block was 6 kg and the third block was 4 kg. The floor sliding coefficient of friction was 0.2. Find the acceleration and the internal forces at the block edges.

Answer 1

Frictional force f = mu (N) (Therefore N = m_1 + m_2 + m_3) g = 20 times 9.8 Newton Therefore f = 0.2 (196) f = 39.2 newton Therefore acceleration a = f - f/(m_1 + m_2 + m-3) = 100 - 39.2/20 = 3.04 m/s^2 normal force b/w 6 kg & 4 kg (Therefore acceleration is some for all bodies, they move together) N_1 - f_1 = m_3 a Therefore N_1 = (4 times 3.04) + f_1 = N_1 = 20 newton's (f_1 = mu m_3 g) = 0.2 times 4 times 9.8 = 7.84 newton f_2 = mu times 6 times 9.8 = 11.76 newton N_2 = f_2 + N_1 + 6a N_1 = 11.76 + 20 + (6 times 3.04) N_2 = 50 newton