In the figure block A mass 145 kg is in equilibrium but it w

In the figure, block A (mass 14.5 kg) is in equilibrium, but it would slip if block B (mass 7.34 kg) were any heavier. For angle ? = 37.5°, what is the coefficient of static friction between block A and the surface below it?

Solution

T_B = m_B g = 7.34 * 9.8 = 71.93 N

Solving the two balance of forces equation for the joint simultaneously

T = T_B/ cos theta

T_A= T sin theta = (T_B/ cos theta ) sin theta = T_B tan theta

T_A = 71.93tan 37.5 = 55.17 N

N = m_A g = 14.5 * 9.8 = 142.1 N

f = T_A = 55.17 N

By definition, the coefficient of static friction is the ratio of the maximum static friction to the normal force

mu_s = f / N = 0.388