Three blocks of equal mass M are arranged as shown on a smooth floor. The lower contact between blocks 1 and 2 is smooth, while friction is available between blocks 2 and 3. A horizontal force F is applied so that all three blocks move together. The minimum coefficient of friction required is:

Question

G1 Physics · Accepted Answer

All three masses have common acceleration a = F/(3M). The top block of mass M is accelerated only by friction, so required friction is Ma = F/3. Since N = Mg at that interface, μₘᵢₙ = (F/3)/(Mg) = F/(3Mg).