Which Block Starts Sliding First?
Two identical blocks share the same friction limit. The answer therefore depends only on which block receives the larger horizontal normal reaction from the rod.
Identify the two contact forces
Let \(R_A\) and \(R_B\) denote the magnitudes of the horizontal normal forces exerted by the rod on blocks A and B, respectively.
Compare the normal reactions
The force is increased gradually, so before any slipping begins the rod is in horizontal equilibrium. Hence,
Since \(F>0\), the normal reaction on block B is necessarily greater than the normal reaction on block A.
Compare with limiting friction
The contacts with the rod are horizontal, so the vertical normal reaction from the floor is \(mg\) for each block. Because the blocks and their contact with the floor are identical, their maximum static frictions are equal.
Required friction: \(f_A=R_A\)
Maximum available: \(f_{A,\max}=\mu_s mg\)
Required friction: \(f_B=R_B\)
Maximum available: \(f_{B,\max}=\mu_s mg\)
Both blocks have the same friction ceiling, but block B has the larger friction demand. Therefore B reaches limiting friction first as \(F\) increases.
\(R_B=\mu_s mg\), whereas \(R_A=R_B-F<\mu_s mg\). Thus A can still remain at rest when B is on the verge of sliding.