Friction: Tricky Questions Solved (2)

Question

A block of mass $m_1 = 4$ kg rests on top of a block of mass $m_2 = 6$ kg, which rests on a frictionless horizontal floor. Coefficient of static friction between the blocks is $\mu_s = 0.4$. A horizontal force $F$ is applied to the lower block ($m_2$). Find the maximum value of $F$ such that the two blocks move together. Take $g = 10$ m/s².

Solution — Step by Step

Why This Works

The key insight: friction is what drags the upper block along. When you push the lower block, the only thing accelerating the upper block is static friction at the contact surface. If the required friction exceeds $\mu_s m_1 g$, the surface gives up and the upper block slips behind.

We compute $a_{\text{max}}$ from the upper block’s perspective, then apply it to the whole system. Don’t apply Newton’s law to the lower block alone first — the friction on it equals the friction on the upper block (third-law pair), but in the opposite direction, and that complicates the algebra unnecessarily.

Alternative Method

Treat the upper block alone. Required force on it = $m_1 a$. Maximum is $\mu_s m_1 g$, giving $a_{\max} = \mu_s g = 4$ m/s². Then $F_{\max} = (m_1+m_2) a_{\max} = 40$ N. Notice $a_{\max}$ doesn’t depend on $m_1$ at all — only on $\mu_s$ and $g$.

Common Mistake