Q.A block of mass $2\ \text{kg}$ is placed on top of a block of mass $3\ \text{kg}$, which itself rests on a frictionless floor. The coefficient of static friction between the two blocks is $0.3$. Find the maximum horizontal force that can be applied to the lower block so that the two blocks move together, without the upper block slipping. (Take $g = 9.8\ \text{m/s}^2$.)

Given: $m_1 = 2\ \text{kg}$ (top block), $m_2 = 3\ \text{kg}$ (bottom block, on a frictionless floor), $\mu = 0.3$ between the two blocks, $g = 9.8\ \text{m/s}^2$.\n\nThe only horizontal force that can act on the top block is friction from the bottom block beneath it (there is nothing else touching it horizontally), so the maximum acceleration the top block can be given, without slipping relative to the bottom block, is limited by the maximum available friction between them: $f_{\max} = \mu m_1 g = 0.3 \times 2 \times 9.8 = 5.88\ \text{N}$ $a_{\max} = \frac{f_{\max}}{m_1} = \frac{5.88}{2} = 2.94\ \text{m/s}^2$\n\nFor the two blocks to move together, without slipping, at this maximum common acceler …

Find the maximum acceleration the top block can have from the maximum friction it can receive, $a_{\max} = \mu g$, then apply that same a …

• Using the total normal reaction on the floor ($(m_1+m_2)g$) instead of just the top block's weight when computing the maximum friction between the two blocks. …