Answer to Question #96025 in Classical Mechanics for Jake

Question

2. A wood block of mass rests on a larger wood block of mass that rests on a wooden table. What is the minimum horizontal force, F, applied to the lower block that will cause it to slide out from under the upper block?

Answer 1

Denote the upper-block mass by $m$, the lower-block mass by $M$, let $\mu_s$ and $\mu_d$ denote, respectively, the coefficients of static and dynamic friction between two wooden surfaces, and let $g$ denote the acceleration of gravity. In order that the lower block start sliding over the table, the horizontal force $F$ must exceed the maximal static friction force between this block and the table, which is $(M + m) g \mu_s$, so one of the conditions is $F > (M + m) g \mu_s$. The two blocks will move without mutual sliding if the force accelerating the upper block does not exceed the maximal static friction force between the blocks, the condition of which is $m a \leq m g \mu_s$, or $a \leq g \mu_s$, where $a$ is the acceleration. The acceleration in this case is found from the equation $(M + m) a = F - (M + m) g \mu_d$, whence $a = F/(M + m) - g \mu_d$. The previous condition $a \leq g \mu_s$ then becomes $F/(M + m) - g \mu_d \leq g \mu_s$, or $F \leq (M + m) g ( \mu_s + \mu_d)$. If this inequality is violated, the upper block will be sliding over the lower one. Hence, the answer is that the force has to exceed the value $F_{\rm min} = (M + m) g ( \mu_s + \mu_d)$.