Question

7 mark for review a block of mass $m_0$ is at rest between two vertical surfaces, each of which exerts a constant normal force $n_0$ on the block. the coefficient of static friction between each vertical surface and the block is $mu_s$. a second block of mass $m$ is then placed on top of the original block as shown in the figure. what is the largest value $m$ can have before the block of mass $m_0$ begins to slide downward? a $\frac{mu_sn_0}{g}-m_0$ b $\frac{2mu_sn_0}{g}-m_0$ c $\frac{mu_sn_0}{g}+m_0$ d $\frac{2mu_sn_0}{g}+m_0$

Explanation:

Step1: Analyze the maximum static - friction force

The maximum static - friction force on each side of the block of mass $M_0$ is $f_s=\mu_sN_0$. Since there are two vertical surfaces, the total maximum static - friction force acting on the block of mass $M_0$ is $F_{total\ f_s}=2\mu_sN_0$.

Step2: Set up the equilibrium equation

In the vertical direction, the total maximum static - friction force must balance the total weight of the two - block system when the block of mass $M_0$ is on the verge of sliding. Let the acceleration due to gravity be $g$. The total weight of the two - block system is $(M_0 + m)g$. At the verge of sliding, $2\mu_sN_0=(M_0 + m)g$.

Step3: Solve for $m$

Rearrange the equation $2\mu_sN_0=(M_0 + m)g$ to solve for $m$. First, expand the right - hand side: $2\mu_sN_0 = M_0g+mg$. Then, isolate $m$: $mg=2\mu_sN_0 - M_0g$. So, $m=\frac{2\mu_sN_0}{g}-M_0$.

Answer:

B. $\frac{2\mu_sN_0}{g}-M_0$