Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 186: 102

$F_{min}=\mu_sg(2m+M)$

We can find the minimum horizontal force as follows: $\Sigma F_y=-mg-Mg+N=0$ $\implies N=(m+M)g$ Similarly $\Sigma F_x=-\mu_smg+F_{min}-\mu_sN=0$ $\implies F_{min}=\mu_s(mg+N)$ $\implies F_{min}=\mu_s(mg+(m+M)g)$ $\implies F_{min}=\mu_sg(2m+M)$