A pen of mass ‘m’ is lying on a piece of paper of mass M placed on a rough table. If the cofficient of friction between the pen and paper and, the paper and the table are (\mu_1) and (\mu_2) respectively, what is the minimum horizontal force with which the paper has to be pulled for the pen to start slipping?


For pen to start slipping, maximum horizontal acceleration for pen and paper to start slipping is (f=\mu_1)g
Therefore, (a=\mu_1 g) is the common acceleration for pen and paper.
If (f_1) and (f_2) be the frictional forces for pen and paper respectively, the net force for the system
$$ F=f_1+f_2+Ma
Now, (a=\mu_1 g)
Hence, $$ F=(m+M)(\mu_1+\mu_2)g$$