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?

**Solution: **

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

F=\mu_1mg+\mu_2mg+Ma$$

Now, \(a=\mu_1 g\)

Hence, $$ F=(m+M)(\mu_1+\mu_2)g$$