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