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