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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 coefficients of friction between the pen and paper and the paper and the table are $\mu_1$ and $\mu_2$, respectively. Then, the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by
$(m+M)\left(\mu_1+\mu_2\right) g$
$\left(m \mu_1+M \mu_2\right) g$
$\left(m \mu_1+(m+M) \mu_2\right) g$
$m\left(\mu_1-\mu_2\right) g$
Solution
(a)
Limiting value of friction on pen is
$f_1=\mu_1 m g$
So, for pen to start slipping, force acting over pen is
$F_1 \geq f_{1,}=\mu_1 m g$
So, minimum acceleration of pen is
$a_{\min }=\frac{F_1}{m}=\mu_1 g$
Hence, minimum acceleration of paper is $a_{\min }=\mu_1 g$
Now, consider free body diagram of paper
We have,
$F_{\text {net }} =M a_{\min }=F-f_1-f_2$
$\Rightarrow \quad F= M a_{\min }+f_1+f_2$
$= M \mu_1 g+\mu_1 m g+\mu_2(m+M) g$
$=(M+m)\left(\mu_1+\mu_2\right) g$
