A pen of mass m is lying on a piece of paper | vedclass

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Question

(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

Vedclass · Accepted Answer

$(m+M)\left(\mu_1+\mu_2\right) g$

Vedclass · Answer

(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,

$F_{	ext {net }} =M a_{\min }=F-f_1-f_2$

$\Rightarrow \quad F= M a_{\min }+f_1+f_2$

$= M \mu_1 g+\mu_1 m g+\mu_2(m+M) g$

$=(M+m)\left(\mu_1+\mu_2ight) g$

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