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

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Question

(A) The limiting friction between the pen and the paper is $f_1 = \mu_1 mg$.For the pen to start slipping,the force acting on the pen must be at least

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The limiting friction between the pen and the paper is $f_1 = \mu_1 mg$.For the pen to start slipping,the force acting on the pen must be at least $f_1$. This force is provided by the friction from the paper,which is equal to the force of friction on the pen.Thus,the acceleration $a$ of the pen is $a = \frac{f_1}{m} = \mu_1 g$.For the pen to slip,the paper must accelerate at least with this acceleration $a = \mu_1 g$.Now,consider the free body diagram of the paper of mass $M$. The forces acting on the paper are the applied force $F$,the friction from the pen $f_1$ (acting backwards),and the friction from the table $f_2$ (acting backwards).The normal force on the table is $N = (M+m)g$,so $f_2 = \mu_2(M+m)g$.Applying Newton's second law to the paper:$F - f_1 - f_2 = Ma$$F = Ma + f_1 + f_2$Substituting the values:$F = M(\mu_1 g) + \mu_1 mg + \mu_2(M+m)g$$F = (M+m)\mu_1 g + (M+m)\mu_2 g$$F = (M+m)(\mu_1 + \mu_2)g$.

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