Two blocks of masses M_1 and M_2 are connecte | vedclass

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(B) For the system to be in equilibrium:For the block of mass $M_1$ with additional mass $m$ on it:The normal force $N = (M_1 + m)g$.The limiting fric

Vedclass · Accepted Answer

$\frac{M_2}{\mu} - M_1$

Vedclass · Answer

(B) For the system to be in equilibrium:For the block of mass $M_1$ with additional mass $m$ on it:The normal force $N = (M_1 + m)g$.The limiting frictional force $f = \mu N = \mu(M_1 + m)g$.For equilibrium,the tension $T$ in the string must be equal to the frictional force:$T = \mu(M_1 + m)g$.For the block of mass $M_2$ hanging vertically:For equilibrium,the tension $T$ must be equal to the weight of the block:$T = M_2 g$.Equating the two expressions for tension:$\mu(M_1 + m)g = M_2 g$.Dividing both sides by $\mu g$:$M_1 + m = \frac{M_2}{\mu}$.Therefore,the additional mass $m$ required is:$m = \frac{M_2}{\mu} - M_1$.

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