The coefficient of friction between blocks A | vedclass

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(C) Let $a$ be the common acceleration of the system consisting of blocks $A$ and $B$.The total force applied is $F$,and the total mass is $(M+m)$.Usi

Vedclass · Accepted Answer

$\frac{(M+m) g}{\mu}$

Vedclass · Answer

(C) Let $a$ be the common acceleration of the system consisting of blocks $A$ and $B$.The total force applied is $F$,and the total mass is $(M+m)$.Using Newton's second law for the whole system: $F = (M+m)a$,which gives $a = \frac{F}{M+m}$.Now,consider the free body diagram of block $B$ of mass $m$.For block $B$ not to slip down,the upward frictional force $f$ must balance the downward gravitational force $mg$.So,$f = mg$.The frictional force is given by $f = \mu N$,where $N$ is the normal force exerted by block $A$ on block $B$.The normal force $N$ provides the acceleration $a$ to block $B$,so $N = ma$.Substituting $a$: $N = m \left( \frac{F}{M+m} \right)$.Now,equate the frictional force to the weight: $\mu N = mg$.$\mu \left( m \frac{F}{M+m} \right) = mg$.Solving for $F$: $F = \frac{(M+m)g}{\mu}$.Thus,the correct option is $C$.

The coefficient of friction between blocks $A$ and $B$ is $\mu$. The minimum force $F$ with which $A$ must be pushed such that $B$ will not slip down is

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