Two identical weights of mass M are linked by | vedclass

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(A) Let the acceleration of the system be $a$. The total mass of the system is $(2M + m)$. The net driving force is the weight of the extra mass $m$,w

Vedclass · Accepted Answer

$\frac{2mMg}{2M + m}$

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(A) Let the acceleration of the system be $a$. The total mass of the system is $(2M + m)$. The net driving force is the weight of the extra mass $m$,which is $mg$.Using Newton's second law for the whole system: $mg = (2M + m)a$,so $a = \frac{mg}{2M + m}$.Now,consider the free body diagram of the small mass $m$. It is moving downwards with acceleration $a$. The forces acting on it are its weight $mg$ (downwards) and the normal reaction force $N$ (upwards) exerted by the mass $M$.Applying Newton's second law to mass $m$: $mg - N = ma$.Rearranging for $N$: $N = mg - ma$.Substituting the value of $a$: $N = mg - m\left(\frac{mg}{2M + m}\right)$.$N = mg \left(1 - \frac{m}{2M + m}\right) = mg \left(\frac{2M + m - m}{2M + m}\right)$.$N = \frac{2Mm g}{2M + m}$.

Two identical weights of mass $M$ are linked by a thread wrapped around a frictionless pulley with a fixed axis. $A$ small weight of mass $m$ is placed on one of the weights. What is the reaction force between $m$ and $M$?

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