A constant force F is applied in the horizont | vedclass

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(B) Let the acceleration of the system be $a$. The total mass is $(M + m + M')$.Using Newton's second law,$F = (M + m + M')a$,so $a = \frac{F}{M + m +

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$N > N'$

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(B) Let the acceleration of the system be $a$. The total mass is $(M + m + M')$.Using Newton's second law,$F = (M + m + M')a$,so $a = \frac{F}{M + m + M'}$.Considering the free body diagram of $M'$:The only horizontal force acting on $M'$ is the contact force $N'$ from $m$.Thus,$N' = M' a = \frac{M' F}{M + m + M'}$.Considering the free body diagram of $M$:The forces acting on $M$ are the applied force $F$ and the contact force $N$ from $m$ in the opposite direction.Thus,$F - N = M a$,which gives $N = F - M a = F - \frac{M F}{M + m + M'} = F \left(1 - \frac{M}{M + m + M'}\right) = F \left(\frac{m + M'}{M + m + M'}\right)$.Comparing $N$ and $N'$:$N = \frac{(m + M') F}{M + m + M'}$ and $N' = \frac{M' F}{M + m + M'}$.Since $m > 0$,it follows that $(m + M') > M'$,therefore $N > N'$.

$A$ constant force $F$ is applied in the horizontal direction as shown. If the contact force between $M$ and $m$ is $N$ and the contact force between $m$ and $M'$ is $N'$,then:

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