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(C) Let the tension in the string be $T$. For the particle of mass $m$,the equation of motion is $T - mg = ma$. For the heavy body of mass $M$,the equ

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$4\,mg$

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(C) Let the tension in the string be $T$. For the particle of mass $m$,the equation of motion is $T - mg = ma$. For the heavy body of mass $M$,the equation of motion is $Mg - T = Ma$. Adding these two equations,we get $(M - m)g = (M + m)a$,which gives $a = \left(\frac{M - m}{M + m}\right)g$. Substituting $a$ into the first equation: $T = m(g + a) = m\left(g + \frac{M - m}{M + m}g\right) = m\left(\frac{Mg + mg + Mg - mg}{M + m}\right)g = \frac{2Mmg}{M + m}$. Since the body is very heavy,$M >> m$,so $T \approx \frac{2Mmg}{M} = 2mg$. The total downward force on the pulley is the sum of the tensions in both segments of the string,which is $2T$. Therefore,$2T = 2(2mg) = 4mg$.

$A$ particle of small mass $m$ is joined to a very heavy body of mass $M$ by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley is

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