If the system is in equilibrium,what should b | vedclass

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(B) Consider the system consisting of the man of mass $m$ and the platform of mass $M$. The total downward force acting on the system is the weight of

Vedclass · Accepted Answer

$\frac{1}{2}(M + m)g$

Vedclass · Answer

(B) Consider the system consisting of the man of mass $m$ and the platform of mass $M$. The total downward force acting on the system is the weight of the man and the platform,which is $(M + m)g$. The rope is connected to the platform and is also held by the man. Since the rope passes over a fixed pulley,the tension $T$ acts upwards on the platform and also upwards on the man's hand. Thus,there are two segments of the rope pulling the system upwards,each with tension $T$. For the system to be in equilibrium,the total upward force must equal the total downward force: $2T = (M + m)g$ Therefore,the tension in the rope is: $T = \frac{(M + m)g}{2}$

If the system is in equilibrium,what should be the tension in the rope?

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