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(B) Let $T$ be the tension in the rope. Since the man pulls both sides of the rope,there are $4$ segments of the rope pulling the system (man + platfo

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$300$

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(B) Let $T$ be the tension in the rope. Since the man pulls both sides of the rope,there are $4$ segments of the rope pulling the system (man + platform) upwards.Total mass of the system $M = m_{man} + m_{platform} = 75 + 25 = 100\,kg$.The equation of motion for the system is: $4T - Mg = Ma$.$4T - 100 	imes 10 = 100 	imes 2$.$4T - 1000 = 200 \implies 4T = 1200 \implies T = 300\,N$.Now,consider the free body diagram of the man. The forces acting on the man are: tension $2T$ from the two ropes he is holding,the normal reaction $N$ from the weighing machine,and his weight $m_{man}g$ acting downwards.Equation of motion for the man: $2T + N - m_{man}g = m_{man}a$.$2(300) + N - 75(10) = 75(2)$.$600 + N - 750 = 150$.$N - 150 = 150$.$N = 300\,N$.

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