A person stands on a platform-and-pulley system, as shown in figure. The masses of the platform and person are M and m respectively. The rope and pulley is massless. The man can pull on the rope so as to produce one of the situations shown in column I. In column II, T is tension & N is normal reaction between the man and the platform. Match the possible situations. Column I Column II (A) At equilibrium (P) N = (2m + M) g (B) When the system falls down freely (Q) T = (m + M) g (C) When accelerating upwards (R) N < \frac{mg}{2} (D) When accelerating downwards (S) N > (2m + M)g (T) T

Engineering

Physics

Newtons First Law of Motion

Question

A person stands on a platform-and-pulley system, as shown in figure. The masses of the platform and person are M and m respectively. The rope and pulley is massless. The man can pull on the rope so as to produce one of the situations shown in column I. In column II, T is tension & N is normal reaction between the man and the platform. Match the possible situations.

Column I	Column II
(A) At equilibrium	(P) N = (2m + M) g
(B) When the system falls down freely	(Q) T = (m + M) g
(C) When accelerating upwards	(R) $N < \frac{mg}{2}$
(D) When accelerating downwards	(S) N > (2m + M)g
	(T) T < (m + M)g

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$\frac{\begin{array}{l} 2 T = N + Mg \\ N = T + mg \end{array}}{T = (M + m) g} : equ .$

N < 0 ⇒ 2T – Mg < 0 T < $\frac{Mg}{2}$ : Break off.

(C) $\frac{\begin{array}{l} 2 T - N - Mg = ma \\ N - T - mg = ma \end{array}}{\begin{array}{l} T = (m + M) (g + a) \\ N = (2 m + M) (g + a) \end{array}}$

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