Three masses M = 100 , kg,m_{1} = 10 , kg,and | vedclass

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(C) Let the acceleration of the $100 \, kg$ block be $a_{1}$ towards the right.For the $m_{2} = 20 \, kg$ block,it is moving upward with acceleration

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$3120$

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(C) Let the acceleration of the $100 \, kg$ block be $a_{1}$ towards the right.For the $m_{2} = 20 \, kg$ block,it is moving upward with acceleration $a_{y} = 2 \, m/s^{2}$ relative to the $100 \, kg$ block. The tension $T$ in the string acts upward,and gravity $20g$ acts downward. The pseudo force $20a_{1}$ acts horizontally to the left.Applying Newton's second law in the vertical direction for $m_{2}$:$T - m_{2}g = m_{2}a_{y}$$T - 20(10) = 20(2)$$T - 200 = 40 \Rightarrow T = 240 \, N$.Now,consider the $m_{1} = 10 \, kg$ block on top of the $100 \, kg$ block. It is connected to the $20 \, kg$ block via the string. The horizontal acceleration of $m_{1}$ relative to the $100 \, kg$ block is $a_{x} = 2 \, m/s^{2}$ (since the string is inextensible,the horizontal acceleration of $m_{1}$ must match the vertical acceleration of $m_{2}$).Applying Newton's second law for $m_{1}$ in the horizontal direction:$T = m_{1}a_{1} \Rightarrow 240 = 10a_{1} \Rightarrow a_{1} = 24 \, m/s^{2}$.Finally,consider the whole system $(M + m_{1} + m_{2})$ moving with acceleration $a_{1}$:$F = (M + m_{1} + m_{2})a_{1}$$F = (100 + 10 + 20) 	imes 24$$F = 130 	imes 24 = 3120 \, N$.

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