Two blocks of mass $10 k g$ and $30 k g$ are placed along a vertical line. The first block is raised through a height of $7 c m$ .By what distance should the second mass be moved to raise the center of mass by $1 c m ?$

2 c m

upward

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1 c m

upward

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2 c m

downward

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1 c m

downward

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Solution

The correct option is D $1 c m$ downward
Center of mass= $\frac{(m_{1} y_{1} + m_{2} y_{2})}{(m_{1} + m_{2})}$
$\begin{matrix} 1 = \frac{(10 \times 7 + 30 y_{2})}{(10 + 30)} 1 = \frac{(70 + 30 y_{2})}{(40)} 40 = 70 + 30 y_{2} 30 y_{2} = - 30 y_{2} = - 1 \end{matrix}$
So, $m_{2} = 30$ kg body should be displaced $1$ cm downward.

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Two blocks of mass 10kg and 30kg are placed along a vertical line. The first block is raised through a height of 7cm.By what distance should the second mass be moved to raise the center of mass by 1cm?

The correct option is D 1cm downwardCenter of mass=(m1y1+m2y2)(m1+m2)1=(10×7+30y2)(10+30)1=(70+30y2)(40)40=70+30y230y2=−30y2=−1So, m2=30 kg body should be displaced 1 cm downward.

Two blocks of mass $10 k g$ and $30 k g$ are placed along a vertical line. The first block is raised through a height of $7 c m$ .By what distance should the second mass be moved to raise the center of mass by $1 c m ?$