Question

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

Answer 1

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》To find :

By what distance should the second mass be moved to raise the centre of mass 1cm, $y_{2}$ = ?

》Given :

$m_{1}$ = 10 kg

$m_{2}$ = 30 kg

$y_{1}$ = 7 cm

》Solution :

According to question, two masses $m_{1}$ and $m_{2}$ are kept in vertical line.

Also, the centre of mass is raised by 1 cm.

So,

Centre of mass = $\frac{m_{1}y_{1}+m_{2}y_{2}}{m_{1}m_{2}}$

⟹ 1 = $\frac{10\times 7+30y_{2}}{10+30}$

⟹ 1 = $\frac{70+30y_{2}}{40}$

⟹ 40 = $70+30y_{2}$

⟹ $30y_{2}$ = $-30$

⟹ $y_{2}$ = $-1$

So, the $m_{2}$  should be displaced 1 cm downward to raise the centre of mass 1 cm.

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Thanks !!..

Answer 2

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Solution :- Two mass of m1 and m2 kept on vertical line of mass m1 = 10 kg and m2 = 30kg

The block raised through 7cm The centre of mass raised 1cm

•°• 1 = m1y1 + m2y2 / m1 + m2 1 = 10 * 7 + 30* y2/ 10 + 30 40 = 70 + 30y2 - 30 = 30y2 y2 = - 1cm

The 30kg body should be displaced 1 cm downward in order to raise the centre of mass by 1cm .

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