A frictionless pulley is attached to one arm | vedclass

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(B) When the clamp is in place,the system is static. The total downward force on the arm is the sum of the weights: $T_{initial} = (m_1 + m_2)g$.When

Vedclass · Accepted Answer

$\frac{{{{({m_1} - {m_2})}^2}}}{{{m_1} + {m_2}}}$

Vedclass · Answer

(B) When the clamp is in place,the system is static. The total downward force on the arm is the sum of the weights: $T_{initial} = (m_1 + m_2)g$.When the clamp is removed,the masses $m_1$ and $m_2$ move under gravity. The tension $T$ in the string for an Atwood machine is given by $T = \frac{2m_1m_2g}{m_1+m_2}$.Since the string passes over the pulley,the total downward force exerted on the arm of the balance is $2T$ (one tension for each side of the string pulling down on the pulley): $T_{final} = 2 	imes \frac{2m_1m_2g}{m_1+m_2} = \frac{4m_1m_2g}{m_1+m_2}$.The change in force required to restore balance is $\Delta F = T_{initial} - T_{final} = (m_1+m_2)g - \frac{4m_1m_2g}{m_1+m_2}$.Simplifying this expression: $\Delta F = g \left[ \frac{(m_1+m_2)^2 - 4m_1m_2}{m_1+m_2} ight] = g \left[ \frac{m_1^2 + m_2^2 + 2m_1m_2 - 4m_1m_2}{m_1+m_2} ight] = \frac{(m_1-m_2)^2 g}{m_1+m_2}$.Thus,the change in counter mass required is $\frac{(m_1-m_2)^2}{m_1+m_2}$.

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