Question

A pulley is attached to one arm of a balance and a string passed round, it carries two masses ${m_1}$ and ${m_2}$. The balance is counterpoised and the pulley is clamped so that ${m_1}$ and ${m_2}$ do not move. How much counterweight is to be reduced or increased to restore balance if the clamp is released?

(A) $\dfrac{{g{{\left( {{m_1} - {m_2}} \right)}^2}}}{{\left( {{m_1} + {m_2}} \right)}}$ to be reduced
(B) $\dfrac{{g{{\left( {{m_1} - {m_2}} \right)}^2}}}{{\left( {{m_1} + {m_2}} \right)}}$ to be increased
(C) $\dfrac{{g\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}$ to be reduced
(D) $\dfrac{{g\left( {{m_1} - {m_2}} \right)}}{{\left( {{m_1} + {m_2}} \right)}}$ to be increased

Answer 1

Hint: Tension increases when the force increases axially. Counterweight provides equal weight to that of the pulley in order to maintain the balance. The difference in the tension of the pulley in initial and final condition gives the weight to be added or removed from the counterweight in order to maintain the stability of the balance.

Formula used:
By the Newton’s second law of motion,
$F = m \times a$
Where,
$F$ is the force of the object
$m$ is the mass of the object
$a$ is the acceleration of the object.

Complete step by step answer:
Given, Pulley attached to the balance is with the masses ${m_1}$ and ${m_2}$.
A pulley at the one end of a balance is clamped initially. Hence, the tension in the pulley is the sum of force exerted by the two masses \[{m_{_1}}\] and ${m_2}$ suspended from the pulley.
$T = {T_1} + {T_2}$
Substituting the tension with the product of mass and acceleration due to gravity,
$T = {m_1}g + {m_2}g$
Taking the common term from the equation,
$T = g\left( {{m_1} + {m_2}} \right)$
The masses start to move when the clamp is released from the pulley. So higher mass ${m_2}$ moves down and lower mass ${m_1}$ moves up. Due to this, tension changes.
The net force of the mass ${m_2}$ is equal to the algebraic sum of the force due to gravity and the tension in the rope,
${F_{{g_2}}} - T = {F_2}$
Where, ${F_{{g_2}}}$ is the force due to gravity of mass ${m_2}$, $T$ is the tension and ${F_2}$ is the net force of the mass ${m_2}$.
${m_2}g - T = {m_2}a$
Finding acceleration, a by moving everything to RHS,
$a = \dfrac{{\left( {{m_2}g - T} \right)}}{{{m_2}}}\,...............\left( 1 \right)$
Now,
$T - {m_1}g = {m_2}a$
Finding acceleration, a by moving everything to RHS,
$a = \dfrac{{\left( {T - {m_1}g} \right)}}{{{m_1}}}\,..............\left( 2 \right)$
By solving equation (1) and (2),
$\dfrac{{\left( {{m_2}g - T} \right)}}{{{m_2}}} = \dfrac{{\left( {T - {m_1}g} \right)}}{{{m_1}}}$
Taking $T$ on one side and the other terms in another side,
$T = \dfrac{{2g\left( {{m_2}{m_1}} \right)}}{{{m_1} + {m_2}}}$
Pulley is assumed to be massless, so the tension is $2T$. Difference between the initial and final tension is
$D = 2T - \left( {{m_1} + {m_2}} \right)g$
Where,
$D$ is the difference between the initial and final tension.
Substituting the value of the final $T$ in the above equation,
$D = 2\left\{ {\dfrac{{2g\left( {{m_2}{m_1}} \right)}}{{{m_1} + {m_2}}}} \right\} - \left( {{m_1} + {m_2}} \right)g$
By simplifying,
$D = \dfrac{{ - g{{\left( {{m_1} - {m_2}} \right)}^2}}}{{{m_1} + {m_2}}}$
This difference in the tension gives the weight of counterweight that is to be increased or decreased. The negative sign indicates that counterweight is to be decreased.

Thus, the option (A) is correct.

Note:
In the net force equation, the tension force of the rope is subtracted from the force due to gravity. Because the force of the mass due to gravity acts downwards but the tension force of the rope acts upwards. So, both the forces are acting in different directions, then they are subtracted.