Question

Two weights ${w_1}$ and ${w_2}$ are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with acceleration g, the tension in the string will be:

A) $\dfrac{{4{w_1}{w_2}}}{{{w_1} + {w_2}}}$.

B) $\dfrac{{2{w_1}{w_2}}}{{{w_1} + {w_2}}}$.

C) $\dfrac{{{w_1} - {w_2}}}{{{w_1} + {w_2}}}$.

D) $\dfrac{{{w_1}{w_2}}}{{2\left( {{w_1} + {w_2}} \right)}}$.

Answer 1

Hint: The tension always acts away from the body and therefore the tension in the string is equal to the weight of the mass its hanging it with but if the string is at move then the acceleration comes into action and then we need to make the free body diagram and make the equations.

Complete step by step solution:

It is given in the problem that two weights ${w_1}$ and ${w_2}$are suspended from the ends of a light string passing over a smooth fixed pulley and we need to find the tension in the string.

Let us make the free body diagram of the lift and its components. Here we consider ${w_2} > {w_1}$ because there will be no acceleration if the string has the same weight on both of its sides.

The lift is moving with acceleration g upwards and the string is moving with acceleration a.

For the motion of mass ${m_1}$ we have, 

$ \Rightarrow {w_2} + {m_2}g - T = {m_2}a$

$ \Rightarrow {w_2} + {w_2} - T = {m_2}a$

$ \Rightarrow 2{w_2} - T = {m_2}a$………eq. (1)

For the motion of mass ${m_2}$ we have,

$ \Rightarrow T - {m_1}g - {w_1} = {m_1}a$

$ \Rightarrow T - {w_1} - {w_1} = {m_1}a$

$ \Rightarrow T - 2{w_1} = {m_1}a$………eq. (2)

Adding equation (1) and equation (2) we get,

$ \Rightarrow 2{w_2} - T + T - 2{w_1} = {m_2}a + {m_1}a$

$ \Rightarrow 2{w_2} - 2{w_1} = a\left( {{m_2} + {m_1}} \right)$

Since, ${w_1} = {m_1}g$ and ${w_2} = {m_2}g$therefore ${m_1} = \dfrac{{{w_1}}}{g}$and${m_2} = \dfrac{{{w_2}}}{g}$.

$ \Rightarrow 2{w_2} - 2{w_1} = \left( {\dfrac{{{w_2}}}{g} + \dfrac{{{w_1}}}{g}} \right) \times a$

$ \Rightarrow \left( {\dfrac{{{w_2}}}{g} + \dfrac{{{w_1}}}{g}} \right) \times a = 2{w_2} - 2{w_1}$

$ \Rightarrow a = \dfrac{{\left( {2{w_2} - 2{w_1}} \right)}}{{\left( {\dfrac{{{w_2}}}{g} + \dfrac{{{w_1}}}{g}} \right)}}$

$ \Rightarrow a = \dfrac{{g \times \left( {2{w_2} - 2{w_1}} \right)}}{{{w_1} + {w_2}}}$………eq. (3)

Replacing the value of the acceleration in equation (2) we get,

$ \Rightarrow T - 2{w_1} = {m_1}a$

$ \Rightarrow T - 2{w_1} = \dfrac{{{m_1}g\left( {2{w_2} - 2{w_1}} \right)}}{{{w_1} + {w_2}}}$

$ \Rightarrow T = \dfrac{{{m_1}g\left( {2{w_2} - 2{w_1}} \right)}}{{{w_1} + {w_2}}} + 2{w_1}$

Since ${m_1}g = {w_1}$ we get,

$ \Rightarrow T = \dfrac{{{w_1}\left( {2{w_2} - 2{w_1}} \right)}}{{{w_1} + {w_2}}} + 2{w_1}$

$ \Rightarrow T = \dfrac{{\left( {2{w_1}{w_2} - 2{w_1}^2} \right)}}{{{w_1} + {w_2}}} + 2{w_1}$

$ \Rightarrow T = \dfrac{{\left( {2{w_1}{w_2} - 2{w_1}^2} \right) + 2{w_1}\left( {{w_1} + {w_2}} \right)}}{{{w_1} + {w_2}}}$

$ \Rightarrow T = \dfrac{{\left( {2{w_1}{w_2} - 2{w_1}^2} \right) + \left( {2{w_1}^2 + 2{w_1}{w_2}} \right)}}{{{w_1} + {w_2}}}$

$ \Rightarrow T = \dfrac{{2{w_1}{w_2} + 2{w_1}{w_2}}}{{{w_1} + {w_2}}}$

$ \Rightarrow T = \dfrac{{4{w_1}{w_2}}}{{{w_1} + {w_2}}}$.

The tension in the string is equal to$T = \dfrac{{4{w_1}{w_2}}}{{{w_1} + {w_2}}}$. 

Therefore, option (A) is correct.

Note: The Free body diagram is very important as the equations we can make is from the free body diagrams only. The lift is moving upwards with an acceleration g but the string on the pulley is moving with an acceleration a.