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 at an acceleration $g$, the tension in the suspension of pulley will be:

• A. $\frac{4w_1{w}_2}{(w_1+w_2)}$
• B. $\frac{2w_1{w}_2}{(w_1+w_2)}$
• C. $\frac{w_1-w_2}{(w_1+w_2)}$
• D. $\frac{w_1{w}_2}{\left\{2\right(w_1+w_2\left)\right\}}$

Answer 1

The correct option is A $\frac{4w_1{w}_2}{(w_1+w_2)}$

The weights of the blocks are and also we known that,

Weight = mg, let the mass of block having weight be and the mass of the block having weight $w_2$ be $m_2$ . So we have,

$w_1=m_{1}g$

$m_1=\frac{w_1}{g}$ .....................(i)

This is the mass of first block.

Similarly the mass of second block will be

$m_2=\frac{w_2}{g}$

The acceleration is given upward and is equal to $g$ . So according to the situation one block will move upward and the other will move downward.

Let acceleration of the blocks be $“a”$ and tension $“T”$

Subtracting both the equation the value of acceleration is

$a=\frac{2g(m_2-m_1)}{m_2+m_1}$ .................(ii)

From the equation (i) and (ii)

$T=m_{1}g+m_{1}a+m_{1}g=m_1(2g+a)$

$\frac{w_1}{g}(2g+\frac{2g(m_2-m_1)}{m_2+m_1})$

The tension in the string will be

$T=\frac{4w_1{w}_2}{w_1+w_2}$