Question

two weights w1 and w2 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 string will be..

Answer 1

The tension in the string is $\bold{\frac{4 w_{1} w_{2}}{w_{1}+w_{2}}}$

SOLUTION:

The weights of the blocks are $w_{1}$ and $w_{2}$ also we known that,

Weight = mg, let the mass of block having weight $w_{1}$ be $m_{1}$ and the mass of the block having weight w_2be m_2 . So we have,

$w_{1}=m_{1} g$

$m_{1}=\frac{w_{1}}{g}$  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”

From equilibrium we have

$\begin{array}{l}{m_{2} g-T=m_{2}(a-g) \ldots \ldots \ldots . \text { (i) }} \\ {T-m_{1} g=m_{1}(a+g) \ldots \ldots \ldots \text { (ii) }}\end{array}$

Subtracting both the equation the value of acceleration is

$a=\frac{2 g\left(m_{2}-m_{1}\right)}{\left(m_{2}+m_{1}\right)} \ldots \ldots \ldots(\text { iii })$

From the equation (i) , (ii) and (iii)

$\begin{array}{l}{T=m_{1} g+m_{1} a+m_{1} g=m_{1}(2 g+a)} \\ {\frac{w_{1}}{g}\left(2 g+\frac{2 g\left(m_{2}-m_{1}\right)}{m_{2}+m_{1}}\right.}\end{array}$

The tension in the string will be

$T=\frac{4 w_{1} w_{2}}{w_{1}+w_{2}}$

Answer 2

Explanation:

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