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Question: Two weights \(W _ { 1 }\) and \(W _ { 2 }\) are suspended from the ends of a light string passin...

Two weights W1W _ { 1 } and W2W _ { 2 } are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration g, the tension in the string will be

A

4W1W2W1+W2\frac { 4 W _ { 1 } W _ { 2 } } { W _ { 1 } + W _ { 2 } }

B

2W1W2W1+W2\frac { 2 W _ { 1 } W _ { 2 } } { W _ { 1 } + W _ { 2 } }

C

W1W2W1+W2\frac { W _ { 1 } W _ { 2 } } { W _ { 1 } + W _ { 2 } }

D

W1W22(W1+W2)\frac { W _ { 1 } W _ { 2 } } { 2 \left( W _ { 1 } + W _ { 2 } \right) }

Answer

4W1W2W1+W2\frac { 4 W _ { 1 } W _ { 2 } } { W _ { 1 } + W _ { 2 } }

Explanation

Solution

When the system is at rest tension in string

T=2m1m2(m1+m2)gT = \frac { 2 m _ { 1 } m _ { 2 } } { \left( m _ { 1 } + m _ { 2 } \right) } g

If the system moves upward with acceleration g then T=2m1m2m1+m2(g+g)T = \frac { 2 m _ { 1 } m _ { 2 } } { m _ { 1 } + m _ { 2 } } ( g + g ) =4m1m2m1+m2g= \frac { 4 m _ { 1 } m _ { 2 } } { m _ { 1 } + m _ { 2 } } g or T=4w1w2w1+w2T = \frac { 4 w _ { 1 } w _ { 2 } } { w _ { 1 } + w _ { 2 } }