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Question

Find the reasing of the spring balance shown in figure. The elevator is going up with an acceleration of g /10, the pulley and the string are light and the pulley is smooth.

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Solution

Let the acceleration of the 3 kg mass relative to the elevator is 'a' in the downward direction.

From the free body diagram,

T -1.5 g - 1.5 (g /10)+ 3a = 0

and T - 3g -3 (g /10) + 3a = 0

T= 1.5 g + 1.5 g + 1.5 (g /10) + 1.5a ...(i)

and T = 3 g +3 (g /10) + 3a = 0 ...(ii)

Now equation (i) ×2,

6g + 3 (g /10) + 3a = 2T

Equation (ii) ×1

3g + 3 (g /10)+ 3a = T

Substracting the above two equations , we get ,

T= 6a

Putting T = 6a in equation (ii), we get

6a = 3g + 3 (g /10) -3a

9a=33g10

9=(9.8)3310=32.34

a=32.349=3.59

T=6a=6×3.59=21.55N

T1=2T=2×21.55=43.1N

Mass=Weightg

=43.19.8=4.394.4kg


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