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Question

Find the reading of the spring balance shown in figure (5−E6). The elevator is going up with an acceleration g/10, the pulley and the string are light and the pulley is smooth.


Solution

Let the left and right blocks be A and B, respectively.
And let the acceleration of the 3 kg mass relative to the elevator be 'a' in the downward direction.



From the free-body diagram,
mAa=T-mAg-mAg10    ...1
mBa=mBg+mBg10-T    ...2

Adding both the equations, we get:
amA+mB=mB-mAg+mB-mAg10

Putting value of the masses,we get:
9a=33g10ag=1130    ...3
Now, using equation (1), we get:
T=mAa+g+g10
The reading of the spring balance = 2Tg=2gmAa+g+g10
2×1.5ag+1+110=31130+1+110= 4.4 kg

Physics
HC Verma - I
Standard XI

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