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Question

A meter stick is hung from two spring balances A and B of equal lengths that are located at 20 cm and 70 cm mark of the meter stick. Weights of 2 N are placed at 10 cm and 40 cm marks, while a weight of 1.0 N is placed at the 90 cm mark. The weight of the uniform meter stick is 1.5 N. Determine the scale readings of the two balances A and B.

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Solution

From Newton's laws
T1+T2=(1.5+1)NT1+T2=2.5N(1)
Balancing the torque about point A
30(1.5)+70(1)=50T1T1=4.5+75=11.55=2.3N
T2=2.52.3=0.2N
Here, T1&T2 are corresponding readings of spring balances on the right and left.

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