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, Adding both the equations, we get: $a\left({m}_{A}+{m}_{B}\right)=\left({m}_{B}-{m}_{A}\right)g+\left({m}_{B}-{m}_{A}\right)\frac{g}{10}$ Putting value of the masses,we get: Now, using equation (1), we get: $T={m}_{A}\left(a+g+\frac{g}{10}\right)$ The reading of the spring balance = $\frac{2T}{g}=\frac{2}{g}{m}_{A}\left(a+g+\frac{g}{10}\right)$ PhysicsHC Verma - IStandard XI

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