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.

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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,
$m_{A} a = T - m_{A} g - \frac{m_{A} g}{10} . . . (1)$
$m_{B} a = m_{B} g + \frac{m_{B} g}{10} - T . . . (2)$

Adding both the equations, we get:
$a (m_{A} + m_{B}) = (m_{B} - m_{A}) g + (m_{B} - m_{A}) \frac{g}{10}$

Putting value of the masses,we get:
$9 a = \frac{33 g}{10} \Rightarrow \frac{a}{g} = \frac{11}{30} . . . (3)$
Now, using equation (1), we get:
$T = m_{A} (a + g + \frac{g}{10})$
The reading of the spring balance = $\frac{2 T}{g} = \frac{2}{g} m_{A} (a + g + \frac{g}{10})$
$\Rightarrow 2 \times 1.5 (\frac{a}{g} + 1 + \frac{1}{10}) = 3 (\frac{11}{30} + 1 + \frac{1}{10}) = 4.4 kg$

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