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When an empty lift is moving down with an acceleration of $\frac{g}{4} m s^{- 2}$ , the
tension in the cable is 9000N. When the lift is moving up with an

acceleration of $\frac{g}{4} m s^{- 2}$ , the tension in the cable is

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16,000 N

12,000 N

18,000 N

15,000 N

answer is A.

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Detailed Solution

$a = \frac{g}{4}$

$m g - 9000 = \frac{m g}{4}$

$m g - \frac{m g}{4} = 9000$

$\frac{3 m g}{4} = 9000$

$m = 1200 k g$

Question Image

$a = \frac{g}{3}$

$T - m g = m a$

$T = m g + m a$

$= 12000 + \frac{12000}{3}$

$= 12000 (1 + \frac{1}{3})$

$= 12000 \times \frac{4}{3}$

$T = 16000 N$

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