Question

Find the reasing of the spring balance shown in figure. The elevator is going up with an acceleration of g /10, the pulley and the string are light and the pulley is smooth.

Answer 1

Let the acceleration of the 3 kg mass relative to the elevator is 'a' in the downward direction.

From the free body diagram,

T -1.5 g - 1.5 (g /10)+ 3a = 0

and T - 3g -3 (g /10) + 3a = 0

$\Rightarrow$ T= 1.5 g + 1.5 g + 1.5 (g /10) + 1.5a ...(i)

and T = 3 g +3 (g /10) + 3a = 0 ...(ii)

Now equation (i) $\times 2$,

6g + 3 (g /10) + 3a = 2T

Equation (ii) $\times 1$

3g + 3 (g /10)+ 3a = T

Substracting the above two equations , we get ,

T= 6a

Putting T = 6a in equation (ii), we get

6a = 3g + 3 (g /10) -3a

$\Rightarrow 9a=\frac{33g}{10}$

$\Rightarrow 9=\frac{\left(9.8\right)33}{10}=32.34$

$\Rightarrow a=\frac{32.34}{9}=3.59$

$\therefore T=6a=6\times 3.59=21.55N$

$\therefore T_1=2T=2\times 21.55=43.1N$

$\therefore Mass=\frac{Weight}{g}$

$=\frac{43.1}{9.8}=4.39\approx 4.4kg$