Question

The reading of the spring balance for the system shown in the figure is (in $kg$). The elevator is going up with an acceleration of $\frac{g}{10}{\text{m/s}}^2$.
(Take $g=10{\text{m/s}}^2$)

Answer 1

Using the concept of pseudo force in the frame of elevator we have,

Acceleration of of the elevator,

$a=\frac{\text{Supporting force-Opposing force}}{\text{Total mass}}$

$\Rightarrow a=\frac{3(g+\frac{g}{10})-1.5(g+\frac{g}{10})}{3+1.5}$

$\Rightarrow a=\frac{1.5\times 11g}{10\left(4.5\right)}$

$\therefore a=\frac{11g}{30}$

For $1.5\text{kg}$ mass:

$T=m(g^{\prime }+a)$

$=1.5(\frac{11g}{10}+\frac{11g}{30})$

$=\frac{3}{2}\times 11g\times \frac{4}{30}$

$=22\text{N}$

$\therefore F=2T=44\text{N}$

Thus, the reading of the spring balance is,

$\text{Equivalent mass}=\frac{44}{g}\text{kg}=4.4\text{kg}$