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 $\frac{g}{10}m/s^2$.
(Take $g=10m/s^2$)

Answer 1

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

$a=\frac{\text{Supporting force-Opposing force}}{\text{Total mass}}$
$=\frac{3(g+\frac{g}{10})-1.5(g+\frac{g}{10})}{3+1.5}=\frac{1.5\times 11g}{10\left(4.5\right)}$
$a=\frac{11g}{30}$

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