Question

For the system shown in the figure, the cylinder on the left at $L$ has a mass of $600\, kg$ and a cross-sectional area of $800\, cm ^{2}$. The piston on the right, at $S$, has cross-sectional area $25\, cm ^{2}$ and negligible weight. If the apparatus is filled with oil $\left(\rho=0.75 \,g / cm ^{3}\right)$. The force $F$ (in $N$) required to hold the system in equilibrium is____. Take $g=10\, m / s ^{2}$

Answer 1

In equilibrium $\frac{600 \times 10}{800 \times 10^{-4}}$
$=\frac{F}{25 \times 10^{-4}}+h \rho g$
$\Rightarrow \frac{F}{25 \times 10^{-4}}=\frac{60}{8} \times 10^{4}-8 \times\left(0.75 \times 10^{3}\right) \times 10$
$\frac{F}{25 \times 10^{-4}}=1.5 \times 10^{4}$
$\Rightarrow F=37.5\, N$