Question

In a simple hydraulic press, the cross-sectional area of the two cylinders are $5 \times 10^{-4} \,m ^{2}$ and $10^{-2} \,m ^{2}$, respectively. A force of $20 \,N$ is applied at the small plunger. The pressure produced in the cylinder of larger cross-sectional area is $m \times 10^{4} \, N / m ^{2}$. Find $m$.

Answer 1

In a hydraulic press, a force $F_{1}$ applied to the smaller plunger creates a pressure $\left(F_{1} / A_{1}\right)$ in the liquid and this pressure is transmitted equally throughout the liquid and acts on the larger plunger.
The thrust acting on the larger plunger upwards due to this pressure is $F_{2}=A_{2}\left(\frac{F_{1}}{A_{1}}\right)$.
Hence, the pressure produced in the cylinder of larger cross-sectional area is
$\frac{F_{1}}{A_{1}}=\frac{20}{5 \times 10^{-4}}=40000\, N / m ^{2}$