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 $20N$ 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

Answer: 4

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
$\frac{F_1}{A_1}=\frac{20}{5\times 10^{-4}}=40000N/m^2$