Question

A Hydraulic presswith a larger piston of diameter 30cm at a height 1m above the smaller piston of diameter 5cm is present.The mass on the Smaller piston is 10kg . What is the force exerted on the Load by the larger piston . (Density of Oil = 750kgm^-3

Answer 1

Pressure:

It is the force acting normally per unit area.

Pressure on the larger piston is equal to the pressure on the smaller piston + liquid pressure:

$$\begin{gathered} \frac{F_1}{A_1}=\frac{F_2}{A_2}+\rho gh \\ \frac{F_1}{\pi {r_1}^2}=\frac{mg}{\pi {r_2}^2}+\rho gh \\ {F}_1=\pi {r_1}^2\left(\frac{mg}{\pi {r_2}^2}+\rho gh\right) \\ =\pi {\left(15\times {10}^{-3}\right)}^2\left[\frac{10\times 9.8}{\pi \times {\left(2.5\times {10}^{-3}\right)}^2}+750\times 9.8\times 1\right] \\ =3533.2\mathrm{N} \end{gathered}$$