Question

The hydraulic jack in our lab has a small piston of radius 5.40 mm and a large piston of radius 15.9 mm. How much input force is exerted on the small input piston to produce a force of 3850 N at the large (output) cylinder?

Answer 1

ANSWER:

444 N

STEP-BY-STEP EXPLANATION:

We can calculate the force, by means of Pascal's principle, which is stated as follows

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

We need to know F1, which would be the force exerted on the small input piston, we solve for F1:

$\begin{gathered} F_1=\frac{F_2\cdot A_1}{A_2} \\ \text{ Replacing:} \\ F_1=\frac{3850\cdot2\pi\cdot5.4^2}{2\pi\cdot15.9^2} \\ F_1=444.07\cong444\text{ N} \end{gathered}$

The input force is exerted on the small input piston is 444 N