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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?

1 Answer

3 votes

ANSWER:

444 N

Explanation:

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


(F_1)/(A_1)=(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=(F_2\cdot A_1)/(A_2) \\ \text{ Replacing:} \\ F_1=(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

User Janusz Przybylski
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