Question

A hydrolic press consist of a large piston with cross sector area,Ai=200cm2 and the small piston with crossectional area A2=2cm2, if a force of 200N is applied to the small piston,find the force F on the large piston

Answer 1

Answer on Question 66269, Physics, Mechanics, Relativity

Question:

A hydraulic press consist of a large piston with cross-sectional area $A_{1} = 200~cm^{2}$ and the small piston with cross-sectional area $A_{2} = 2~cm^{2}$. If a force of $200N$ is applied to the small piston, find the force on the large piston.

Solution:

By the hydraulic press formula we have:

$$\frac {F _ {1}}{A _ {1}} = \frac {F _ {2}}{A _ {2}},$$

here, $A_{1}$ is the cross-sectional area of the large piston, $A_{2}$ is the cross-sectional area of the small piston, $F_{1}$ is the force acting on the large piston, $F_{2}$ is the force acting on the small piston.

Then, from this formula we can find the force acting on the large piston:

$$F _ {1} = F _ {2} \cdot \frac {A _ {1}}{A _ {2}} = 200 N \cdot \frac {200~cm ^ {2}}{2~cm ^ {2}} = 20000 N.$$

Answer:

$$F _ {1} = 20000 N.$$

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