Question

In a hydraulic system, piston 1 has a surface area of 100 cm2, and piston 2 has a surface area of 900 cm2. A force of 150 N is exerted on piston 1 of the hydraulic lift. What force will be exerted on piston 2?

Answer 1

Since it is a hydraulic system, we will assume the pressure are equal on each piston.
This translates to force=pressure * area, or
P=F1/A1=F2/A2
Solving for F2
F2=F1*(A2/A1)=150N *(900/100) = 1350 N

Answer 2

In the hydraulic system, the pressure is equal on each piston. The force exerted on the second piston is 1350 N.

Assume, In the hydraulic system, the pressure is equal on each piston.

P1 = P2

So,

Force = Pressure x Area

$\bold {P=\dfrac {F1}{A1 }=\dfrac {F2}{A2}}$

Where,

F1 - Force exerted on first piston = 150 N

A1 = area of first piston = 100 $\bold {cm^2}$

A2 = area of the second piston = 900 $\bold {cm^2}$

Put the values and solve for F2,

$\bold {F2=F1\times (\dfrac {A2}{A1})}\\\\\bold {F2 = 150N \times (\dfrac {900}{100})}\\\\\bold {F2 = 1350 N}$

Therefore, the force exerted on the second piston is 1350 N.

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