The small piston of a hydraulic lift, has an area of 0.01m2. If a force of 250N is applied to the small piston if it has an area of 0.05m2.

Critical Thinking

Given that,

Small piston Hydraulic piston has an area

A1 = 0.01m²

If the force applied is 250N is applied to the small piston at an area of 0.05 m²

Then,

F2 = 250 N and A2 = 0.05m²

Then, applying pascal principle,

Pressure at small area = pressure are bigger area

P1 = P2

F1 / A1 = F2 / A2

F1 / 0.01 = 250 / 0.05

F1 / 0.01 = 5000

Cross multiply

F1 = 5000 * 0.01

F1 = 50 N

50N

Explanation:

Pressure = Force/Area

Where P1 = P2

F1/A1 = F2/A2

Given that F1 = ? A1 = 0.01m^2

F2 = 250N A2 = 0.05m^2

F1/0.01m^2 = 250N/0.05m^2

F1 * 0.05 = 0.01 * 250

F1 = 0.01*250/0.05

F1 = 2.5/0.05

F1 = 50N