Question

What force is applied on a piston of area of cross-section 2 cm2 to obtain a force of 150 N on the piston of area of cross-section 12 cm2 in a hydraulic machine?

Answer 1

Force applied is - 25 N

Pascal's law states that "a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere."

Pressure is defined as "the force acting on an object perpendicular to its surface, and per unit area."

For pressure, force has to act at 90°.

Thus, pressure is given by the formula-

P = f/a

P denotes pressure, f denotes force and a denotes area.

According to Pascal's law, hydraulic machine will have equal amount of pressure.

Hence, P1 = P2

F1/A1 = F2/A2

150/12 = F2/2

F2 = 25 N

Thus, force required to get results according to question is - 25 N

Answer 2

Force on the narrow piston = 25 N

Explanation:

Area of cross-section of narrow piston = 2 cm2 = 2 * 10^-4 m2

Area of cross-section of wider piston = 12 cm2 = 12 * 10^-4 m2

In a hydraulic machine, we know that the pressure is equal at both the pistons.

So pressure at narrow piston = pressure at wider piston

P1 = P2

Pressure = Force / Area, so we can write it as:

F1 / A1 = F2 / A2

F1 / ( 2 * 10^-4) = 150 / ( 12 * 10^-4)

F1 = 150 * 2 / 12

F1 = 25 N  

Force on the narrow piston = 25 N