Question

The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. Assume the pistons each have negligible weight. (a) Determine the force that must be applied to the small piston. (b) Determine the pressure applied to the fluid in the lift.

Answer 1

a)

600 N

b)

1.2 x 10⁵ Pa

Step-by-step explanation:

(a)

d₁ = diameter of small piston = 8 cm = 0.08 m

d₂ = diameter of large piston = 40 cm = 0.40 m

F₂ = force applied to large piston = 15000 N

F₁ = force applied to small piston = ?

Using pascal's law

`(F_(1))/((0.25)\pi d_(1)^(2)) = (F_(2))/((0.25)\pi d_(2)^(2))`

Inserting the values

`(F_(1))/((0.08)^(2)) = (15000)/((0.40)^(2))`

F₁ = 600 N

b)

Pressure applied is given as

`P = (F_(1))/((0.25)\pi d_(1)^(2))`

`P = ((600))/((0.25)(3.14) (0.08)^(2))`

P = 1.2 x 10⁵ Pa