Question

In a hydraulic lift, the diameter of the input piston is 10.0 cm and the diameter of the output piston is 50.0 cm. (a) How much force must be applied to the input piston so that the output piston can lift a 250N object? (b) If the object is lifted a distance of 0.3 m, then how far is the input piston moved?

Answer 1

a) Force must be applied to the input piston = 10N

b) The input piston moved by 7.5 m

Step-by-step explanation:

a) For a hydraulic lift we have equation

`(F_1)/(A_1)=(F_2)/(A_2)`

F₁ = ?

d₁ = 10 cm = 0.1 m

F₂ = 250 N

d₂ = 50 cm = 0.5 m

Substituting

`(F_1)/((\pi * 0.1^2)/(4))=(250)/((\pi * 0.5^2)/(4))\\\\F_1=10N`

Force must be applied to the input piston = 10N

b) We have volume of air compressed is same in both input and output.

That is A₁x₁ = A₂x₂

A is area and x is the distance moved

x₂ = 0.3 m

Substituting

`(\pi * 0.1^2)/(4)* x_1=(\pi * 0.5^2)/(4)* 0.3\\\\x_1=7.5m`

The input piston moved by 7.5 m