Question

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross-section having a radius of 3.11 cm. This pressure is transmitted by a liquid to a second piston of radius 18.3 cm. What force must the compressed air exert in order to lift a car weighing 11200 N?

Answer 1

The force that must the compressed air exert in orden to lift a car weighing 11200 N is 323.47 N

Step-by-step explanation:

Given:

ri = radius of first piston = 3.11 cm

ro = radius of second piston = 18.3 cm

wc = weight of the car = 11200 N

The inward pressure is:

`(F_(i) )/(A_(i) ) =(F_(o) )/(A_(o) ) \\F_(i) =(A_(i) )/(A_(o) ) F_(o) \\F_(i)=(\pi r_(i)^(2) )/(\pi r_(o)^(2) ) F_(o) \\F_(o)=w_(c) \\F_(i)=(\pi r_(i)^(2) )/(\pi r_(o)^(2) ) w_(c)`

Replacing:

`F_(i) =(3.11^(2) )/(18.3^(2) ) *11200=323.47N`

Answer 2

323.47 N

Step-by-step explanation:

Applying pascals Principle,

F₁/A₁ = F₂/A₂...................... Equation 1

Where F₁ = Force exerted on the small piston, A₁ = Area of the small piston, F₂ = Force exerted on the second piston, A₂ = Area of the second piston

Note: Since the piston has a circular cross-section,

Then,

A₁ = πR₁².................... Equation 2

A₂ = πR₂².................... Equation 3

Where R₁ = Radius of the first piston, R₂ = Radius of the second piston

Substitute equation 2 and equation 3 into equation 1

F₁/πR₁² = F₂/πR₂²

F₁/R₁² = F₂/R₂²................. Equation 4

make F₁ the subject of the equation

F₁ = R₁²(F₂/R₂²)................. Equation 5

Given: F₂ = 11200 N, R₁ = 3.11 cm, R₂ = 18.3 cm

Substitute into equation 5

F₁ = 11200(3.11²/18.3²)

F₁ = (9.6721/334.89)11200

F₁ = 323.47 N.

Hence the force that must be exerted by the compressed air = 323.47 N