Question

In a car lift used in a service station, compressed air exerts a force on a small piston that has a circular cross section and a radius of 5.00 cm. The pressure is transmitted by a liquid to a piston that has a radius of 15.0 cm. a) What force must the compressed air exert to lift a car weighing 13,300N? b) What air pressure produces this force?
Answer: a) 1.48 x 103 N b) 1.88 x 105 Pa

How do you show the work on this?

Answer 1

Answer:

a) F₁ = 1.48 x 10³ N

b) P = 1.88*10⁵ Pa

c) The work  is equal in both pistons

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F₁) on a small area piston (A₁), then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F₂) can be exerted that is proportional to the area(A₂) of the piston.

Pressure is defined as the force per unit area:

$P=\frac{F}{A}$  Formula (1)

P₁=P₂

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$ Formula (2)

Data

r₁= 5 cm = 0.05 m

r₂= 15 cm = 0.15 m

F₂=  13300N

Area of the pistons (A₁,A₂)

A=π*r² : Area of the circle

A₁ = π*(0.05)²=7.85*10⁻³ m²

A₂= π*(0.15)²= 70.69*10⁻³ m²

a) Force that compressed air must exert to lift a car weighing 13300 N

We replace data in the formula (2)

$\frac{F_{1} }{A_{1} } =\frac{F_{2} }{A_{2} }$

$F_{1} = \frac{13300*7.85*10^{-3} }{70.69*10^{-3} }$

F₁ =  1.48 x 10³ N

b) Air pressure produced by F₁

We replace data in the formula (1)

$P=\frac{F}{A}$

F₁ =  1.48 x 10³ N , A₁ = 7.85*10⁻³ m²

$P=\frac{1.48*10^{3} }{7.85*10^{-3} }$

P= 1.88*10⁵ Pa

c)The volume of liquid displaced by the small piston is distributed in a thin layer on the large piston, so that the product of the force by the displacement (the work) is equal in both pistons.