Question

A hydraulic jack has a small piston area A1 = 2 m^2 and the large piston area A2 = 20 m^2. A 1500 Kg car needs to be lifted. a) How much force equivalent needed on the effort end F1? P1 = F1/A1, P1 = P2, (F = mass x g) b) What is the mechanical advantage of the system?

Answer 1

1471.5 Newton

10

Step-by-step explanation:

Small piston area = A₁ = 2 m²

Large piston area A₂ = 20 m

m = Mass of car = 1500 kg

g = Acceleration due to gravity = 9.81 m/s²

Force

F = mg = 1500×9.81 = 14715 N

Force applied by car is 14715 N

a) Pascal's law

`(F_1)/(A_1)=(F_2)/(A_2)\\\Rightarrow F_1=F_2(A_1)/(A_2)\\\Rightarrow F_1=14715(2)/(20)=1471.5`

Force required is 1471.5 Newton

b) Mechanical advantage

`(F_2)/(F_1)=(14715)/(1471.5)=10`

Mechanical advantage is 10