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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?

Respuesta :

Answer:

1471.5 Newton

10

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

[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\Rightarrow F_1=F_2\frac{A_1}{A_2}\\\Rightarrow F_1=14715\frac{2}{20}=1471.5[/tex]

Force required is 1471.5 Newton

b) Mechanical advantage

[tex]\frac{F_2}{F_1}=\frac{14715}{1471.5}=10[/tex]

Mechanical advantage is 10

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