Question

A car jack works when a mechanic uses a force 3.8 N, on a small piston of radius 10 cm. What force can he exert on a larger piston of radius 0.22 m?

Answer 1

Answer:

18.392 N

Explanation:

Given that,

Smaller force acting on the car, F₁ = 3.8 N

Radius of smaller radius, r₁ = 10 cm = 0.1 m

The radius of the larger piston, r₂ = 0.22 m

We need to find the force exerted on the larger piston. Let it is F₂.

Using the relation,

$\dfrac{F_1}{F_2}=\dfrac{A_1}{A_2}\\\\\dfrac{F_1}{F_2}=\dfrac{\pi r_1^2}{\pi r_2^2}\\\\\dfrac{F_1}{F_2}=\dfrac{r_1^2}{r_2^2}\\\\F_2=\dfrac{F_1r_2^2}{r_1^2}\\\\F_2=\dfrac{3.8\times (0.22)^2}{(0.1)^2}\\\\F_2=18.392\ N$

Hence, the force exerted on the larger piston is 18.392 N.