Question

Question #21241

Answer 1

The force applied at smaller piston = `20` `kg`

Explanation:

By Pascal's law of multiplication of thrust we have

`F_2 = F_1 " x " (A_2/A_1)`
where force exerted on larger piston (`F_2`) = 1600 `kg`
force exerted on smaller piston (`F_1`) = unknown
area of larger piston (`A_2`) = 800 `cm^2`
area of larger piston (`A_1`) = 10 `cm^2`

`F_1 = F_2 " x " (A_1/A_2)`
= `1600 " x "10/800`
= `20` `kg`