If the input piston of a hydraulic jack has a radius of 0.005 m, what force is necessary to apply to it to lift a 1500-kg car on the output piston whose radius is 0.5 m?

Answer 1

Nine different identities. You poor soul.

Answer 2

p= F₁/A₁= F₂/A₂
F₁/π R₁²= F₂/π R₂²
F₁ = F₂ R₁²/R₂²= m₂g •R₁²/R₂²=
=1500•9.8 •0.005²/0.5²=1.47 N