Two piston of a hydraulic lift have diameters of 60 cm and 5 cm. What is the force exerted by the larger piston when 50 N is placed on the smaller piston ?

Explanation

Since, the diameter of the pistons are given, we can calculate the radius of the piston
`r = D/2`
Area of smaller piston, `A_1 = pi(5/2)^(2) = pi(2.5)^(2)`
Area of larger piston, `A_2 = pi(60/2)^2 = pi(30)^(2)`
`F_2 = (A_2)/(A_1) xx F_1 = (50 N) xx (30/2.5)^2 = 7200 N`
This means, with the force of 50 N, the force of 7200 N, can be lifted.