Question

A hydraulic lift is to be used to lift a truck weighing 2700 kg. The diameters of the piston where the lifting force is applied is 1 cm. The diameter of the piston on which the truck is to be placed is 16 cm ? Determine the minimum force needed to lift the truck. 2. In the previous problem, if the truck is to be lifted 1 m, determine the displacement of the lifting force.

Answer 1

Fn = P * A = pressure * area

A = π R^2 = π (D/2)^2

A2 / A1 = (D2 / D1)^2 = (16 / 1)^2 = 256

The applied force is then 1/256 the force required to lift truck.

(Note - we shall use 2700 kg as the weight of the truck which is not actually correct since W = M g - but we will find the mass of the applied force)

F = 2700 / 256 = 10.5 kg actual force applied

F = 10.5 kg * 9.8 m/s^2 = 103 N (force exerted by 10.5 kg)

For part II displacement is given as 1 meter

The actual work that would be done is

W = F * d = 103 N * 1 m = 103 Joules