Question

A hydraulic lift is used to jack a 1070 kg car 12 cm off the floor. The diameter of the output piston is 18 cm, and the input force is 250 N.

(a) What is the area of the input piston?

(b) What is the work done in lifting the car 12 cm?

(c) If the input piston moves 13 cm in each stroke, how high does the car move up for each stroke?

(d) How many strokes are required to jack the car up 12 cm? (Include fractions of a stroke in your answer).

(e) Calculate the combined work input of all of the strokes.

Answer 1

Area of the Output piston = 3.14 * (0.18/2)^2 m^2
Area of the Output piston = 0.0254 m^2

(a)
F/A1 = F/A2
250/A1 = 1070 * 9.8/0.0254
A1 = 6.055 * 10^-4 m^2
Area of the input piston, A1 = 6.055 * 10^-4 m^2

(b)
Work done = m*g*h
W = 1070*9.8*0.12 J
W = 1258.3 J
Work done to lift the car, W = 1258.3 J

(c)
Work done by Input Piston = Force * Distance
Work done Input Piston = 250 * 0.13 = 32.5 J

Work done by Input Piston = Force * Distance moved by Car
Distance moved by Car = Work done by Input Piston/Force
Distance moved by Car = 32.5/(1070*9.8)
Distance moved by Car = 0.31 cm

(d)
No of Strokes needed = 12 cm/0.31 cm
No of Strokes needed = 38.7 Strokes

(e)
Combined Work input = 38.7 * 32.5 J
Combined Work input = 1258.0 J