A hydraulic lift is used to jack a 920 kg car 14 cm off the floor. The diameter

Question

A hydraulic lift is used to jack a 920 kg car 14 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? m^2

(b) What is the work done in lifting the car 14 cm? (J)

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

(d) How many strokes are required to jack the car up 14 cm? (Include fractions of a stroke in your answer).

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

Explanation / Answer

(a) Fi / Ai = Fo / Ao

=> Ai = Fi Ao / Fo = 250 * * 0.092 / (920 * 9.81) = 7.05 * 10-4 m2

(b) Work done, W = Fod = (920 * 9.81) * 0.14 = 1263.5 J

(c) Car movement each stroke = (Ai / Ao) * 0.13 = [(7.05 * 10-4 / ( * 0.092)] * 0.13 = 0.0036 m

(d) Number of strokes required = 0.14 / 0.0036 = 38.88

(e) Combined work output, W = 250 * 38.88 * 0.13 = 1263.6 J

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