A hydraulic lift is used to jack a 870 kg car 9 cm off the floor. The diameter o

Question

A hydraulic lift is used to jack a 870 kg car 9 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?
m2
(b) What is the work done in lifting the car 9 cm?
J
(c) If the input piston moves 13 cm in each stroke, how high does the car move up for each stroke?
m
(d) How many strokes are required to jack the car up 9 cm? (Include fractions of a stroke in your answer).

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

Explanation / Answer

Here is my reasoning: 1. The output force = m*g = 870kg*9.8m/s^2 = 8530N; the input force = 250N..so the ratio of forces is 8530N/250N = 34.1.. a)S o the area of the input = area of the output/34.1= (pi*9^2)/34.1 = 7.46cm^2 b)W = m*g*y = 870kg*9.8m/s^2*0.9m = 7673J c)The volume of fluid moved is constant so A*h)input = A*h)output.. If h input = 13cm, then the h output = Ain/Aout*hin = hin/34.1 = 13cm/34.1 = 0.381cm d) Now 11cm = n*0.381cm => n = 9/0.381 = 23.6= 24 strokes e)W input = 24*0.381x10^-2*250N = 22.86J

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