In an hydraulic lift, the radii of the small and large pistons are R_1, and R_2,

Question

In an hydraulic lift, the radii of the small and large pistons are R_1, and R_2, respectively. A car of weight W is to be lifted by the force of the large piston. The fluid used in the lift is incompressible. What minimum force F_1 must be applied to the small piston to just begin lifting the car? Assume the two pistons are at the same vertical height. Write F_1 in terms of W and the areas of the two pistons. b) When the small piston is pushed down a distance y_1, how far up (y_2) does the large piston move? Write y_2 in terms of y_1 and the areas of the two pistons. c) If R_1 = 2.50 cm, R_2 = 10.0 cm, and W = 10.0 kN, calculate the mechanical advantage of the lift, which is the ratio W/F_1 (the amount of weight lifted divided by the force required to lift the weight). d) If y_1 = 10.0 cm, how far is the car lifted? e) How much work is done on the small piston? How much work does the large piston do on the car? Should these two works be the same? Why or why not?

Explanation / Answer

a) F1/a1=w/a2

F1=W/a2×a1

b) input volume =output volume

A1y1=A2y2

y2=A1y1/A2

C) F1/a1=w/a2

W/F1=a2/a1=pi×R2^2/pi×R1^2=0.1^2/0.025^2(R2=10cm=0.1m andR1=2.5cm=0.025m)

W/F1=16

d) A1y1=A2y2

pi×R1^2×y1=pi×R2^2×y2

0.025^2×0.1=0.1^2×y2

y2=0.00625m

e) work done on small piston F1×y1 =work done on large piston W×y2 acc to law of conservation of energy input energy is equal to output energy

W×y2=10×10^3×0.00625=62.5J

F1=w/16×y1=62.5J

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