You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a forc

Question

You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a force on a smaller piston of not more than 550N .a) What should be the specifications on the dimensions of the pistons?Asmall piston/Alarge piston = ???b) How far down will you need to push the piston in order to lift the car 50cm ?h = ???

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Thiên Thanh 4 years 2021-08-24T12:55:25+00:00 1 Answers 119 views 0

Answers ( )

  1. kiet
    1
    2021-08-24T12:56:58+00:00
    Reply

    Answer:

    (a) Area(small piston)/Area(large piston) = 0.037

    (b) h = 1336.36 cm = 13.36 m

    Explanation:

    (a)

    The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,

    Stress (small piston) = Stress (large piston)

    Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)

    Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)

    Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)

    Area(small piston)/Area(large piston) = 0.037

    (b)

    The work is also transmitted equally to the large piston. So,

    Work(small piston) = Work(Large Piston)

    Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)

    (550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)

    h = 735000 N.cm/550 N

    h = 1336.36 cm = 13.36 m

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