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Hydraulic lift?

Hydraulic lift has two pistons; small one of cross-sectional area of 4cm^2 and large one cross-sectional area 250cm^2. Designed to raise 3500kg car, what minimum force must be applied to the small piston?

Update:

Also, if the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?

Update 2:

I know the answers are: 549N and 1.37x10^6 Pa. Just need to know how to get there.

4 Answers

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  • Favorite Answer

    How do you know what the answers are if you don't know how to get there? If you wanted some one to confirm your work that I could understand, but your answers are correct, so odds are you did it right

    the force on the piston is easy 3500kgf = 34,300N, 34,300N * (4 / 250) = 548.8N or 549N

    Another answerer gave a correct equation to find the pressure in kgfcm^2 ; 4/250 x 3500/4 kg

    since the 4cm^2 area of the small piston cancels, I'd use 34,300N / 250cm^2 = 137.2Ncm^2 to get Pa multiply by 1x10^4, therefore 1.37 X 10^6Pa

  • 1 decade ago

    Force = 3500 x 9.8 x (4/250) N = 5488 N

    Pressure = 5488 / 4 x 10^(-4) Pa = 1.22 x 10^(-2) Pa.

  • Anonymous
    1 decade ago

    4/250 x 3500kg

    4/250 x 3500/4 kg

  • 1 decade ago

    i think 70 N

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