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Homework Help: Hydraulic lift

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data

    In a hydraulic lift the radii of the pistons are 1.75cm and 10.5cm. A car weighing 18.5kN is to be lifted by the force of the large piston.

    a.) what force must be applied to the small piston?

    b.) when the small piston is pushed 1.00cm, how far is the car lifted?

    c.) find the mechanical advantage of the lift (this is the ratio of the large force to the small force)


    3. The attempt at a solution

    a.) F= (r/R)^2 (F)
    = (1.75 cm/10.5 cm)^2 (18.5 kN)
    = .514 or 514 N

    b.) Win(f)= Wout(F)= fh=FH=H= (f/F)h= (pi)r^2/(piR^2=(r/R)^2
    = (1m/R)^2 = (100cm/10.5cm)^2
    = 90.7cm

    c.) MA= F/f= 18.5kN/1.75cm
    = 10.57


    I have no clue if I am doing this right.. can anyone help?
     
    Last edited: Jul 20, 2010
  2. jcsd
  3. Jul 19, 2010 #2

    Doc Al

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    Staff: Mentor

    Good, but careful with units. (Those are kN, not N.)

    FH = fh is correct. What's the ratio of the forces?

    What's the ratio of the forces?
     
  4. Jul 19, 2010 #3
    I'm not sure how to find that?
     
  5. Jul 19, 2010 #4

    Doc Al

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    Staff: Mentor

    Use what you found in part a.
     
  6. Jul 19, 2010 #5
    is it 1/10?
     
  7. Jul 19, 2010 #6

    Doc Al

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    Staff: Mentor

    Nope. You were given one force and for part a you found the other force. What's their ratio?
     
  8. Jul 19, 2010 #7
    14.5/101
     
  9. Jul 20, 2010 #8
    I just realized my problem had the wrong numbers- so I edited the problem...

    For part b.) is my ratio now 18.5/514????
     
  10. Jul 20, 2010 #9
    (18.5/514)^2 ??

    and then I think c.) is MA=F/f = 18.5/.514
    = 35.9

    Can anyone help?
     
  11. Jul 20, 2010 #10
    Trying to figure this out:

    b.) (r/R)^2
    so, therefore it should be (1.75/10.5)^2
    which = .0278

    c.) MA= F/f = 18.5/.514 = 35.9

    ????
     
  12. Jul 20, 2010 #11

    Doc Al

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    Staff: Mentor

    That's fine. The ratio of forces is given by (R/r)^2 = (10.5/1.75)^2 = 6^2 = 36.

    So for b, the large piston will raise by 1/36 of the distance the small piston is lowered.

    And for c, the mechanical advantage is just that ratio of forces.
     
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