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Homework Help: Pascal's Principle?

  1. Dec 13, 2007 #1

    (a) If a 22 pound baby sits atop a 1 ft^2 piston that is connected to a 10 ft^2 platform via sealed, water filled pipes, how heavy and massive of a car can the baby's weight lift (answer in pounds, Newtons, and Kilograms)?

    (b) If the baby's initial height on the piston is 10 ft, how high will the car be raised after the baby's piston is completely compressed?

    3. a)
    F1/A1=F2/A2 or
    P1 = P2

    Let baby sitting on 1ft^2 platform be P1 --> 22lbs/144in^2 = .153 lbs/in^2

    Let car on platform be P2 -->
    weight of car/14400in^2 =P2
    --> weight of car = 14400in^2 X .153 lbs/in^2 = 2,203.2 lbs.

    Powerful baby.

    I assumed pounds in this problem meant pounds force rather than pounds mass. 1 pound force = 4.45 N so 22 lb = 97.9 N, and the mass that would give 97.9 N is 97.9N/9.8m/s^2 = 9.99 kg


    If the baby\'s piston is completely compressed, the volume of water the baby\'s piston held will now be in the other piston - the one holding the car on a platform. First you figure out what volume of water would be in the baby's piston before it was compressed. So you had a cylinder with a cross sectional area of 1 ft^2 and a height of 10 ft, so that is a volume of 1ft^2X10ft = 10ft^3.

    That\'s the volume that will be transferred to the other cylinder, but since that cylinder has a larger cross sectional area, the height the water will rise will be less than 10 ft.

    Volume = Area of base times height

    10ft^3 = 10ft^2 X H -->

    H = 1 ft. The car will be raised 1 ft.

    Are my answers/solutions correct?? Or did I make a mistake? Please help me...thanks in advance!
  2. jcsd
  3. Dec 13, 2007 #2


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    There is a big difference between (10ft)^2 and 10ft^2. And you seem to be using one in a) and another in b). Also in a) you are asked for the stats (in lbs, N and kg) for the car, not the baby. Further, there is no need to convert to in^2. Overall, this is kind of confusing. Can we focus on one problem. Like for example, in a) you say the baby can lift something 100 times it's mass, in b) you say the distance displaced by the car is 1/10 that of the baby. Those two are definitely in conflict.
  4. Dec 14, 2007 #3
    is this correct??

    a>. Applying pascals law , pressure transmitted will be equal through sealed fluid at two pistons
    we get f/a} ={F/A,
    f,F are force exerted at smaller & larger cross section & a ,A are cross sectional areas resp.
    force exerted at larger cross section F =( f/a) x A
    F= (22 x g) x 10 = 220 x g wt pounds = 220 pounds(mass) car will be lifted.

    In Kg : 220 x0.46=101.2 kg , In Newton 101.2 x9.8 = 991.76N
  5. Dec 14, 2007 #4


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    Yes. The ratio of the two forces is equal to the ratio of the areas. Likewise, the ratio of the displacements. If you let v=volume, then volume=AH where H is the change in height. So V=v means AH=ah so H=ah/A.
  6. Dec 14, 2007 #5
    That makes more sense.
    I didn\'t need to convert to inches. I think I wanted to deal with psi, but it isn\'t really necessary since you don\'t require the conversion. And I don’t know why I converted the baby’s weight to newtons and kilograms instead of the car’s! But this time I converted the car’s weight.

    a) So P1=P2 --> 22lb/1ft^2 = x/10ft^2 --> x = 2,200 lbs.

    and 2,200 lbs. X 4.45 N/lb = 9790N

    9790N/9.8 m/s^2 = 998.98 kg
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