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Force must be applied to the small piston to lift the engine?

  1. Jan 25, 2006 #1
    In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sectional area of 0.057m^2 and a large piston with a cross-sectional area of 0.208m^2. An engine weighing 2900N rests on the large piston. What force must be applied to the small piston to lift the engine? If the engine rises 0.235m, how far does the smaller piston move?

    I correctly found the force to be 794.712N. I thought that the smaller piston moves the same length as the engine rising, but I could be wrong. I don't know how to figure that out.
    Last edited by a moderator: Mar 7, 2013
  2. jcsd
  3. Jan 26, 2006 #2


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    By applying Pascal's principle to the hydraulic jack we get that the change in pressure at the input piston is equal to the pressure change at the output piston , therefore the input force is magnified by the ratio of the areas of the pistons at the output piston. The volume of oil displaced at the input piston will be equal to the volume of oil rising at the output piston as the engine is lifted up. From this you can find the relationship between the areas of the pistons and the distances the two pistons move during the lifting process. Notice what this new relationship obtained this way tells us something about the work done during the lifting process.
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