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piston 1 has a diameter of 0.25 in; pistol 2 has a diameter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.

  • physics -

    I will have to guess what your problem is because if you tried to copy and paste a picture it did not work.
    If we are talking a hydraulic problem with force on the small piston lifting a weight on the big one then
    F =(little area/big area) weight
    F = (.25^2/1.5^2) 500
    F = [ 1/(4*2.25) ] 500
    F = 55.56 lb

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