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28 February, 10:04

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

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  1. 28 February, 10:25
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    Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

    Answer:

    225 N

    Explanation:

    From Pascal's principle,

    F/A = f/a ... Equation 1

    Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

    Making f the subject of the equation,

    f = F (a) / A ... Equation 2

    Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

    Substituting into equation 2

    f = 15000 (3/200)

    f = 225 N.

    Hence the downward force that must be applied to small piston = 225 N
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