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Question
What force is applied on a piston of area of cross section 2 cm2 to obtain a force 150 N on the piston of area of cross section 12 cm2 in a hydraulic machine?
Solution
In a hydraulic machine
Pressure on narrow piston = Pressure on wider piston
∴ P1 = P2
∴ `F_1/A_1 = F_2/A_2`
∴ `F_1/(2 xx 10^-4) = 150/(12 xx 10^-4)`
∴ `F_1 = 150/(12 xx 10^-4) xx 2 xx 10^-4`
∴ `F_1 = 150/12 xx 2 = 25` N
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