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Question
A force of 50 kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5 cm and 25 cm respectively.
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Solution
Ratio of diameter of smaller piston to bigger piston = 5 : 25
∴ Ratio of area of smaller piston to bigger piston = 25 : 625
Force applied on smaller piston , F1 = 50 kgf
Let F2 be the force on the bigger piston .
By the principle of hydraulic machine ,
Pressure on narrow piston = pressure on wider piston
or , `F_1/A_1 = F_2/A_2`
or , `F_1/F_2 = A_1/A_2`
or , `50/F_2 = 25/625`
or , `F_2 = 50 xx 625/25 = 1250` kgf
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