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Question
The areas of pistons in a hydraulic machine are 5 cm2 and 625 cm2. What force on the smaller piston will support a load of 1250 N on the larger piston? State any assumption which you make in your calculation.
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Solution
Area of small piston A1 = 5 cm2
Area of wider piston, A2 = 625 cm2
Force on small piston be F1
Force on wider piston or load, F2 = 1250 N
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_1/5 = 1250/625`
or, `F_1 = 1250/625 xx 5`
or `F_1 = 10` N
Assumption: No friction or leakage of liquid happens.
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