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.

Explanation

Force applied to the smaller piston = `F_1` = 50 kgf
Diameter of smaller piston = `5 "cm" xx 10^(-2)` m
Radius of smaller piston = 2.5 cm = `2.5 xx 10^(-2)` m
Area of smaller piston = `A_1=pixx (2.5)^2 xx 10^(-4) m^2`
Force exerted on the larger piston = `F_2`
Diameter of smaller piston = 25 cm = `25 xx 10^(-2)` m
Radius of smaller piston = 12.5 cm = `12.5 xx 10^(-2)` m
Area of smaller piston = `A_1 = pi (12.5)^2 xx 10^(-4) m^2`
Pressure on smaller piston = Pressure on larger piston
Or, `F_1//A_1=F_2//A_2`
Or, `50//[pixx(2.5)^2xx10^(-4)]=F_2//[pixx(12.5)^2xx10^(-4)]`
Or, `F_2=50xx(12.5)^2//(2.5)^2`
`therefore F_2` = 1250 kgf