Question

In a hydraulic press, a load of 2000 kgf is to be balanced on the piston area of 80 cm2 by a force of 100 kgf on piston of smaller area. Find the area of smaller piston.

Answer 1

The area of smaller piston is 40 x 10^-4 m^2.

Explanation:

Given data:

Mass of load = 2000 Kg

Force "F1"= mg = 2000 x 10 = 20,000 N

Area of larger piston "A1" = 80 cm^2 = 80 x 10^-4 m^2

Mass of smaller load = 1000 Kg

Force "F2" = mg = 1000 x 10 = 10000 N

Area of smaller piston "A2" = ?

We know that

F1 / A1 = F2 / A2

F1 x A2 = F2 x A1

A2 =  F2 x A1 / F1

A2 = 10000 x 80 x 10^-4 / 20000

A2 = 80 x 10^-4 / 2

A2 = 40 x 10^-4 m^2

Hence the area of smaller piston is 40 x 10^-4 m^2.

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