In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. the system has a small piston with a cross-sectional area of 0.075 m2 and a large piston with a cross-sectional area of 0.237 m2 . an engine weighing 3400 n rests on the large piston. what force must be applied to the small piston in order to lift the engine? answer in units of n.

Explanation:

Given:

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

is the required effort force.

   F = 1076 N

Explanation:

given,

small piston area, a = 0.075 m²

large piston area, A = 0.237 m²

weight on the large piston, W = 3400 N

force applied on the second piston, F = ?

using pascal law for the force calculation

   F = 0.3165 x 3400

   F = 1076 N

The force applied to the small piston in order to lift the engine is equal to 1076 N.

volts times amps 6 x 0.5 3w