Question

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.

Answer 1

Answer:

$F_s=1075.9493\ N$

Explanation:

Given:

• area of piston on the smaller side of hydraulic lift, $a_s=0.075\ m^2$

• area of piston on the larger side of hydraulic lift, $a_l=0.237\ m^2$

• Weight of the engine on the larger side, $W_l=3400\ N$

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.

$P_s=P_l$

$\frac{F_s}{a_s}=\frac{W_l}{a_l}$

$\frac{F_s}{0.075} =\frac{3400}{0.237}$

$F_s=1075.9493\ N$ is the required effort force.

Answer 2

Answer:

   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

$\dfrac{F}{W}=\dfrac{a}{A}$

$\dfrac{F}{3400}=\dfrac{0.075}{0.237}$

   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.