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

F = 1076 N

Step-by-step 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)/(W)=(a)/(A)`

`(F)/(3400)=(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.

Answer 2

`F_s=1075.9493\ N`

Step-by-step 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`

`(F_s)/(a_s)=(W_l)/(a_l)`

`(F_s)/(0.075) =(3400)/(0.237)`

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