Question

A force of 250N is applied to a hydraulic jack piston wth

a0.1m diameter. If the output piston, which supports the load,has a
diameter of .5m, how much weight can be lifted?

Answer 1

To solve this problem we will apply the Pascal principle. This law establishes that the pressure exerted on an incompressible fluid and in equilibrium within a container of non-deformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Mathematically the relationship is given as

`(F_1)/(A_1) = (F_2)/(A_2)`

Since it is a circular piston the area will be given as

`A = \pi ((d)/(2))^2`

Replacing,

`(F_1)/(\pi ((d_1)/(2))^2 ) = (F_2)/(\pi ((d_2)/(2))^2 )`

`F_2 = ((d_2)/(d_1))^2 F_1`

Our values are given as

`F_1 = 250N\\ d_2 = 0.5m\\d_1 = 0.1m`

Replacing we have that

`F_2 = ((0.5)/(0.1))^2(250)`

`F_2 = 6250N`

Therefore the wieght that can be lifted is 6250N