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?

Respuesta :

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

[tex]\frac{F_1}{A_1} = \frac{F_2}{A_2}[/tex]

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

[tex]A = \pi (\frac{d}{2})^2[/tex]

Replacing,

[tex]\frac{F_1}{\pi (\frac{d_1}{2})^2 } = \frac{F_2}{\pi (\frac{d_2}{2})^2 }[/tex]

[tex]F_2 = (\frac{d_2}{d_1})^2 F_1[/tex]

Our values are given as

[tex]F_1 = 250N\\ d_2 = 0.5m\\d_1 = 0.1m[/tex]

Replacing we have that

[tex]F_2 = (\frac{0.5}{0.1})^2(250)[/tex]

[tex]F_2 = 6250N[/tex]

Therefore the wieght that can be lifted is 6250N