The small piston of a hydraulic press has an area of 3.2 in2. If the applied force is 2.0 lb, what must the area of the large piston be to exert a pressing force of 795 lb?

Respuesta :

Given data:

The area of the hydraulic press is,

[tex]A_1=3.2in^2[/tex]

The applied force is,

[tex]F_1=2\text{ lb}[/tex]

The pressing force on the large piston is

[tex]F_2=795\text{ lb}[/tex]

Given hydraulic press works on Pascal's law principle.

From the Pascal's law, the equation can be given as,

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

Here,

[tex]A_2[/tex]

is the area of the larger piston.

Substituting the values in the above equation, we get:

[tex]\begin{gathered} \frac{2\text{ lb}}{3.2in^2}=\frac{795\text{ lb}}{A_2} \\ A_2=1272in^2 \end{gathered}[/tex]

Thus, the area of the large piston is

[tex]1272in^2[/tex]