Answer:
The force that must be applied on the small piston has a magnitude of 500 newtons.
Step-by-step explanation:
By Pascal's Principle, we know that pressure within any point of a closed hydraulic system is the same. Hence, the hydraulic machine can be described by the following model:
[tex]\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}[/tex] (1)
Where:
[tex]F_{1}, F_{2}[/tex] - Forces applied on the small and large pistons, in newtons.
[tex]A_{1}, A_{2}[/tex] - Areas of the small and large pistons, in square meters.
If we know that [tex]A_{1} = 0.5\,m^{2}[/tex], [tex]A_{2} = 1\,m^{2}[/tex] and [tex]F_{2} = 1000\,N[/tex], then the force to be applied on the small piston is:
[tex]F_{1} = \left(\frac{A_{1}}{A_{2}} \right)\cdot F_{2}[/tex]
[tex]F_{1} = \left(\frac{0.5\,m^{2}}{1\,m^{2}} \right)\cdot (1000\,N)[/tex]
[tex]F_{1} = 500\,N[/tex]
The force that must be applied on the small piston has a magnitude of 500 newtons.
