Question

The area of cross section of the large piston in a hydraulic machine is 1m² and that of small piston is 0.5m².calculate the force that must be applied on the small piston to lift a load of 1000N on the large piston

Answer 1

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:

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ (1)

Where:

$F_{1}, F_{2}$ - Forces applied on the small and large pistons, in newtons.

$A_{1}, A_{2}$ - Areas of the small and large pistons, in square meters.

If we know that $A_{1} = 0.5\,m^{2}$, $A_{2} = 1\,m^{2}$ and $F_{2} = 1000\,N$, then the force to be applied on the small piston is:

$F_{1} = \left(\frac{A_{1}}{A_{2}} \right)\cdot F_{2}$

$F_{1} = \left(\frac{0.5\,m^{2}}{1\,m^{2}} \right)\cdot (1000\,N)$

$F_{1} = 500\,N$

The force that must be applied on the small piston has a magnitude of 500 newtons.