In a car lift,compressed air exerts a force $F_{1}$ on a small piston having a radius of $5.0 \; cm$. This pressure is transmitted to a second piston of radius $15 \; cm$. If the mass of the car to be lifted is $1350 \; kg$,calculate $F_{1}$. What is the pressure necessary to accomplish this task? $(g = 9.8 \; m s^{-2})$

(N/A) According to Pascal's Law,pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
Given:
Radius of small piston,$r_{1} = 5.0 \; cm = 5.0 \times 10^{-2} \; m$
Radius of large piston,$r_{2} = 15 \; cm = 15 \times 10^{-2} \; m$
Mass of the car,$m = 1350 \; kg$
Acceleration due to gravity,$g = 9.8 \; m s^{-2}$
The force exerted by the car on the large piston is $F_{2} = m \times g = 1350 \times 9.8 = 13230 \; N$.
Using the principle of hydraulic lift,$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$,where $A_{1} = \pi r_{1}^{2}$ and $A_{2} = \pi r_{2}^{2}$.
$F_{1} = F_{2} \times \frac{A_{1}}{A_{2}} = F_{2} \times \left(\frac{r_{1}}{r_{2}}\right)^{2}$
$F_{1} = 13230 \times \left(\frac{5}{15}\right)^{2} = 13230 \times \left(\frac{1}{3}\right)^{2} = 13230 \times \frac{1}{9} = 1470 \; N$.
Rounding to two significant figures,$F_{1} \approx 1.5 \times 10^{3} \; N$.
The pressure $P$ required is $P = \frac{F_{1}}{A_{1}} = \frac{1470}{\pi \times (5.0 \times 10^{-2})^{2}} = \frac{1470}{3.14159 \times 25 \times 10^{-4}} \approx 1.87 \times 10^{5} \; Pa \approx 1.9 \times 10^{5} \; Pa$.