Q.In a car lift compressed air exerts a force $F_1$ on a small piston having a radius of $5.0\ \text{cm}$. This pressure is transmitted to a second piston of radius $15\ \text{cm}$. If the mass of the car to be lifted is $1350\ \text{kg}$, calculate $F_1$. What is the pressure necessary to accomplish this task? ($g = 9.8\ \text{m s}^{-2}$).

The key idea is Pascal's Principle, which states that pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.

1. The force exerted by the car on the large piston is $F_2 = mg = 1350\ \text{kg} \times 9.8\ \text{m s}^{-2} = 13230\ \text{N}$.
2. The areas of the small and large pistons are $A_1 = \pi (0.05\ \text{m})^2 = 0.0025\pi\ \text{m}^2$ and $A_2 = \pi (0.15\ \text{m})^2 = 0.0225\pi\ \text{m}^2$, respectively.
3. According to Pascal's Principle, the pressure is uniform throughout the fluid: $P = \frac{F_1}{A_1} = \frac{F_2}{A_2}$. …

A car lift is a classic example of a hydraulic system, which operates on a fundamental principle of fluid mechanics known as Pascal's Principle. This principle states that when pressure is applied to an enclosed incompressible fluid, that pressure is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.

Imagine you have two pistons of different sizes connected by a fluid-filled tube. If you push down on the smaller piston, the pressure you create in the fluid is instantly felt everywhere in that fluid, including under the larger piston. Because the pressure is the same on both pistons, but the larger piston has a much greater area, the force exerted on the larger piston will be proportionally greater. This allows a small input force to generate a large output force, making it possible to lift heavy objects like cars with relatively little effort.

Let's break down the problem step-by-step.

1. Identify Given Information and Convert Units:

   We are given the radii of the two pistons and the mass of the car. It's crucial to work in consistent SI units (meters, kilograms, seconds).

   • Radius of small piston, $r_1 = 5.0\ \text{cm} = 0.05\ \text{m}$
   • Radius of large piston, $r_2 = 15\ \text{cm} = 0.15\ \text{m}$
   • Mass of the car, $m = 1350\ \text{kg}$
   • Acceleration due to gravity, $g = 9.8\ \text{m s}^{-2}$

2. Calculate the Force Exerted by the Car ($F_2$):

   The car's weight is the force that needs to be lifted, and this force acts on the larger piston.

   $F_2 = \text{weight of car} = m g$

   $F_2 = (1350\ \text{kg}) \times (9.8\ \text{m s}^{-2})$

   $F_2 = 13230\ \text{N}$

3. Calculate the Areas of the Pistons:

   The pistons are circular, so their areas are given by $A = \pi r^2$.

   • Area of small piston, $A_1 = \pi r_1^2$

     $A_1 = \pi (0.05\ \text{m})^2$

     $A_1 = \pi (0.0025\ \text{m}^2)$

     $A_1 \approx 0.007854\ \text{m}^2$

   • Area of large piston, $A_2 = \pi r_2^2$

     $A_2 = \pi (0.15\ \text{m})^2$

     $A_2 = \pi (0.0225\ \text{m}^2)$

     $A_2 \approx 0.070686\ \text{m}^2$

4. Apply Pascal's Principle to Find $F_1$:

   According to Pascal's Principle, the pressure transmitted by the fluid is the same at both pistons: $P_1 = P_2$.

   Since pressure $P = F/A$, we have:

   $\frac{F_1}{A_1} = \frac{F_2}{A_2}$

   We want to find $F_1$, so we rearrange the formula:

   $F_1 = F_2 \left(\frac{A_1}{A_2}\right)$

   Substitute the calculated values:

   $F_1 = (13230\ \text{N}) \left(\frac{0.007854\ \text{m}^2}{0.070686\ \text{m}^2}\right)$ …

F2=mg=13509.8=13230N (car weight). A1=pi(0.05)^2, A2=pi*(0.15)^2. F1=F2*(r1/r2)^ …