In car lift compressed air exerts a force `F_1` on a small piston having a radius of 5 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, what is `F_1 ?` What is the pressure necessary to ac complish this task ?

Explanation

To solve the problem, we need to find the force \( F_1 \) exerted by the compressed air on the small piston and the pressure necessary to lift the car.

### Step 1: Calculate the weight of the car
The weight \( F_2 \) of the car can be calculated using the formula:
\[ F_2 = m \cdot g \]
where:
- \( m = 1350 \, \text{kg} \) (mass of the car)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating \( F_2 \):
\[ F_2 = 1350 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 13230 \, \text{N} \]

### Step 2: Calculate the areas of the pistons
The area \( A \) of a piston is given by the formula:
\[ A = \pi r^2 \]
For the small piston (radius \( r_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)):
\[ A_1 = \pi (0.05)^2 = \pi (0.0025) \approx 0.00785 \, \text{m}^2 \]

For the large piston (radius \( r_2 = 15 \, \text{cm} = 0.15 \, \text{m} \)):
\[ A_2 = \pi (0.15)^2 = \pi (0.0225) \approx 0.07069 \, \text{m}^2 \]

### Step 3: Use Pascal's principle to find \( F_1 \)
According to Pascal's principle, the pressure is the same on both pistons:
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]
Rearranging the formula to find \( F_1 \):
\[ F_1 = F_2 \cdot \frac{A_1}{A_2} \]
Substituting the values:
\[ F_1 = 13230 \, \text{N} \cdot \frac{0.00785}{0.07069} \]
Calculating \( F_1 \):
\[ F_1 \approx 13230 \, \text{N} \cdot 0.111 \approx 1475.5 \, \text{N} \]

### Step 4: Calculate the pressure necessary to lift the car
The pressure \( P \) can be calculated using the formula:
\[ P = \frac{F_1}{A_1} \]
Substituting the values:
\[ P = \frac{1475.5 \, \text{N}}{0.00785 \, \text{m}^2} \approx 187,000 \, \text{Pa} \text{ or } 1.87 \times 10^5 \, \text{Pa} \]

### Final Results
- The force \( F_1 \) exerted by the compressed air on the small piston is approximately \( 1475.5 \, \text{N} \).
- The pressure necessary to lift the car is approximately \( 1.87 \times 10^5 \, \text{Pa} \).