The problem describes a hydraulic lift used in a vehicle service station. The smaller piston has a radius $r_1 = 3 \text{ cm}$ and is operated by air pressure. When the smaller piston is pushed down, the larger piston with radius $r_2 = 18 \text{ cm}$ lifts a car with a weight of $36000 \text{ N}$. We are asked to: i. Define pressure. ii. Calculate the pressure exerted on the larger piston due to the car's weight. iii. Calculate the force that needs to be applied on the smaller piston by the air. iv. If 10% of the applied force on each piston is used to overcome friction, calculate the new force that needs to be applied to the smaller piston to lift the car.

1. Problem Description

The problem describes a hydraulic lift used in a vehicle service station. The smaller piston has a radius $r_1 = 3 \text{ cm}$ and is operated by air pressure. When the smaller piston is pushed down, the larger piston with radius $r_2 = 18 \text{ cm}$ lifts a car with a weight of $36000 \text{ N}$. We are asked to:

i. Define pressure.

ii. Calculate the pressure exerted on the larger piston due to the car's weight.

iii. Calculate the force that needs to be applied on the smaller piston by the air.

iv. If 10% of the applied force on each piston is used to overcome friction, calculate the new force that needs to be applied to the smaller piston to lift the car.

2. Solution Steps

i. Pressure Definition:

Pressure is defined as the force acting perpendicularly per unit area.

$P = \frac{F}{A}$, where $P$ is pressure, $F$ is force, and $A$ is area.

ii. Pressure on the Larger Piston:

The force on the larger piston is the weight of the car, $F_2 = 36000 \text{ N}$. The area of the larger piston is $A_2 = \pi r_2^2 = \pi (18 \text{ cm})^2 = \pi (0.18 \text{ m})^2$. Therefore, the pressure on the larger piston is:

$P_2 = \frac{F_2}{A_2} = \frac{36000 \text{ N}}{\pi (0.18 \text{ m})^2}$

$P_2 = \frac{36000}{\pi \times 0.0324} \text{ Pa} \approx 353677.65 \text{ Pa}$

iii. Force on the Smaller Piston:

According to Pascal's principle, the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Therefore, the pressure on the smaller piston is equal to the pressure on the larger piston: $P_1 = P_2$.

The area of the smaller piston is $A_1 = \pi r_1^2 = \pi (3 \text{ cm})^2 = \pi (0.03 \text{ m})^2$.

The force on the smaller piston is:

$F_1 = P_1 A_1 = P_2 A_1 = \frac{36000}{\pi (0.18)^2} \times \pi (0.03)^2 = 36000 \times \frac{(0.03)^2}{(0.18)^2} = 36000 \times \frac{0.0009}{0.0324} = 36000 \times \frac{9}{324} = 36000 \times \frac{1}{36} = 1000 \text{ N}$

iv. Force on Smaller Piston with Friction:

10% of the force is used to overcome friction. Let $F_1'$ be the new force applied to the smaller piston. Only 90% of this force is effective in creating pressure. Let the effective force be denoted $F_{1,eff} = 0.9 F_1'$. This produces pressure $P_1' = F_{1,eff} / A_1$. Since the force at the larger piston also has 10% friction loss, the effective force is $F_{2, eff} = 0.9 \times 36000 = 32400 \text{ N}$. Thus $P_2'=F_{2,eff} / A_2$.

$P_1' = P_2'$, so $\frac{0.9F_1'}{A_1} = \frac{0.9 F_2}{A_2}$. Hence $\frac{F_1'}{A_1} = \frac{F_2}{A_2}$.

$F_1' = \frac{A_1}{A_2}F_2 = \frac{\pi r_1^2}{\pi r_2^2}F_2 = \frac{r_1^2}{r_2^2}F_2 = \frac{(3)^2}{(18)^2}(36000) = \frac{9}{324}(36000) = \frac{1}{36}(36000) = 1000 \text{ N}$. This is incorrect.

The intended meaning of the problem is that we need to find the force applied on the small piston that when we account for the friction force on both pistons it is equal to 36000 N.

Friction on large piston = $0.1*36000 = 3600$ N. This means force to be applied by hydraulic system = $36000 + 3600 = 39600$ N.

$F_{1,eff} / A_1=F_{2,eff}/A_2$

$F_{1,eff}= \frac{A_1}{A_2} F_{2,eff} = \frac{\pi * 3^2}{\pi * 18^2} 39600 = \frac{1}{36} 39600= 1100 \text{ N}$

$0.9F_1' = 1100$

$F_1'=1100/0.9 \approx 1222.22 \text{ N}$

3. Final Answer

i. Pressure is defined as the force acting perpendicularly per unit area.

ii. The pressure exerted on the larger piston due to the car's weight is approximately $353677.65 \text{ Pa}$.

iii. The force that needs to be applied on the smaller piston by the air is $1000 \text{ N}$.

iv. The new force that needs to be applied to the smaller piston to lift the car is approximately $1222.22 \text{ N}$.