A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000\ \text{kg}$. The area of cross-section of the piston carrying the load is $425\ \text{cm}^{2}$. What maximum pressure would the smaller piston have to bear?

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000kg. The area of cross-section

Exercises · 9.8

3000\ \text{kg}

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Concept understanding — Pascals Principle

Pascal's Principle: The Intuition

Imagine you have a sealed plastic bottle completely filled with water, with no air inside. If you squeeze the bottle at one spot, what happens? The water doesn't just compress at that spot — the entire bottle bulges outward everywhere. The pressure you applied at one point travels instantly through the entire liquid.

That's the core idea: in a confined fluid, pressure applied anywhere is transmitted undiminished to every part of the fluid and to the walls of the container.

Now, why does this happen? Fluids (liquids and gases) are made of particles that can slide past each other. When you push on one part of a confined fluid, those particles bump into their neighbours, which bump into their neighbours, and so on — like a chain of dominoes. Because the fluid is nearly incompressible (especially liquids), the push doesn't get absorbed by squishing the fluid itself. It gets passed along perfectly.

Note

This principle works only for confined fluids — fluids trapped inside a container with no way to escape. An open bucket of water doesn't behave this way because the water can just spill out.

The Precise Statement

P_{\text{applied}} = \frac{F}{A}

In simpler words: if you increase the pressure at one point by $\Delta P$ , every other point in the fluid experiences exactly the same increase $\Delta P$ .

The Key Equation

The pressure $P$ is defined as force per unit area:

P = \frac{F}{A}

So if you apply a force $F_1$ over a small area $A_1$ , you create a pressure $P = F_1/A_1$ . That same pressure pushes on a larger area $A_2$ elsewhere, producing a larger force:

F_2 = P \times A_2 = \frac{F_1}{A_1} \times A_2

This is the force multiplication that makes hydraulic systems work.

Tip

Think of it like this: pressure is the "currency" that gets transferred. Force is what you get when you "cash in" that pressure over a bigger area.

The Classic Example: Hydraulic Lift

A car repair shop uses a hydraulic jack. Here's how it works:

You push down on a small piston (area $A_1 = 10\ \text{cm}^2$ ) with a force of $100\ \text{N}$ .
This creates a pressure $P = 100\ \text{N} / 0.001\ \text{m}^2 = 100,000\ \text{Pa}$ .
That same pressure pushes up on a large piston (area $A_2 = 1000\ \text{cm}^2 = 0.1\ \text{m}^2$ ).
The upward force is $F_2 = P \times A_2 = 100,000 \times 0.1 = 10,000\ \text{N}$ .

You applied $100\ \text{N}$ and got $10,000\ \text{N}$ — a 100× force multiplication. But notice: the small piston moves down 100 cm while the large piston moves up only 1 cm. Work done ( $F \times d$ ) is the same on both sides — you don't get energy for free.

Watch out

Pascal's Principle does not violate energy conservation. Force is multiplied, but distance is divided by the same factor. The product $F \times d$ (work) remains equal.

Where It Applies (and Where It Doesn't)

Applies	Does Not Apply
Hydraulic brakes in cars	Open containers (water in a glass)
Hydraulic jacks and lifts	Gases that are not confined
Dental chairs	Fluids under very high compression (rare)

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A hydraulic lift operates on Pascal's Principle, meaning the pressure exerted by the car on the large piston is transmitted undiminished throughout the fluid to the smaller piston. We calculate this pressure by dividing the car's weight by the area of the large piston, resulting in a maximum pressure of $\boxed{6.92 \times 10^5\ \text{Pa}}$ .

A hydraulic lift is a classic application of Pascal's Principle, a fundamental concept in fluid mechanics. 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 the walls of the containing vessel.

Imagine squeezing a toothpaste tube. The pressure you apply at one end is felt throughout the toothpaste, pushing it out the other end. In a hydraulic lift, a small force applied over a small area on one piston creates a certain pressure. This pressure is then transmitted through the hydraulic fluid to a larger piston, where the same pressure acts over a much larger area, generating a significantly larger force capable of lifting heavy objects like cars.

The problem asks for the maximum pressure the smaller piston would have to bear. By Pascal's Principle, the pressure exerted by the fluid on the larger piston (due to the car's weight) is the same pressure that the smaller piston must generate to lift the car. Therefore, we need to calculate the pressure exerted by the car on the larger piston.

P_1 = P_2

Let's break down the calculation:

Identify Given Values and Target:
- Maximum mass of the car ( $m$ ) = $3000\ \text{kg}$
- Area of cross-section of the piston carrying the load ( $A$ ) = $425\ \text{cm}^{2}$
- Acceleration due to gravity ( $g$ ) = $9.8\ \text{m/s}^2$ (standard value, unless specified otherwise)
- We need to find the maximum pressure ( $P$ ) the smaller piston would have to bear, which is equal to the pressure exerted by the car on the larger piston.
Convert Units to SI:

The mass is already in kilograms ( $\text{kg}$ ), which is an SI unit. However, the area is given in square centimeters ( $\text{cm}^2$ ), which needs to be converted to square meters ( $\text{m}^2$ ) for consistency with SI units (where pressure is in Pascals, $\text{Pa}$ , which is $\text{N/m}^2$ ).

We know that $1\ \text{m} = 100\ \text{cm}$ .

Therefore, $1\ \text{m}^2 = (100\ \text{cm})^2 = 10000\ \text{cm}^2$ .

So, $1\ \text{cm}^2 = \frac{1}{10000}\ \text{m}^2 = 10^{-4}\ \text{m}^2$ .

$A = 425\ \text{cm}^2 = 425 \times 10^{-4}\ \text{m}^2 = 0.0425\ \text{m}^2$ . …

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Q.A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg3000\ \text{kg}3000 kg. The area of cross-section of the piston carrying the load is 425 cm2425\ \text{cm}^{2}425 cm2. What maximum pressure would the smaller piston have to bear?

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Q.A hydraulic automobile lift is designed to lift cars with a maximum mass of $3000\ \text{kg}$ . The area of cross-section of the piston carrying the load is $425\ \text{cm}^{2}$ . What maximum pressure would the smaller piston have to bear?

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