Problem 39 - College Physics

Question

A hydraulic lift in a garage has two pistons: a small one of cross-sectional area \(4.00 \mathrm{~cm}^{2}\) and a large one of cross-sectional area \(250 \mathrm{~cm}^{2}\). (a) If this lift is designed to raise a 3500 -kg car, what minimum force must be applied to the small piston? (b) If the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?

Step-by-Step Solution

Answer

Apply 549.36 N to the small piston, requiring air pressure of 1373400 Pa.

Step 1: Understand the Problem

We are given a hydraulic lift with a small piston of cross-sectional area \(4.00 \mathrm{\: cm}^2\) and a large piston of cross-sectional area \(250 \mathrm{\: cm}^2\). A 3500-kg car needs to be lifted using this setup. We need to find the minimum force applied to the small piston (part a) and the air pressure needed to apply this force (part b).

Step 2: Find the Required Force on the Large Piston

First, calculate the force required to lift the car. The force needed to lift the car is equal to its weight, which can be calculated using the formula:\[F_{ ext{large}} = m \cdot g = 3500 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 34335 \text{ N}.\]

Step 3: Apply Pascal's Principle

According to Pascal's Principle, the pressure is the same in any point in the hydraulic fluid. Thus, the ratio of forces is equal to the ratio of cross-sectional areas:\[\frac{F_{ ext{small}}}{A_{ ext{small}}} = \frac{F_{ ext{large}}}{A_{ ext{large}}}.\]Substitute the values:\[F_{ ext{small}} = \frac{F_{ ext{large}} \cdot A_{ ext{small}}}{A_{ ext{large}}} = \frac{34335 \text{ N} \cdot 4.00 \text{ cm}^2}{250 \text{ cm}^2} = 549.36 \text{ N}.\]

Step 4: Determine the Required Pressure

Pressure is defined as force per unit area. To find the minimum air pressure applied to the small piston, use the formula:\[P = \frac{F_{ ext{small}}}{A_{ ext{small}}}.\]Convert the area from \( \text{cm}^2 \) to \( \text{m}^2 \):\[A_{ ext{small}} = 4.00 \text{ cm}^2 = 4.00 \times 10^{-4} \text{ m}^2.\]Substitute into the pressure equation:\[P = \frac{549.36 \text{ N}}{4.00 \times 10^{-4} \text{ m}^2} = 1373400 \text{ Pa} \text{ (Pascals)}.\]

Key Concepts

Pascal's Principle

Pascal's Principle is an essential concept in understanding how hydraulic systems work. It states that when pressure is applied to a confined fluid, the pressure change occurs throughout the entire fluid. This principle is the foundation of hydraulic systems that allow a small force applied at one point to be transformed into a larger force at another point.

In the context of a hydraulic lift, like the one in the exercise, Pascal's Principle means that any increase in pressure in the fluid due to a force applied on the small piston, transmits this pressure equally to the large piston. This allows for a heavy weight, such as a car, to be lifted by a relatively small force exerted on the smaller piston. By maintaining continuous pressure over different areas, hydraulic systems efficiently utilize Pascal's Principle to amplify force.

• This principle ensures the pressure remains constant throughout the hydraulic fluid.
• The transmitted pressure allows for force multiplication.
• It enables lifting heavy loads with smaller input forces.

Force Calculation

Calculating force in hydraulic systems is crucial for understanding how much effort is necessary to lift heavy objects. The force exerted by the larger piston is required to lift the car in this exercise, and it is equal to the weight of the car:

\[F_{\text{large}} = m \cdot g\]

Where:

• \( m \)
• is the mass of the car, \(3500 \text{ kg}\)
• \( g \)
• is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2\)

Substituting these values, the force required to lift the car is \(34335 \text{ N}\).
Using Pascal's Principle, we can then determine the force needed on the smaller piston:

\[F_{\text{small}} = \frac{F_{\text{large}} \cdot A_{\text{small}}}{A_{\text{large}}}\]

This formula shows how a smaller force is translated into a larger force by taking advantage of the differences in areas between the pistons. By applying these calculations, it's clear that a force of \(549.36 \text{ N}\) must be applied to lift the car.

Pressure Calculation

Pressure calculation is vital in hydraulic systems to ensure that the applied force is effective. Pressure is defined as the force per unit area. In this exercise, we determine the necessary air pressure to apply to the small piston using the formula:

\[P = \frac{F_{\text{small}}}{A_{\text{small}}}\]

Here, it's essential to convert the area unit from square centimeters to square meters to calculate accurately.

• Thus, \(A_{\text{small}} = 4.00 \text{ cm}^2 = 4.00 \times 10^{-4} \text{ m}^2\).
• Substituting the values, the pressure required is found to be \(1373400 \text{ Pa}\) (Pascals).

This calculation helps to determine the exact air pressure needed to achieve the force necessary to lift the car. Calculating pressure ensures that the system operates efficiently and safely within its capabilities.

Area Conversion

Area conversion is a crucial step in hydraulic system calculations, as it ensures consistency and accuracy in the mathematical equations used. Many problems involve converting units of area from square centimeters to square meters, which is the standard unit system for calculations in the International System of Units (SI).

In the exercise provided, converting the area of the small piston is necessary for accurate pressure calculations. This process involves multiplying the area in square centimeters by \(10^{-4}\) to convert it to square meters:

• \(4.00 \text{ cm}^2\)
• becomes \(4.00 \times 10^{-4} \text{ m}^2\)

Unit conversion is important because using different units in calculations can lead to errors, causing incorrect conclusions about the hydraulic system's efficiency and safety. Proper conversion ensures results align with expected standards and principles.