Problem 35 - University Physics with Modern Physics

Question

Three identical 8.50-kg masses are hung by three identical springs (\(\textbf{Fig. E6.35}\)). Each spring has a force constant of 7.80 kN/m and was 12.0 cm long before any masses were attached to it. (a) Draw a free-body diagram of each mass. (b) How long is each spring when hanging as shown? (\(Hint\): First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)

Step-by-Step Solution

Answer

Spring 1 is 13.07 cm, Spring 2 is 14.14 cm, and Spring 3 is 15.20 cm long when the masses are attached.

Step 1: Understanding the Problem

We have three identical 8.50-kg masses attached to three identical springs. Each spring has a spring constant \( k = 7.80 \text{ kN/m} = 7800 \text{ N/m} \) and an initial length before stretching \( L_0 = 12.0 \text{ cm} \). We need to find the length of each spring when the masses are attached.

Step 2: Analyzing Free-Body Diagrams

For each mass, the forces acting on it are the gravitational force downward and the spring force upward. For mass 1 (bottom mass), the spring force is equal to the weight of the mass, \( F_{s1} = mg \). For mass 2, the spring force is the weight of mass 2 plus mass 1, \( F_{s2} = 2mg \). For mass 3 (top mass), the spring force is the total weight of all three masses, \( F_{s3} = 3mg \).

Step 3: Calculating the Force on Each Spring

The gravitational force on each mass is \( F_g = mg = 8.50 \text{ kg} \times 9.8 \text{ m/s}^2 = 83.3 \text{ N} \). Using Hooke's Law \( F = kx \), the force needed to stretch each spring can be calculated as: - For spring 1: \( F_{s1} = 83.3 \text{ N} \)- For spring 2: \( F_{s2} = 2 \times 83.3 \text{ N} = 166.6 \text{ N} \)- For spring 3: \( F_{s3} = 3 \times 83.3 \text{ N} = 249.9 \text{ N} \)

Step 4: Finding the Extension for Each Spring

Using \( F = kx \), solve for \( x \), the extension of each spring:- For spring 1: \( x_1 = \frac{83.3 \text{ N}}{7800 \text{ N/m}} = 0.01068 \text{ m} \)- For spring 2: \( x_2 = \frac{166.6 \text{ N}}{7800 \text{ N/m}} = 0.02136 \text{ m} \)- For spring 3: \( x_3 = \frac{249.9 \text{ N}}{7800 \text{ N/m}} = 0.03204 \text{ m} \)

Step 5: Calculating the Total Length of Each Spring

The total length of each spring when stretched is its original length plus the extension:- Spring 1: \( L_1 = 0.12 \text{ m} + 0.01068 \text{ m} = 0.13068 \text{ m} \)- Spring 2: \( L_2 = 0.12 \text{ m} + 0.02136 \text{ m} = 0.14136 \text{ m} \)- Spring 3: \( L_3 = 0.12 \text{ m} + 0.03204 \text{ m} = 0.15204 \text{ m} \)

Key Concepts

Free-Body Diagram

Understanding a free-body diagram is crucial when solving physics problems involving forces. It is a visual tool used to depict all the forces acting on a single object. In our problem, each of the three masses will have its own free-body diagram. Here's why it is important:

- A free-body diagram allows you to identify all forces acting on an object, which typically include gravitational force and spring force in this context. - For each of the masses, you must consider the force of gravity pulling down and the spring force exerting upward pull. For the bottom mass (Mass 1), imagine a diagram showing: - A downward arrow representing the gravitational force. - An upward arrow representing the spring force, which should be equal in magnitude if the system is in equilibrium.
For Mass 2, it experiences the same gravitational force as Mass 1, but now must support its own weight as well as the weight of Mass 1.
On top of that, Mass 3, at the very top, carries its own weight plus the weight of Mass 1 and Mass 2. By drawing a free-body diagram for each mass, you take a systematic approach to solving force-related problems, ensuring all forces are accounted for.

Spring Constant

The spring constant, often denoted as \( k \), is a characteristic of the spring that measures how stiff and resistant to deformation it is. This value tells us how much force is needed to stretch or compress the spring by a unit of length.

- The spring constant is expressed in units of Newtons per meter (N/m).- In our exercise, the spring constant \( k \) is given as 7800 N/m, meaning each spring needs 7800 N of force to be stretched by just one meter.Understanding the spring constant is important because of Hooke's Law:
\[ F = kx \]Here, \( F \) is the force exerted on the spring, \( k \) is the spring constant, and \( x \) is the displacement or change in length from its original position.- It implies that the force needed to stretch or compress a spring is directly proportional to the extension or compression (\( x \)).In this problem, as different weights are added, each spring will stretch to a new length, which is what we calculate using this relationship. The calculation involves rearranging Hooke's Law to find the extension:- \( x = \frac{F}{k} \) where \( F \) is known from the weight of the attached mass.The higher the spring constant, the less stretch occurs for the same force.

Gravitational Force

Gravitational force is the force with which the Earth pulls objects towards its center. This force is crucial in our exercise as it determines the weight of each mass.

Gravitational force **on each mass** can be calculated using the formula:- \( F_g = mg \)Where:

• \( m \) is the mass of the object (8.50 kg in our case).
• \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

Applying this formula:- For each 8.50 kg mass, the gravitational force \( F_g \) becomes \( 8.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 83.3 \, \text{N} \).- This force acts directly downward on each of the masses.As a mass is added below each spring, this gravitational force becomes the primary force which causes each spring to stretch. In the context of this exercise, knowing \( F_g \) helps us calculate how much each spring stretches using Hooke's Law.