Problem 7 - Thomas Calculus

Question

Subway car springs It takes a force of \(21,714\) ib to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of 8 in. to its fully compressed height of 5 in. \begin{equation}\begin{array}{l}{\text { a. What is the assembly's force constant? }} \\ {\text { b. How much work does it take to compress the assembly the }} \\ \quad {\text { first half inch? the second half inch? Answer to the nearest }} \\ {\quad \text { in.-lb. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

a) 7,238 lb/in , b) First half inch: 905 in-lb, Second half inch: 2,714 in-lb.

Step 1: Understanding Hooke's Law

Hooke's Law states that the force required to compress or extend a spring by a distance is proportional to that distance. It's given by the formula: \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the free length of the spring.

Step 2: Calculate the Force Constant

We are given that a force \( F = 21,714 \) lb compresses the spring from 8 in to 5 in, which means a displacement \( x = 3 \) in. Using Hooke's Law \( F = kx \), we rearrange to find \( k = \frac{F}{x} = \frac{21,714}{3} \). Calculate \( k \).

Step 3: Calculating Spring Constant \( k \)

Substitute the values into the equation: \( k = \frac{21,714}{3} = 7,238 \) lb/in. So, the force constant \( k \) is 7,238 lb/in.

Step 4: Calculating Work for First Half Inch

Work done on the spring can be calculated by the formula \( W = \frac{1}{2} k x^2 \). For the first half inch (0.5 in), \( x = 0.5 \). Substitute \( k = 7,238 \) and \( x = 0.5 \) into the formula: \( W = \frac{1}{2} \times 7,238 \times (0.5)^2 \). Calculate \( W \).

Step 5: Calculating Work for First Half Inch (Result)

Now, compute: \( W = \frac{1}{2} \times 7,238 \times 0.25 = 904.75 \) in-lb. So, the work done to compress the first half inch is roughly 905 in-lb.

Step 6: Calculating Work for Second Half Inch

For the second half inch, the displacement is from 0.5 in to 1 in. Using the same formula \( W = \frac{1}{2} k (x_2^2 - x_1^2) \), where \( x_1 = 0.5 \) and \( x_2 = 1 \). Substitute the values: \( W = \frac{1}{2} \times 7,238 \times (1^2 - 0.5^2) \). Calculate \( W \).

Step 7: Calculating Work for Second Half Inch (Result)

Now, compute: \( W = \frac{1}{2} \times 7,238 \times (1 - 0.25) = 2,714.25 \) in-lb. So, the work done for the second half inch is approximately 2,714 in-lb.

Key Concepts

Spring Constant

The spring constant, often symbolized as "k", is a measure of a spring's stiffness. It tells us how much force is needed to compress or extend the spring by a certain amount. In simple terms, it defines the relationship between the force applied to the spring and the resulting displacement. Remember Hooke's Law, which is expressed as \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( x \) is the displacement from its original position.
• \( k \) is the spring constant.

For this subway car spring, we calculated \( k \) using the force of 21,714 lb needed to compress the spring 3 inches: \[ k = \frac{F}{x} = \frac{21,714}{3} = 7,238 \text{ lb/in} \]This means that 7,238 lb of force per inch is needed to compress or elongate the spring.

Work Done on Spring

The work done when compressing or extending a spring is the energy required to displace it. This can be calculated using the formula for work: \[ W = \frac{1}{2} k x^2 \]Substituting the spring constant, \( k \), and the displacement, \( x \), into this formula, allows you to calculate the energy expended. For example:

• To compress the spring by the first half inch (0.5 in), the work done is: \[ W = \frac{1}{2} \times 7,238 \times (0.5)^2 = 904.75 \text{ in-lb} \]
• For the second half inch (from 0.5 in to 1 in), we adjust the displacement like this:\[ W = \frac{1}{2} \times 7,238 \times (1^2 - 0.5^2) = 2,714.25 \text{ in-lb} \]

Thus, it takes more work to compress the spring by each subsequent segment as the spring becomes shorter and stiffer.

Force and Displacement Relationship

The relationship between force and displacement in a spring is a key part of Hooke's Law. This law describes how the force exerted by a spring is proportional to its displacement from the rest position, and the spring constant \( k \) is that proportionality factor. To put it simply:

• If \( x \) is doubled, the force \( F \) required will also double.
• A large \( k \) value means a stiff spring that doesn't compress much under force.
• A small \( k \) value indicates a more easily compressible spring.

Understanding this proportional relationship allows engineers to design systems with precise control over spring behavior, ensuring that they can predict how springs will respond when forces are applied.