The value of "k" for spring constant, is the book answer incorrect?

In summary, the conversation revolves around solving a Hook's law problem for the values of Fi and k. The given data shows a linear relationship between applied force and displacement, with a value of 50lb/in for k. Some individuals have tried different methods to solve the problem, but the simplified equation of Δy = ΔF/k seems to provide the most straightforward solution. This information is provided in the textbook as well.
  • #1
SShep71
5
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Homework Statement
1-3. The spring force F and displacement y for a close-wound tension spring are measured as shown in Fig. P1.3 The spring force and displacement satisfy the linear equation y=((1/k)F-(Fi/k)
, where k is the spring constant and Fi is the preload induced during manufacturing of the spring.

(a) Determine the spring constant k and the pre-load F using the given data in Fig. P1.3. (b) Sketch the graph of the line y(F) and clearly indicate both the spring constant k and preload Fi using the given data.
Relevant Equations
y=((1/k)F-(Fi/k))
There is some discussion currently and I was hoping to get some opinions here. The question is in regard to a Hook's law problem. The text gives the problem as seen below. The text says the answer is 50lb/in. Several people have tried from several different approaches. Factoring the "y" equation to solve for Fi, straight method, etc.

In case the image does not work, here are the specifics:
k= spring constant
Fi=preload at manufacturing
F=spring force
y=displacement

Observations:
F(lbf) y(in)
100 1.0
75 0.5

The students are to determine the "Fi" and "k" values. The book states 50lb/in as the answer
 

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  • #2
SShep71 said:
Observations:
F(lbf) y(in)
100 1.0
75 0.5

The students are to determine the "Fi" and "k" values. The book states 50lb/in as the answer
If the relationship between (total) applied force and (total) displacement is linear, then (using that data) an additional extension of 0.5 inches requires an additional force of 25 lbf. Hence ##k = 50## lbf per inch.
 
  • #3
No disagreement from me, the logic is sound. I guess it's the method of algebraically solving it that seems to have a few "twisted-under britches" in this group.
 
  • #4
SShep71 said:
No disagreement from me, the logic is sound. I guess it's the method of algebraically solving it that seems to have a few "twisted-under britches" in this group.
Note that your relevant equation:

Relevant Equations: y=((1/k)F-(Fi/k))

Can be simplified to:
$$\Delta y = \frac{\Delta F}{k}$$Which may help to straighten out your britches.
 
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1. What is the significance of the "k" value in determining the spring constant?

The "k" value represents the stiffness of a spring and is a measure of the force required to stretch or compress the spring by a certain distance. It is an important factor in understanding the behavior of springs in various systems.

2. How is the "k" value typically calculated?

The "k" value can be calculated by dividing the applied force by the resulting displacement of the spring. This calculation is based on Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed.

3. Can the "k" value vary for different types of springs?

Yes, the "k" value can vary depending on the type of spring, as well as its dimensions and material composition. For example, a steel spring will have a different "k" value than a rubber spring, even if they have the same dimensions.

4. Why might the book answer for the "k" value be incorrect?

The book answer for the "k" value may be incorrect if it is based on theoretical calculations or assumptions, rather than actual measurements. Additionally, the "k" value can change over time due to wear and tear on the spring or changes in environmental conditions.

5. How can the accuracy of the "k" value be ensured?

In order to ensure accuracy, the "k" value should be measured using proper equipment and techniques. It is also important to consider any potential sources of error, such as friction or external forces, and account for them in the calculations.

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