Springs and Hooke's Law -- Unequal forces applied to the two ends....

In summary: A small slice of the rod accelerates due to difference in the tension on the two sides.So if all the parts are to have same acceleration, the difference need to be the same, and not the tension itself.This tells you that the tension varies along the rod, but it does so in a way that is consistent with Hooke's Law.
  • #1
Utilite
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Okay here is my question.
Assume that you are pulling a spring with 3N from one end and 5N from the other one. How can we apply Hooke's Law in this case? And let's assume that the spring isn't stretching anymore just to simplify things, but it is accelerating.
If we divide the spring into eight segments. I feel like the net force on the end of third segment is zero (3N is applied to first segment). If that idea is true then we can apply Hooke's Law to each side.
But I couldn't prove that idea and the springs got me really confused.
 
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  • #2
Utilite said:
And let's assume that the spring isn't stretching anymore just to simplify things, but it is accelerating.
What do you mean by that? Is the spring stretching or not?
 
  • #3
nasu said:
What do you mean by that? Is the spring stretching or not?
It isn't stretching but it is moving. Like when you stretch a spring hold it and move it around.
 
  • #4
But what keeps the spring stretched?
 
  • #5
So, you are talking about a rigid rod with a force of 5 N applied at one end and a force of 3 N applied in the opposite direction at the other end, and you are asking how the tension in the rod varies with position along the rod from one end to the other. Is that correct?
 
  • #6
Chestermiller said:
So, you are talking about a rigid rod with a force of 5 N applied at one end and a force of 3 N applied in the opposite direction at the other end, and you are asking how the tension in the rod varies with position along the rod from one end to the other. Is that correct?
Yes that is kind of what i am asking. And also we know that the rigid rod stretched a bit as some force is applied. Can we calculate the stretching using Hooke's Law?
 
  • #7
nasu said:
But what keeps the spring stretched?
I don't know doesn't the spring stretch when you pull it with some force on different ends. (We first stretched it by pulling it then it stabilized then it started moving around) Perhaps it doesn't it just accelerates(?). I got really confused with springs.
 
  • #8
Utilite said:
Yes that is kind of what i am asking. And also we know that the rigid rod stretched a bit as some force is applied. Can we calculate the stretching using Hooke's Law?
Yes. But first, let's look at the rod. Let M be the mass of the rod. In terms of M, what is the acceleration of the rod? Are all parts of the rigid rod experiencing the same acceleration?
 
  • #9
Chestermiller said:
Yes. But first, let's look at the rod. Let M be the mass of the rod. In terms of M, what is the acceleration of the rod? Are all parts of the rigid rod experiencing the same acceleration?
yes definitely. the rod is accelerating with 2N/M. since it is rigid every part is accelerating with same acceleration but how is that possible if tension changes from one point to another
 
  • #10
Utilite said:
yes definitely. the rod is accelerating with 2N/M. since it is rigid every part is accelerating with same acceleration but how is that possible if tension changes from one point to another
Well, let's just see. Suppose we divide the rod into 8 parts like you did in your first post. What is the mass of each part? Now, suppose we do a free body diagram on the part immediately adjacent to the 5 N end. The tension of the right end of this segment is 5 N, and call the tension on the left end of this segment T. What is the 2nd law force balance on this segment?
 
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  • #11
A small slice of the rod accelerates due to difference in the tension on the two sides.
So if all the parts are to have same acceleration, the difference need to be the same, and not the tension itself.
What does this tell you about the way the tension varies along the rod?

Edit
Sorry, Chet posted a second before. :)
 
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  • #12
Chestermiller said:
Well, let's just see. Suppose we divide the rod into 8 parts like you did in your first post. What is the mass of each part? Now, suppose we do a free body diagram on the part immediately adjacent to the 5 N end. The tension of the right end of this segment is 5 N, and call the tension on the left end of this segment T. What is the 2nd law force balance on this segment?
Yes now I get it it is just like two masses connected with a string. I was thinking about the tension on a very little dL as just from one side. But now it is actually from two sides. Then what do we mean as tension on some point?
 
  • #13
Utilite said:
Yes now I get it it is just like two masses connected with a string. I was thinking about the tension on a very little dL as just from one side. But now it is actually from two sides. Then what do we mean as tension on some point?
I don't follow your question. The tension is distributed over the local cross section of the rod, at each location.
 
  • #14
Chestermiller said:
I don't follow your question. The tension is distributed over the local cross section of the rod, at each location.
I am sorry I couldn't state my question properly but it doesn't matter now. I got what I was looking for you have been very helpful thank you.
 

1. What is Hooke's Law?

Hooke's Law is a scientific law that states the relationship between the force applied to a spring and the resulting displacement of the spring. It states that the force applied is directly proportional to the displacement of the spring, as long as the spring remains within its elastic limit.

2. What is the elastic limit of a spring?

The elastic limit of a spring is the maximum amount of force that can be applied before the spring permanently deforms. Beyond this point, the spring will not return to its original shape and the relationship between force and displacement will no longer be linear.

3. Can unequal forces be applied to the two ends of a spring?

Yes, unequal forces can be applied to the two ends of a spring. In this case, the spring will experience a net force, causing it to stretch or compress. However, the relationship between the force and displacement will still follow Hooke's Law as long as the elastic limit is not exceeded.

4. How do you calculate the force exerted on a spring?

The force exerted on a spring can be calculated using the equation F = kx, where F is the force in Newtons, k is the spring constant (a measure of the stiffness of the spring), and x is the displacement of the spring in meters.

5. What are some real-life applications of Hooke's Law?

Hooke's Law has many real-life applications, including in the design of springs for various mechanical devices, such as car suspensions, door hinges, and trampolines. It is also used in the fields of physics and engineering to understand the behavior of materials under stress and to design structures that can withstand external forces.

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