Solve Spring & Pulley Homework: Two Masses 2 & 1 kg with Equal Spring Constants

In summary: Extension in case 1 was 2g/k, so the ratio of extensions case 1:case 2::3:2Yes, the ratio of extensions would remain the same.
  • #1
f(x)
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Homework Statement



Consider a pulley, which has two masses of 2kg each attached with a massless rope followed by a spring (on one side only). Consider another system of a pulley, same as the previous except that the mass connected to the rope with the spring is of 1 kg. Find the ratio of extensions of the two springs, assuming the spring constants to be equal in both cases.

2.What i think

I tried making a freebody diagram of the spring...it has a force equal to tension (2g in first case) acting upwards, and another tension force acting down. Both the forces try to stretch the spring, so total force on it is 4g in first case and 3g in second. I assume system to be in equilibrium in both cases.
Thus extension should be 4:3.
But this is not the right answer. Rather i feel i m very much off the track on this one. Kindly suggest how do i solve this.
Any help is appreciated
 
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  • #2
The spring extension will be proportional to the tension in the rope/spring. In the first case, the tension in the rope equals mg, since the system is in equilibrium. The tension equals mg, not 2mg! To resist the downward pull of the hanging mass (weight = mg) the tension must equal mg; to resist the upward pull, via the pulley, of the other mass (weight = mg) the tension must equal mg.

The spring is pulled from both sides by a force equal to the tension--the two forces don't "add up".

For the second case you first have to solve for the tension in the rope/spring by using Newton's 2nd law. In this case, the system is not in equilibrium.
 
  • #3
Doc Al said:
The spring is pulled from both sides by a force equal to the tension--the two forces don't "add up".

Yeah that's what i want to know, why don't they add up ?
 
  • #4
Maybe answering these questions will help your understanding of tension:

(1) Imagine a rope attached to a wall. You pull on it with a force of 100N. What's the tension in the rope? What force does the wall exert on the rope?

(2) Now imagine that you and a friend grab a rope at each end and pull with a force of 100N. What's the tension in the rope?

Is there a difference between these two cases?
 
  • #5
Doc Al said:
Maybe answering these questions will help your understanding of tension:

(1) Imagine a rope attached to a wall. You pull on it with a force of 100N. What's the tension in the rope? What force does the wall exert on the rope?

(2) Now imagine that you and a friend grab a rope at each end and pull with a force of 100N. What's the tension in the rope?

Is there a difference between these two cases?

Tension is 100 N in both cases
 
  • #6
f(x) said:
Tension is 100 N in both cases
Exactly. Now apply that same thinking to your problem.
 
  • #7
OK, i understand that the spring is being pulled by equal force, each of which is mg. But how to i get the extension of the spring ? Is it equal to mg/k in each direction, adding up to give 2mg/k totally ?
 
  • #8
No. If you pull on the spring with a force of 100N, it must pull back on you with the same force. The force the spring exerts is related to how much it stretches, via Hooke's law. Note that the spring pulls just as hard--100N--at both ends, but the tension is only 100N.

Part of the point is that you can't just pull a spring (or rope) at one end--you have to pull at both ends to maintain a tension.
 
Last edited:
  • #9
So is the extension of the spring mg/k ?

And in part 2, does the presence of the spring make any difference , since it pulls both the ends with the same tension ?
 
  • #10
f(x) said:
So is the extension of the spring mg/k ?
Yes, in the first case, but not in the second. To find the extension in the second case, first find the tension.

And in part 2, does the presence of the spring make any difference , since it pulls both the ends with the same tension ?
I'm not sure what you mean by part 2 (do you mean case 2?), but if I understand you correctly: No, the spring doesn't effect the tension, it just responds to it. (Lets ignore any oscillitory effects.)
 
  • #11
For the 2nd case,
2g-T=2a
T-g=a
Adding, a=g/3.
Putting this in eqn 2, T=4g/3
So, the force on the spring is 4g/3, hence extension is 4g/3k
Extension in case 1 was 2g/k, so the ratio of extensions case 1:case 2::3:2
Is this correct ?
Would the ratio remain same if i interchange the 2kg and 1kg mass ?
 
  • #12
f(x) said:
For the 2nd case,
2g-T=2a
T-g=a
Adding, a=g/3.
Putting this in eqn 2, T=4g/3
So, the force on the spring is 4g/3, hence extension is 4g/3k
Extension in case 1 was 2g/k, so the ratio of extensions case 1:case 2::3:2
Is this correct ?
All good.
Would the ratio remain same if i interchange the 2kg and 1kg mass ?
Yes. The only difference will be the direction of the acceleration, but the tension in the rope/spring will be the same.
 
  • #13
Yay ! Thanks a million Doc Al , specially for the quick replies.
I'll rework the problem and ask if I still have any doubts .
Thx once again
 

1. How do I determine the acceleration of the masses in this system?

To determine the acceleration of the masses, you can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this system, the net force is equal to the difference in tension in the spring on each side of the pulley. By setting up and solving equations for each mass, you can calculate their accelerations.

2. What is the relationship between the masses and the spring constants in this system?

The masses and the spring constants are directly proportional in this system. This means that as the masses increase, the spring constants also increase, and vice versa. This relationship can be seen in the equations used to solve for the accelerations, where the masses and spring constants are multiplied together.

3. Can I use different units for the masses and spring constants in this system?

Yes, as long as the units are consistent and match on both sides of the equations. For example, if the masses are given in kilograms, the spring constants should also be in units of kilograms to ensure that the equations are balanced and accurate.

4. How do I know if my solution for the accelerations is correct?

You can check your solution by plugging the values back into the original equations and ensuring that they are balanced. You can also use the principles of conservation of energy and momentum to verify your solution.

5. Are there any real-world applications of this system?

Yes, this system is commonly used in various machines and devices, such as elevators and escalators, to transfer energy and motion between two masses. Understanding how to solve for the accelerations and tensions in this system is important in designing and maintaining these machines.

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