# Spring attached to two hanging objects

In summary, the conversation discusses determining the relationship of forces among individual objects connected by a spring and finding a final equation to describe the overall relationship of the objects to one another. The equations for Hooke's law and the forces acting on each object are mentioned, and there is confusion about the equations for equilibrium and acceleration. The question is unclear on what equation is needed to represent the relationship between the objects and the spring.

## Homework Statement

Draw a picture of a spring connected over pulleys to two hanging objects of equal
weight at equilibrium. Determine the relationship of forces among each individual object, then find a final equation that describes the overall relationship of the objects to one another.

## Homework Equations

F = ma

Hooke's law for springs:
Fs = -kx

## The Attempt at a Solution

Body-diagram of the model described above is attached to this post. I know the equation for describing the force of a spring is by Hooke's Law:

Fs = -kx

Forces acting on just the spring (for both directions):

Fy = may = 0

Fx = T-kx = max

Forces acting on M1 & M2:

Fx = max = 0

Fy = T-mg = may = 0

T-mg = 0

T = mg

Relationship of spring with M1 & M2:

So if T = mg, we can substitute T for mg in the equation. I know we need to subtitute the equation so the relationship can be determined on one component and for this, I did it for the x axis. Here's what I have so far:

Fx = max

-kx-M1-M2=max

From here, I'm not sure what to do. On the one hand, I know that the masses on both ends of the spring on going to keep extending the spring out. And I know that the masses are going to be proportional to one another as the spring continues to extend. However, I'm not sure what to make of the max portion of the equation. I know that M1 and M2 are equivalent to the force of mg but when the m in the mzx is divided, that would cancel out the "m" variables of M1 and M2. Any help from here is appreciated.

#### Attachments

• Free Body Diagram of spring attached to two masses.jpg
11.8 KB · Views: 1,955
Fs = -kx

Forces acting on just the spring (for both directions):

Fy = may = 0

Fx = T-kx = max

Forces acting on M1 & M2:

Fx = max = 0

Fy = T-mg = may = 0

T-mg = 0

T = mg

Relationship of spring with M1 & M2:

So if T = mg, we can substitute T for mg in the equation. I know we need to subtitute the equation so the relationship can be determined on one component and for this, I did it for the x axis. Here's what I have so far:

Fx = max

-kx-M1-M2=max

From here, I'm not sure what to do. On the one hand, I know that the masses on both ends of the spring on going to keep extending the spring out. And I know that the masses are going to be proportional to one another as the spring continues to extend. However, I'm not sure what to make of the max portion of the equation. I know that M1 and M2 are equivalent to the force of mg but when the m in the mzx is divided, that would cancel out the "m" variables of M1 and M2. Any help from here is appreciated.

The force you're calling tension, you should be calling something else. What other force is acting on the masses aside from gravity?

You saying Fy = may = 0 is correct, but what are the forces that cancel out so that happens? Expand your Fy equation.

"However, I'm not sure what to make of the max portion of the equation."
You stated before that the objects were in equilibrium, so isn't max = 0?

szimmy said:
The force you're calling tension, you should be calling something else. What other force is acting on the masses aside from gravity?

You saying Fy = may = 0 is correct, but what are the forces that cancel out so that happens? Expand your Fy equation.

"However, I'm not sure what to make of the max portion of the equation."
You stated before that the objects were in equilibrium, so isn't max = 0?

Sorry, I probably should have been more clear in my confusion. I'm trying to figure an equation for the relationship between the two hanging mass objects from the spring and showing how they impact the extension of the spring. I attempted before to show it on one axix (the x-axis) but wasn't sure if it I had been on the right track or not with it.

-kx-M1-M2=max
I don't know what you mean by that equation. From the rest of the post, M1 and M2 are labels, not masses or forces. Each has mass m. What is ax the acceleration of?

I was trying to figure out or understand if there can be a possible "acceleration" for when the hanging weights from either end of the spring in causing it to extend out? This was a general problem given that we had to solve from my lab manual as pre-lab question and I guess I didn't understand as to whether or not a possible acceleration existed. From the diagram, M1 and M2 are the hanging weights on either end of the spring and I know from this, each mass would have a force of weight (w) which is equivalent to their mass sizes and gravity (m*g). Is there an acceleration for the hanging weights on both ends of the spring in terms of the spring's extension?

To be honest, I'm not sure what the question is asking for. It says equilibrium, so there's no acceleration. You can write one equation for each mass and one for the spring, but I don't know what they want as one equation to encapsulate everything. I can think of a couple of possibilities, but they're rather artificial. Maybe they just mean an equation relating spring extension to the masses and gravity.
Anyway, you didn't answer my questions: what do M1 and M2 stand for in that equation, and how are you defining ax?

haruspex said:
To be honest, I'm not sure what the question is asking for. It says equilibrium, so there's no acceleration. You can write one equation for each mass and one for the spring, but I don't know what they want as one equation to encapsulate everything. I can think of a couple of possibilities, but they're rather artificial. Maybe they just mean an equation relating spring extension to the masses and gravity.
Anyway, you didn't answer my questions: what do M1 and M2 stand for in that equation, and how are you defining ax?

M1 and M2 stand for m1g and m2g in my equation. For the x axis, I am defining it along the horizontal axis to incorporate the force of the spring in the equation. I restated what I originally defined as tension from the pull of the string to just the force of the pull of the string (Fstring). In going back to look over the problem again, to show Fy = may = 0 along the y axis, I made m1g and m2g equivalent to Fstring but I don't know if this is correct or not. From there, it would make sense to show an equation that relates spring extension to the masses and gravity but I'm not sure if I'm on the right track with that or not. Thanks so much for your help with this problem!

M1 and M2 stand for m1g and m2g in my equation. For the x axis, I am defining it along the horizontal axis to incorporate the force of the spring in the equation. I restated what I originally defined as tension from the pull of the string to just the force of the pull of the string (Fstring). In going back to look over the problem again, to show Fy = may = 0 along the y axis, I made m1g and m2g equivalent to Fstring but I don't know if this is correct or not. From there, it would make sense to show an equation that relates spring extension to the masses and gravity but I'm not sure if I'm on the right track with that or not. Thanks so much for your help with this problem!
Yes, that's the right track, but given your explanations the equation you had, -kx-M1-M2=max, is definitely wrong.
In the OP, you correctly wrote, for the masses: Fy = T-mg = may = 0
For the spring, you wrote
Fy = may = 0
Fx = T-kx = max
What's m there - the mass of the spring? It's very confusing when you use the same symbol for different things. OK, you've made the context clear in each case, so no confusion so far, but at some point you will combine equations from different contexts, and then it becomes impossible to follow.
Does the spring have mass of any significance? Anyway, ax is again 0, right? So T = kx.

## 1. What is the purpose of the spring in a system with two hanging objects?

The spring serves as a connecting element between the two hanging objects. It allows for the transfer of force and energy between the objects, resulting in the oscillation of the system.

## 2. What factors affect the behavior of a spring attached to two hanging objects?

The behavior of the spring is affected by its elasticity, mass of the hanging objects, and the gravitational force acting on the system. Other factors such as air resistance and friction may also play a role.

## 3. How does the spring constant affect the motion of the system?

The spring constant, or stiffness, determines how much force is needed to stretch or compress the spring. A higher spring constant results in a stiffer spring, leading to faster oscillation and a shorter period of motion.

## 4. Can the spring become permanently deformed in a system with two hanging objects?

Yes, if the force applied to the spring exceeds its elastic limit, it can become permanently deformed. This can affect the behavior of the system and lead to inaccurate results.

## 5. How does the length of the spring affect the amplitude of oscillation in a system with two hanging objects?

A longer spring will have a larger amplitude of oscillation compared to a shorter spring. This is due to the increased distance the hanging objects can move away from their equilibrium position before the spring reaches its maximum extension or compression.

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