Understanding Axial Force in Spring System

[James.L]

Homework Statement

Hi

Say I have a spring system as in the attached figure (it is in equilibrium). I am a little confused about the "axial force" concept after reading an example in my book. It goes as follows:

"We have the three springs, where the outer most ones are attached to a wall. We look at the four nodes. Now, at node 2 we apply an axial external force F2 pointing right. The force F3 is zero, since there is no external force at this node".

They then go on to calculate the reaction forces F1 and F4, and naturally they point left. So they set F3=0 since "there is no external force at this node", but there is no external force at node 1 and 4 either, but yet F1 and F4 are different from zero? This I don't seem to understand. Isn't there a reaction force at node 3 too?

Cheers,
James.

 

[vela]

They then go on to calculate the reaction forces F1 and F4, and naturally they point left. So they set F3=0 since "there is no external force at this node", but there is no external force at node 1 and 4 either, but yet F1 and F4 are different from zero? This I don't seem to understand. Isn't there a reaction force at node 3 too?

Axial simply means the force is in the horizontal direction in this problem.

The walls, since they don't move, must exert a force on the springs. When you stretch spring 4, for instance, both ends will exert a force on whatever the spring is connected to. Newton's 3rd law tells you that the wall therefore must exert a force on the spring.

 

[James.L]

Axial simply means the force is in the horizontal direction in this problem.

Thanks for that.

The walls, since they don't move, must exert a force on the springs. When you stretch spring 4, for instance, both ends will exert a force on whatever the spring is connected to. Newton's 3rd law tells you that the wall therefore must exert a force on the spring.

Ok, then let's look at node 3. The second spring will push on node 3, so spring #2 will push #3, which in turn will push back on #2. Then why do they say the force on node 3 is zero? Is it because the net force on node 3 is zero?

If that is the case, then one could also say the net force on node 4 is zero, because even though the spring is pulling the wall, the wall is also pulling the spring.

 

[vela]

Ok, then let's look at node 3. The second spring will push on node 3, so spring #2 will push #3, which in turn will push back on #2. Then why do they say the force on node 3 is zero? Is it because the net force on node 3 is zero?

Yes. Springs 2 and 3 will each exert a force on node 3, and they will adjust their lengths so that the net force turns out to be 0 if the system is in equilibrium.

For node 2, you have the forces from springs 1 and 2. Those two forces won't cancel because of the external force F2, but the sum of the three forces will be equal to 0.

If that is the case, then one could also say the net force on node 4 is zero, because even though the spring is pulling the wall, the wall is also pulling the spring.

You have to be a little careful how you state things here. The wall pulling on the spring and the spring pulling on the wall are an action-reaction pair. You can't sum them because they act on different objects.

What you want to do is talk about a point right at where the spring is attached to the wall, i.e. node 4. The spring pulls on it from one side and the wall pulls on it from the other. Since the point doesn't accelerate, the net force on it must be 0, which tells you the two forces are equal and opposite.

 

[James.L]

So all net forces at the nodes are zero, since we are in equilibrium. Then can you tell what forces my book is actually referring to?

 

[vela]

I assume F_i refers to external forces on the nodes, that is, forces on the nodes not due to the springs.

 

[James.L]

I see, so when they go back and find F1 and F4 (i.e. the reaction forces), then they are finding the force acting on spring 1 and 4 by the walls?

Thanks.

 

[vela]

Yes.

 

[James.L]

Thanks, I understand it now.

Best wishes,
James.