How Does Longitudinal Stress Affect a Body's Internal Tension?

[Physicsnb]

Homework Statement

Say a force of F N is applied on both ends of a rod of Length L . According to me The stress on the body should be 2F/A (Because forces are applied on both ends). However in reality its just F/A . Where am I getting this wrong?

Relevant Equations

Stress= Force/Area

Stress = F(restoring)/ Cross sectional area
Stress = 2F/A

 

[haruspex]

Homework Statement: Say a force of F N is applied on both ends of a rod of Length L . According to me The stress on the body should be 2F/A (Because forces are applied on both ends). However in reality its just F/A . Where am I getting this wrong?
Relevant Equations: Stress= Force/Area

Stress = F(restoring)/ Cross sectional area
Stress = 2F/A

This is a common misunderstanding.
A compressive or tensile force on a rod is not exactly a force. It is more like a pair of equal and opposite forces at every point along the rod. If we say the tension is F we mean that each of those forces is F.

 

[Orodruin]

Stress is a measure of the internal forces inside an object, not the external forces applied to it. In this case, it would be the force per area of a cross section of the rod. If you take an arbitrary cross section and consider the FBD for one of the parts, it should be pretty clear that the internal force across that cross section is F, not 2F.

 

[Physicsnb]

Stress is a measure of the internal forces inside an object, not the external forces applied to it. In this case, it would be the force per area of a cross section of the rod. If you take an arbitrary cross section and consider the FBD for one of the parts, it should be pretty clear that the internal force across that cross section is F, not 2F.

Can you please explain it with a fbd?

 

[Physicsnb]

This is a common misunderstanding.
A compressive or tensile force on a rod is not exactly a force. It is more like a pair of equal and opposite forces at every point along the rod. If we say the tension is F we mean that each of those forces is F.

Still the force should be 2F? Isn't it?

 

[haruspex]

Still the force should be 2F? Isn't it?

As I tried to explain in post #2, a tension is not really a force, so the relationship between the two is a matter of definition. If the pull on each end of a string is F then, by definition, the tension is F. It could have been defined as 2F, but it isn't.

 

[256bits]

Homework Statement: Say a force of F N is applied on both ends of a rod of Length L . According to me The stress on the body should be 2F/A (Because forces are applied on both ends). However in reality its just F/A . Where am I getting this wrong?
Relevant Equations: Stress= Force/Area

Stress = F(restoring)/ Cross sectional area
Stress = 2F/A

What would happen if a force is applied to only one end?
Would the tension in the rod then be F??

 

[haruspex]

What would happen if a force is applied to only one end?
Would the tension in the rod then be F??

Not sure what you are expecting as answer, so I'll give mine.
If the force is only applied at one end, the rod will accelerate. For this to make sense, we must allow the rod to have mass. This means the tension where the force is applied will be the same as if F had been applied at both ends (i.e. F, under the standard convention). The tension will diminish from there, down to zero at the free end. If the rod is uniform then the tension will diminish uniformly.

 

[256bits]

Not sure what you are expecting as answer, so I'll give mine.
If the force is only applied at one end, the rod will accelerate. For this to make sense, we must allow the rod to have mass. This means the tension where the force is applied will be the same as if F had been applied at both ends (i.e. F, under the standard convention). The tension will diminish from there, down to zero at the free end. If the rod is uniform then the tension will diminish uniformly.

!00% correct.
Not sure why the OP was not let to think and answer.

A rod of force F on each end would not accelerate, and would have a tension of F within.

If there was a tension of 2F within ( or any selected tension of arbitrary value ), a FDB would have 2F on one end where the cut is, and F on the the other end. The FBD would have to show an acceleration.

The same argument could be shown for a two pieces of rod joined by a string, with applied force F on the ends.
If the tension within the string is 2F, one could replace the string with an applied force of 2F on the applicable end of the rod where the string had been joining them. Now each of the rods has a force F on one end, and an force 2F on the other end. The two rods would accelerate away from each other.

 

[haruspex]

Not sure why the OP was not let to think and answer.

Because a) I thought it rather beyond the OP's level and b) although tangentially related to the OP's question, I did not see it as leading to a resolution of it.

If there was a tension of 2F within ( or any selected tension of arbitrary value ), a FDB would have 2F on one end

As I posted, I view the fact that the tension resulting from a pair of opposing forces of magnitude F on each end is also ascribed the value F, not 2F nor any other multiple, as a convention. If the convention had arisen that it is 2F, physics and engineering would still have worked perfectly well, so it is not something you can prove.

Remember that tension is not a vector, so cannot be thought of as a force. It is like a negative pressure, with compression being positive pressure.
Edit… that's not quite right either. It works in 1D, but in higher dimensions tension and compression act bidirectionally along a line, so they have orientation but not sense.

 

[256bits]

If the convention had arisen that it is 2F, physics and engineering would still have worked perfectly well

Quite probably mathematically, but KISS might be a valid format to follow.
I just think everyone should just use F=ma everywhere.

 

[Frabjous]

Think of it this way. Connect one end of the bar to a wall. Pull the other end of the bar with a force F. The force on the wall is F. Equal an opposite forces mean that there is a force of the bar at the interface of -F.

A little more formally, you have a traction $\vec T = \frac 1 A \vec F$, respectively on each surface. Since the force is normal to each surface, we can write the stress as $\sigma = \vec T \cdot \vec n$ where $\vec n$ is the normal vector of the surface. Since $\vec T$, $\vec F$, and $\vec n$ have opposite signs on both surfaces, the stresses are identical.

 

[haruspex]

Quite probably mathematically, but KISS might be a valid format to follow.

Sure, but the challenge is to explain to a student why we say the tension is F when there is a force of F pulling at each end. It is a very natural convention, but not a provable fact.

everyone should just use F=ma everywhere

Sorry, I don't get the relevamce.

 

[haruspex]

we can write the stress as $\sigma = \vec T \cdot \vec n$ where $\vec n$ is the normal vector of the surface.

That is still just a convention. If the convention were that the stress is $2\vec T \cdot \vec n$ physics would not fall apart.

 

[Frabjous]

That is still just a convention. If the convention were that the stress is $2\vec T \cdot \vec n$ physics would not fall apart.

Technically correct, but you would also have to change all of the force laws in continuum mechanics so that they would no longer look like F=ma which is silly.

 

[haruspex]

Technically correct, but you would also have to change all of the force laws in continuum mechanics so that they would no longer look like F=ma which is silly.

Interesting… can you post an example?

 

[Frabjous]

Interesting… can you post an example?

Here is a trivial one. Think of a gas pushing a piston. There is now a factor of two in the equation.

 

[haruspex]

Here is a trivial one. Think of a gas pushing a piston. There is now a factor of two in the equation.

Let's remove this from the thread. I'll PM you.
The usual convention is surely the most convenient. I'm just trying to provide an honest answer to the OP: the tension is F, not 2F, because of the way we define the measure of tensions.

 

[Lnewqban]

Homework Statement: Say a force of F N is applied on both ends of a rod...

That is not a force, but two forces acting along a common line and in opposite directions.

It seems that this concept of forces and free body diagrams did not become clear to you in your previous thread about compression of a scale.

https://www.physicsforums.com/threa...achine-normal-vs-gravitational-force.1058285/

Could you define a force?

 

[Physicsnb]

Think of it this way. Connect one end of the bar to a wall. Pull the other end of the bar with a force F. The force on the wall is F. Equal an opposite forces mean that there is a force of the bar at the interface of -F.

A little more formally, you have a traction $\vec T = \frac 1 A \vec F$, respectively on each surface. Since the force is normal to each surface, we can write the stress as $\sigma = \vec T \cdot \vec n$ where $\vec n$ is the normal vector of the surface. Since $\vec T$, $\vec F$, and $\vec n$ have opposite signs on both surfaces, the stresses are identical.

Thanks for your reply .
I understood the 1st para of your explanation buy failed to completely understand the 2nd part.
Here's what I think the 2nd part meant (please correct me if im wrong):
Traction = F/A
Force acting on free end , A = F
Force acting on the constrained end B= F(in opp direction
Since Traction= F/A= stress
So, Traction = stress.
Here's the part the I didn't understand :
Why was normal vector n put into the equations.
Why's the force applied by the wall(action reaction pair)on free end and force on the end B applied by force applied originally ignored in the equation?
(I'm new to the concept of traction . I've never learnt much about it , just read in online now. So I might be aware of all concepts of tractions)

 

[Physicsnb]

That is not a force, but two forces acting along a common line and in opposite directions.

It seems that this concept of forces and free body diagrams did not become clear to you in your previous thread about compression of a scale.

https://www.physicsforums.com/threa...achine-normal-vs-gravitational-force.1058285/

Could you define a force?

I feel I now know about FBD and forces .
What's force?
Force is defined as a push or pull which causes or tries to cause a change in its velocity.
(I got to know about my error in the previous question. I added mg on the weighing machine which is actually acting on the body)
But am I doing something wrong related to fbd and forces here? Please let me know

 

[Physicsnb]

!00% correct.
Not sure why the OP was not let to think and answer.

A rod of force F on each end would not accelerate, and would have a tension of F within.

If there was a tension of 2F within ( or any selected tension of arbitrary value ), a FDB would have 2F on one end where the cut is, and F on the the other end. The FBD would have to show an acceleration.

The same argument could be shown for a two pieces of rod joined by a string, with applied force F on the ends.
If the tension within the string is 2F, one could replace the string with an applied force of 2F on the applicable end of the rod where the string had been joining them. Now each of the rods has a force F on one end, and an force 2F on the other end. The two rods would accelerate away from each other.

I meant to say that the forces on all parts of rod should be 2F.

 

[Physicsnb]

Stress is a measure of the internal forces inside an object, not the external forces applied to it. In this case, it would be the force per area of a cross section of the rod. If you take an arbitrary cross section and consider the FBD for one of the parts, it should be pretty clear that the internal force across that cross section is F, not 2F.

Can you show me am fbd showing that ?
I tried to draw an fbd for that.
I applied force F on both ends of a cross section of the rod one on the top and bottom .
Is that correct

 

[Orodruin]

Can you show me am fbd showing that ?
I tried to draw an fbd for that.
I applied force F on both ends of a cross section of the rod one on the top and bottom .
Is that correct

Please show your attempt.

 

[Orodruin]

I meant to say that the forces on all parts of rod should be 2F.

The net force on any part of the rod is zero or the rod would be accelerating.

 

[Lnewqban]

I feel I now know about FBD and forces .
What's force?
Force is defined as a push or pull which causes or tries to cause a change in its velocity.

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/5-1-forces/

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/5-4-mass-and-weight/

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/5-6-common-forces/

https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/5-7-drawing-free-body-diagrams/

 

[Lnewqban]

...
But am I doing something wrong related to fbd and forces here? Please let me know

Please, see also:
https://en.wikipedia.org/wiki/Strength_of_materials

https://en.wikipedia.org/wiki/Stress_(mechanics)

https://en.wikipedia.org/wiki/Ultimate_tensile_strength

 

[jbriggs444]

Technically correct, but you would also have to change all of the force laws in continuum mechanics so that they would no longer look like F=ma which is silly.

You would only have to multiply all of the components of the Cauchy stress tensor by two.

And then divide by two when taking the product of the Cauchy stress tensor with a directed area element.

[Taking the product of the stress tensor with an area element is how you calculate a force from a stress. It is a slight generalization of the manner in which a force is calculated from a pressure]

 

[Frabjous]

You would only have to multiply all of the components of the Cauchy stress tensor by two.

And then divide by two when taking the product of the Cauchy stress tensor with a directed area element.

[Taking the product of the stress tensor with an area element is how you calculate a force from a stress. It is a slight generalization of the manner in which a force is calculated from a pressure]

You miss my point. Right now, everything looks like F=ma. Following this convention, some things will look like F/2=ma. I stand by saying this would be a silly convention.

It is more than the force equations. We would also need to have ½PdV work, unless one wants multiple pressure definitions.

 

[jbriggs444]

You miss my point. Right now, everything looks like F=ma. Following this convention, some things will look like F/2=ma. I stand by saying this would be a silly convention.

No. You misunderstand the proposal.

We are not messing with the definition of force. We are messing with the definition of stress, tension and pressure.

It is more than the force equations. We would also need to have ½PdV work, unless one wants multiple pressure definitions.

Right. Under the proposal, $F=\frac{1}{2}P\ dV$ and $F=\frac{1}{2}Pa$. Meanwhile, atmospheric pressure, for instance, is doubled. As a result, forces and accelerations are unchanged.

No one is claiming that this is useful. Or even natural. Only that it is a possible convention.

 

[Frabjous]

No. You misunderstand the proposal.

We are not messing with the definition of force. We are messing with the definition of stress, tension and pressure.

Right. Under the proposal, $F=\frac{1}{2}P\ dV$ and $F=\frac{1}{2}Pa$. Meanwhile, atmospheric pressure, for instance, is doubled. As a result, forces and accelerations are unchanged.

No one is claiming that this is useful. Or even natural. Only that it is a possible convention.

You quoted me in post 28. I said that it was technically correct, but silly. I am not sure what you are trying to correct.

Knowing that continuum stresses and strains “look like” forces and displacements makes life simpler. One has to memorize fewer equations.

Or are we in violent agreement with each other.

 

[Physicsnb]

Please show your attempt.

 

[Orodruin]

It is unclear from your diagram what forces are from the wall and what is the internal tension force. All forces have the same size F, but that comes later from the equilibrium equations. You should be calling the tension force something else, like T. I also suggest that you make the FBD of the part of the rod below some mid-point of the rod so that you have the internal force T rather than the force from the wall on the rod.

 

[Lnewqban]

The top blue and black arrows are reversed.
The force exerted by the bar on the wall (black arrow) is pulling on the wall rather than pushing in.

 

[Physicsnb]

It is unclear from your diagram what forces are from the wall and what is the internal tension force. All forces have the same size F, but that comes later from the equilibrium equations. You should be calling the tension force something else, like T. I also suggest that you make the FBD of the part of the rod below some mid-point of the rod so that you have the internal force T rather than the force from the wall on the rod.

Can you please show me the corrected version?