How Does Force Distribution Affect Tension in an Elastic Plank?

[andyrk]

Two opposite forces F₁ = 120 N and F₂ = 80 N act on an elastic plank of modulus of elasticity Y = 2 x 10^11 N/m². and length l = 1 m placed over a smooth horizontal surface. The cross-sectional area of the plank is S = 0.5 m². The change in the length of the plank is x *10^-11m Find the value of x.

I understood most of the solution but I didn't get one part of it, which says:
Tension at any point x in the plank T(x)= F₂ + ((F₁ - F₂)/l)*x

Why do we include the term F₂? Shouldn't T(x) without the term F₂ be the tension at the point?

 

[andyrk]

Anybody there?

 

[pongo38]

I am not quite sure from the question what are the orientations of the forces. Are they or are they not perpendicular to the cross section? On the face of it, the plank is not in static equilibrium, and so my favourite definition of the axial force (the algebraic sum of all the forces on one side of a section - or the other side) is problematic to apply.

 

[andyrk]

Yes, the forces are perpendicular to the cross section.

 

[andyrk]

Can someone please help me on this? One possible explanation I could think of was that when we rearrange the expression for tension it becomes as follows-
T(x) = F₂(l-x)/x + F₁x/l. This can be explained as - Force the on the differential element at a length x from F₁ would be F₁x/l because force on the differential element right next to the application of F₁ (at length x = 0) would be F₁, which is essentially at one end of the plank (length = l). So for unit length it would be F₁/l and for x length it would be F₁x/l. Similarly, for F₂, the force is being applied on the other end of the plank. And so, its distance from the differential element in consideration would be l-x. And using the same method as before, the force on this element due to F₂ would be F₂(l-x)/l.

But my question is that why do we add the two forces together? Shouldn't we subtract them as they are in opposite direction? Or does this have something to with the definition of tension here?

 

[Chestermiller]

Let F1 be the tensile force at x = l, and let F2 be the tensile force at x = 0 (pulling in the opposite direction from F1). Is the plank going to be in equilibrium? If not, write an overall force balance on the plank. After you do this, I'll get back to you with the next step.

Chet

 

[andyrk]

Let F1 be the tensile force at x = l, and let F2 be the tensile force at x = 0 (pulling in the opposite direction from F1). Is the plank going to be in equilibrium? If not, write an overall force balance on the plank. After you do this, I'll get back to you with the next step.

Chet

The net force would be F1 - F2, so that the block would be moving in the direction of force F1.

 

[andyrk]

Hello! Is anybody there? Am I screaming on a deserted island or something? :(

 

[Chestermiller]

Hello! Is anybody there? Am I screaming on a deserted island or something? :(

Chill man. Some of us have to sleep too.

$F_1-F_2=ma$

where m is the mass of the plank.

Now, what I would like you to do is draw a free body diagram of the portion of the plank between the cross section at x = 0 and the cross section at arbitrary location x (< l) along the plank. Call the tension at this cross section T(x). What fraction of the mass of the plank lies between these two cross sections? From this result, what is the mass of plank material between the cross sections at x = 0 and the arbitrary location x (in terms of the total mass of the plank m)? Are all portions of the plank accelerating at the same rate (as given by the equation above)?

Chet

 

[andyrk]

Chill man. Some of us have to sleep too.

$F_1-F_2=ma$

where m is the mass of the plank.

Now, what I would like you to do is draw a free body diagram of the portion of the plank between the cross section at x = 0 and the cross section at arbitrary location x (< l) along the plank. Call the tension at this cross section T(x). What fraction of the mass of the plank lies between these two cross sections? From this result, what is the mass of plank material between the cross sections at x = 0 and the arbitrary location x (in terms of the total mass of the plank m)? Are all portions of the plank accelerating at the same rate (as given by the equation above)?

Chet

Yep. That would be $mx/l$. Yes, all the portions are accelerating at the same rate (so that they are moving the same distance per unit time). If this wouldn't have been true, then the plank would have broken apart (since some particles are moving faster than others, so the distance between them shouldn't remain constant). But wait, shouldn't that be the case only when the body is a rigid body? The question clearly says that the body is elastic, so it is not a necessity that the particles have a constant distance between each other. So the particles 'can' have different accelerations. Whether they have different accelerations or not is another question, something which I am not sure about at present. The force that the element at $x$ would be experiencing would be $(F_1-F_2)x/l$.

 

[Chestermiller]

Yep. That would be $mx/l$. Yes, all the portions are accelerating at the same rate (so that they are moving the same distance per unit time). If this wouldn't have been true, then the plank would have broken apart (since some particles are moving faster than others, so the distance between them shouldn't remain constant). But wait, shouldn't that be the case only when the body is a rigid body? The question clearly says that the body is elastic,and it is not a necessity that the particles have a constant distance between each other. So the particles 'can' have different accelerations. Whether they have different accelerations or not is another question, something which I am not sure about at present.

To do this problem, you need to assume that somehow a steady state has been established so that all portions of the bar are moving with the same acceleration. This is what they expect you to assume.

The force that the element at $x$ would be experiencing would be $(F_1-F_2)x/l$.

Whoa. Slow down pardner. This result is not correct.

In your free body diagram, you know that the mass of your body is mx/l, you know that the tensile force at one end is T(x), you know the tensile force at the other end is F2 in the negative x direction, and you know that the acceleration for the section of plank being considered is the same as that for the overall plank a. So now please write a force balance equation using Newton's second law for the section of your body between x = 0 and x.

Chet

 

[andyrk]

To do this problem, you need to assume that somehow a steady state has been established so that all portions of the bar are moving with the same acceleration. This is what they expect you to assume.

Whoa. Slow down pardner. This result is not correct.

In your free body diagram, you know that the mass of your body is mx/l, you know that the tensile force at one end is T(x), you know the tensile force at the other end is F2 in the negative x direction, and you know that the acceleration for the section of plank being considered is the same as that for the overall plank a. So now please write a force balance equation using Newton's second law for the section of your body between x = 0 and x.

Chet

Whoa. Okay. It would be: T(x) - F2 = mx/l*a. Is that correct?

 

[andyrk]

But what about F1? Shouldn't F1 also act at the body between x = 0 and x?

 

[Chestermiller]

Whoa. Okay. It would be: T(x) - F2 = mx/l*a. Is that correct?

Yes! Yes! Yes!

But now substitute ma from the overall force balance into this equation and see what you get. I think you will be pleasantly surprised.

Chet

 

[andyrk]

Yes! Yes! Yes!

But now substitute ma from the overall force balance into this equation and see what you get. I think you will be pleasantly surprised.

Chet

That would make it as follows-
$F_1-F_2 = ma ⇒ a = (F_1-F_2)/m$
So, $T(x) - F_2 = mx(F_1-F_2)/l * m = x/l *(F_1-F_2)$

 

[Chestermiller]

That would make it as follows-
$F_1-F_2 = ma ⇒ a = (F_1-F_2)/m$
So, $T(x) - F_2 = mx(F_1-F_2)/l * m = x/l *(F_1-F_2)$

Nice job.

Chet

 

[andyrk]

you know that the tensile force at one end is T(x), you know the tensile force at the other end is F2 in the negative x direction

But why isn't F1 also included at the end which is being pulled by tensile force T(x)? Shouldn't it then be T(x) + F1 - F2 = (mx/l)*a?

 

[Chestermiller]

But why isn't F1 also included at the end which is being pulled by tensile force T(x)? Shouldn't it then be T(x) + F1 - F2 = (mx/l)*a?

No. The force F1 is only being applied to the right face of the overall plank, and nowhere else. What we are doing here is equivalent to breaking the plank into two separate segments and asking "what tension would we have to manually apply to the right hand face of the segment on the left in order for that segment to have the same acceleration as it did in the intact plank (i.e., if we had not made the break)?" This is the same as the tension that the right hand segment exerts on the left hand segment at the cross section under consideration in the intact plank.

Chet

 

[andyrk]

Then even F2 is being applied to the left face of the overall plank. Why do we include it in for a mass element at x from the right while computing its overall balanced force equation so as to apply Newton's Second law?

 

[Chestermiller]

Then even F2 is being applied to the left face of the overall plank. Why do we include it in for a mass element at x from the right while computing its overall balanced force equation so as to apply Newton's Second law?

We are analyzing the segment of plank between x=0 (where F2 is applied) and location x (where we want to determine the tension).

Chet

 

[andyrk]

We are analyzing the segment of plank between x=0 (where F2 is applied) and location x (where we want to determine the tension).

Chet

Yep. That makes some sense. You're the man. :)

 

[andyrk]

But wait a minute. This leads to further issues. Now, writing the equation of Young's Modulus for a small element $dx$,it would be like: $Y = (T/A)/(dl/dx)$ where $dl$ is the change in length of the $dx$ element. So, why does is the force acting on this $dx$ element $T$ and not the difference of two tensions at point $x$ and $x+dx$? That's because by definition of F in Young's Modulus, is the net force acting on a body. So why is there only a single force on the dx/differential body?

 

[Chestermiller]

But wait a minute. This leads to further issues. Now, writing the equation of Young's Modulus for a small element $dx$,it would be like: $Y = (T/A)/(dl/dx)$ where $dl$ is the change in length of the $dx$ element. So, why does is the force acting on this $dx$ element $T$ and not the difference of two tensions at point $x$ and $x+dx$? That's because by definition of F in Young's Modulus, is the net force acting on a body. So why is there only a single force on the dx/differential body?

The change in tension over the differential length is negligible. It's just like integrating y = kx from x to x + dx. You just write dI = (kx)dx.

Chet

 

[andyrk]

The change in tension over the differential length is negligible. It's just like integrating y = kx from x to x + dx. You just write dI = (kx)dx.

Chet

If it is negligible does that mean that the net force on the dx element is 0? (Since the tensions act in opposite directions?)

 

[Chestermiller]

If it is negligible does that mean that the net force on the dx element is 0? (Since the tensions act in opposite directions?)

No. It's only negligible in integrating to get the change in length. It's first order in getting the differential mass times the acceleration. So, in integrating it's negligible, but, in differentiating, it is not negligible.

This is really a math question, not a physics question. You need to go back and review your calculus.

Chet

 

[andyrk]

No. It's only negligible in integrating to get the change in length. It's first order in getting the differential mass times the acceleration. So, in integrating it's negligible, but, in differentiating, it is not negligible.

This is really a math question, not a physics question. You need to go back and review your calculus.

Chet

Sorry, but I am not able to understand anything you said above. What I am trying to say is, that tension on a single point say x = x is T(x) (they are equal) in opposite directions. So similarly, tension acting on a differential element is acting in 2 opposite directions. And since you said that the change is negligible, so that simply means that they are almost equal. So that means they almost cancel each other off. So if they cancel each other off then what is the net force F on this differential element that is appearing in the Young's Modulus Equation?

 

[Chestermiller]

Sorry, but I am not able to understand anything you said above. What I am trying to say is, that tension on a single point say x = x is T(x) (they are equal) in opposite directions. So similarly, tension acting on a differential element is acting in 2 opposite directions. And since you said that the change is negligible, so that simply means that they are almost equal. So that means they almost cancel each other off. So if they cancel each other off then what is the net force F on this differential element that is appearing in the Young's Modulus Equation?

Sorry. I misunderstood your question.

OK. How would you apply this same rationale to a bar of finite length that is not accelerating so that the opposing forces on the two ends are equal? Would you say that, since the forces are equal, the bar is not under tension, so that the bar doesn't stretch? If you have a rubber band and apply the same force to both its ends (in opposite directions), does it not stretch?

Chet

 

[andyrk]

Sorry. I misunderstood your question.

OK. How would you apply this same rationale to a bar of finite length that is not accelerating so that the opposing forces on the two ends are equal? Would you say that, since the forces are equal, the bar is not under tension, so that the bar doesn't stretch? If you have a rubber band and apply the same force to both its ends (in opposite directions), does it not stretch?

Chet

So does that mean that F in Young's modulus is not the net force? Then what is it?

 

[Chestermiller]

So does that mean that F in Young's modulus is not the net force? Then what is it?

People at your level don't usually ask questions like this because it gets into a level of complexity that you may not be ready for. I'm going to try to explain it simple terms first, and hope that that works for you. If x is the location of a cross section along the bar, then, when applying Hooke's law involving Young's modulus Y, F is the magnitude of the force exerted by the material at x⁺ (i.e., the material to the right of the cross section) on the material at x^- (i.e., the material to the left of the cross section). This gives rise to a strain ε at cross section x given by the equation:

$$ε(x)=\frac{F(x)}{AY}$$

Chet