# Trivial question regarding Newton's Third Law and Elasticity

## Homework Statement:

Find the total elongation of the bar, if the bar is subjected to axial forces as shown in the figure. The cross-sectional areas is A and modulus of elasticity is E.

## Relevant Equations:

##Y = \frac{F/A}{\Delta L/L}##
Newton's Third Law

1. Drawing Free Body Diagrams for all components we get :

2. Following this we can find total elongation using ##\Delta L = \frac {1}{AY}(F_1*L_1 + F_2 *L_2+ F_3 *L_3) ##

My questions :
a) I am assuming that the internal forces (3t) are neglected in the FBD because of Newton's third law whereby the force and the opposing force would cancel one another.(?)
b) But if this is the case then why is the 1t force considered on BC?
c) The FBD shows that for a stretching force of 5t an equal and opposite restoring force of 5t occurs (is this correct)?

Thank you!

haruspex
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Your BC FBD is wrong. The 3t and 1t shown as internal must be, in fact, externally applied at the respective joints.

Also, does the fact that the parts are not rigid have any effect on the FBD?

Chestermiller
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Since the body is accelerating to the left, the tension at C is greater than 1t.

haruspex
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View attachment 243623
Also, does the fact that the parts are not rigid have any effect on the FBD?
No, no. Only the BC diagram was wrong. All shown forces are externally applied, but there are also internal forces between the blocks. The whole is in equilibrium, so each block is in equilibrium.

haruspex
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Since the body is accelerating to the left, the tension at C is greater than 1t.
As I wrote in post #2, I believe the shown forces are externally applied. The system is in equilibrium.

No, no. Only the BC diagram was wrong. All shown forces are externally applied, but there are also internal forces between the blocks. The whole is in equilibrium, so each block is in equilibrium.
I see so :
1.The block is in equilibrium.
2. All forces shown are 'external' and must be considered in calculating the total elongation?
3. Not really sure why the FBDs AB (5t, 5t) and CD (1t, 1t) ?
4. BC would be (-3t since compression?, 1t)?

Chestermiller
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As I wrote in post #2, I believe the shown forces are externally applied. The system is in equilibrium.
Oh. Sorry, I missed that. So the tensions are changing discontinuously at the locations where the external loads are applied.

BvU
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3. Not really sure why the FBDs AB (5t, 5t) and CD (1t, 1t) ?
equilibrium ! And: see 4:
4. BC would be (-3t since compression?, 1t)?
No ! @haruspex #5:
at B: external -3 to the left, but internal ... (you fill in)
at C: external -1 to the left, but internal ... (you fill in)

If you would really have BC (-3t, 1t) it would shoot off to the right

JC2000

I am listing my reasoning (is it correct?) :
1.From the responses above I gathered that since the whole block is in equilibrium the parts are also in equilibrium (who would have thought!).
2. Thus for AB if it is being elongated to the left by 5t then 5t must act in the other direction as well.
3. CD is being elongated to the right by 1t and thus must undergo a equal and opposite force. (Not sure what happens to the 3t and 1t forces here)
4. At sea when it comes to BC really...

Chestermiller
Mentor
To get the tension in the region between B and C, put a fictitious break between B and C, and see what force you need to apply to the. portion of the bar to the left of the break (and/or to the right of the break) such that the left portion (and/or the right portion) remains in equilibrium.

JC2000
haruspex
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At sea when it comes to BC
For BC you can combine all the the externally applied forces on one side into a single net applied force, and do the same for the other side.
Or, as @Chestermiller suggests, create unknowns for the internal forces between the blocks at B and C and solve the balance equations.

JC2000
To get the tension in the region between B and C, put a fictitious break between B and C, and see what force you need to apply to the. portion of the bar to the left of the break (and/or to the right of the break) such that the left portion (and/or the right portion) remains in equilibrium.
From the question I know the a force of 3t and 1t acts on BC towards the right. Also 'opposite forces' from AB and BD of 5t and 1t respectively act on BD in opposite directions. Thus the total force to the right for BC would be 3t + t + 5t while the total force to the left would be 1t. Thus a net force of 8t would act on BC to the right.
Since BC is in equilibrium a force of 8t exists to the left?

OR

Do the 5t and 1t forces on BC get cancelled by equal and opposite forces from BC and thus can be neglected.
This would mean a net force of 4t to the right exists on BC and for equilibrium a force of 4t to the left is required?

Chestermiller
Mentor
From the question I know the a force of 3t and 1t acts on BC towards the right. Also 'opposite forces' from AB and BD of 5t and 1t respectively act on BD in opposite directions. Thus the total force to the right for BC would be 3t + t + 5t while the total force to the left would be 1t. Thus a net force of 8t would act on BC to the right.
Since BC is in equilibrium a force of 8t exists to the left?

OR

Do the 5t and 1t forces on BC get cancelled by equal and opposite forces from BC and thus can be neglected.
This would mean a net force of 4t to the right exists on BC and for equilibrium a force of 4t to the left is required?
To the left of the break , you would have forces of -5 and + 3, which add up to -2. To the right of the break, you have forces of +1 and +1, which add up to + 2 . So on the part to the left of the break, you would have to apply a force of +2, and, on the part to the right of the break, you would have to apply a force of -2.

JC2000

To the left of the break , you would have forces of -5 and + 3, which add up to -2. To the right of the break, you have forced of +1 and +1, which add up to + 2 . Do on the part to the left of the break, you would have to apply a force of +2, and, on the part to the right of the break, you would have to apply a force of -2.
1. So my attempt (the two possibilities in #13) was completely off the mark?
2. Is the diagram I have made now based on your comment correct?
3. If so, why is 3t towards the left (isn't the applied force to the right?) (which would mean a force of 2t to the left and a force of 2t to the right which results in equilibrium)
4. So all the external forces that act on ABCD also act on BC?
5. Since the FBD's show that each block is acted upon by two forces, would each of the six have an effect on elongation? Why not?

Thank you for bearing with this!

haruspex
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Gold Member
2020 Award
View attachment 243652

1. So my attempt (the two possibilities in #13) was completely off the mark?
2. Is the diagram I have made now based on your comment correct?
3. If so, why is 3t towards the left (isn't the applied force to the right?) (which would mean a force of 2t to the left and a force of 2t to the right which results in equilibrium)
4. So all the external forces that act on ABCD also act on BC?
5. Since the FBD's show that each block is acted upon by two forces, would each of the six have an effect on elongation? Why not?

Thank you for bearing with this!
The diagrams in post #10 were correct, but perhaps those were not yours?
@Chestermiller was using signs to indicate directions under the convention that right is positive. So in the diagram you should either draw all arrows to the right, but show the values as signed, or draw the forces with negative values to the left and show the values in the diagram unsigned.
E.g. at B, either:
-5t and +3t, both to the right,
Or
5t to the left and 3t to the right.

JC2000
The diagrams in post #10 were correct, but perhaps those were not yours?
Yes, I was trying to see if I could understand it by going in the reverse...
I understand the convention now, so 3t would be pointing right and 5t would be pointing left, as you said, 1t's are correct...

haruspex