Chain Link Problem? [Tension Forces; Newton's Third Law]

[harujina]

Homework Statement

A chain of three links, each with a mass 0.2 kg, is being pulled up by a person lifting the top link with 8.88 N of force and the chain accelerates upward. Calculate three forces that are acting on the middle link while the chain is accelerating.

Homework Equations

ƩF = ma
Fg (weight) = mg

The Attempt at a Solution

I drew a free body diagram showing tension force pulling up by 8.88 N and the weight, Fg, pulling the chain down. I calculated Fg = mTg (total mass of chain) = (0.6 kg)(9.8m/s^2) = 5.88 N.
By the way, is it right for me to assume the mass of the middle link = total mass of the chain (0.2 kg*3 = 0.6 kg)?

So, I have found two forces, I believe... and I'm not too sure on what the third force would be.
I've learned about the internal forces of each object if they are acting on one another but I'm not sure if that's it, or how to calculate it.

Oh, and I found acceleration by a = ƩF/m = 3 N/0.6 kg = 5 m/s^2, if what I have done above is correct.

 

[tiny-tim]

hi harujina!

I drew a free body diagram showing tension force pulling up by 8.88 N and the weight, Fg, pulling the chain down.

no

a free body diagram shows all the external forces acting (directly) on one body, and that body is the middle link only

… is it right for me to assume the mass of the middle link = total mass of the chain (0.2 kg*3 = 0.6 kg)?

no, the weight (as an external force for the free body diagram) is the external force acting directly on the link, ie the weight of the link only

I've learned about the internal forces of each object if they are acting on one another but I'm not sure if that's it …

internal forces on one body can be external forces on a smaller body

a free body diagram for a large body will not show the forces between the parts, because they are internal

a free body diagram for one part will show the forces between the parts, because they are external

 

[harujina]

hi harujina! no

a free body diagram shows all the external forces acting (directly) on one body, and that body is the middle link onlyno, the weight (as an external force for the free body diagram) is the external force acting directly on the link, ie the weight of the link onlyinternal forces on one body can be external forces on a smaller body

a free body diagram for a large body will not show the forces between the parts, because they are internal

a free body diagram for one part will show the forces between the parts, because they are external

hi :)

I remember my teacher taught us that you could draw a fbd of combined objects in order to determine acceleration, like what I have done above. Is this incorrect?

Ok, so before I adjust anything I want to make sure I understand this because it's confusing me a lot. A fbd for a large body will not show forces between the parts, but in this case aren't the chains all equal in mass? So would the tension force just equally divide amongst the three links?

 

[haruspex]

I remember my teacher taught us that you could draw a fbd of combined objects in order to determine acceleration, like what I have done above.

That's fine, and you got the right acceleration, but next you need an FBD for the middle link only.

the mass of the middle link = total mass of the chain

Did you leave something out there?

 

[harujina]

That's fine, and you got the right acceleration, but next you need an FBD for the middle link only.

Did you leave something out there?

Wait, would applied force on the middle link be 8.88 N/3 (the three links) = 2.96 N?
and mass of the middle link would just be 0.2 kg?

On another note, thank you so much for your constant help, haruspex! I really appreciate it.

 

[tiny-tim]

hi harujina!

Oh, and I found acceleration by a = ƩF/m = 3 N/0.6 kg = 5 m/s^2, if what I have done above is correct.

where does the 3N come from?

I remember my teacher taught us that you could draw a fbd of combined objects in order to determine acceleration, like what I have done above. Is this incorrect?

that's correct: you can draw a free body diagram for a large body (and it will not show t the internal forces)

however, the question asks "Calculate three forces that are acting on the middle link while the chain is accelerating", so you need a free body diagram for the middle link on its own

A fbd for a large body will not show forces between the parts, but in this case aren't the chains all equal in mass? So would the tension force just equally divide amongst the three links?

no

if you draw a proper free body diagram for the middle link, you'll see why

have you drawn it? what does it show?

 

[haruspex]

Wait, would applied force on the middle link be 8.88 N/3 (the three links) = 2.96 N?
and mass of the middle link would just be 0.2 kg?

On another note, thank you so much for your constant help, haruspex! I really appreciate it.

You are asked for three forces. Draw the FBD. What three forces are there?
(... and you're welcome )

 

[harujina]

hi harujina!


where does the 3N come from?


that's correct: you can draw a free body diagram for a large body (and it will not show t the internal forces)

however, the question asks "Calculate three forces that are acting on the middle link while the chain is accelerating", so you need a free body diagram for the middle link on its own



no

if you draw a proper free body diagram for the middle link, you'll see why

have you drawn it? what does it show?

You are asked for three forces. Draw the FBD. What three forces are there?
(... and you're welcome )

I calculated total force to be 3N by ƩF = Fa - Fg = 8.88 N - 5.88 N.

Ok, to answer the question I will use link A = top link, link B = middle link, and link C = bottom link.
As for the fbd of the link B, I don't know if this is right at all but would it be...
Fa (applied force of 8.88 N; pulling up), aFb (force exerted by link A on link B; also pulling up) and cFb (force exerted by link C on link B; pushing down)?

[EDIT]: Wait, but there's also Fg (weight; pulling down) isn't there

 

[haruspex]

Ok, to answer the question I will use link A = top link, link B = middle link, and link C = bottom link.
As for the fbd of the link B, I don't know if this is right at all but would it be...
Fa (applied force of 8.88 N; pulling up), aFb (force exerted by link A on link B; also pulling up) and cFb (force exerted by link C on link B; pushing down)?

[EDIT]: Wait, but there's also Fg (weight; pulling down) isn't there

How is B aware of the 8.88N force?

 

[tiny-tim]

I calculated total force to be 3N by ƩF = Fa - Fg = 8.88 N - 5.88 N.

oh of course … yes, that's the correct acceleration, then

Fa (applied force of 8.88 N; pulling up), aFb (force exerted by link A on link B; also pulling up) and cFb (force exerted by link C on link B; pushing down)?

[EDIT]: Wait, but there's also Fg (weight; pulling down) isn't there

"Fa (applied force of 8.88 N; pulling up)" is wrong: it's not a direct force on the middle link, so it doesn't get mentioned at all

Fa would be shown on an fbd for the top link, or for the whole body of three links, but not for the middle link (or the bottom link)

and yes, an Fg should be on the fbd for the middle link …

how much should it be?​

 

[harujina]

How is B aware of the 8.88N force?

"Fa (applied force of 8.88 N; pulling up)" is wrong: it's not a direct force on the middle link, so it doesn't get mentioned at all

Fa would be shown on an fbd for the top link, or for the whole body of three links, but not for the middle link (or the bottom link)

and yes, an Fg should be on the fbd for the middle link …

how much should it be?​

Oh, I'm starting to get it now.
Fg (of middle link)= mg; = (0.2 kg)(9.8m/s^2) = 1.96 N; correct?
So now if I want to determine aFb and cFb, would I have to draw fbd's for the top and bottom link as well?

 

[tiny-tim]

Oh, I'm starting to get it now.
Fg (of middle link)= mg; = (0.2 kg)(9.8m/s^2) = 1.96 N; correct?

correct

So now if I want to determine aFb and cFb, would I have to draw fbd's for the top and bottom link as well?

yup!

hint: start with the bottom one … it's easier! ​

 

[harujina]

correct

yup!

hint: start with the bottom one … it's easier! ​

Thank you so much! Can you just see if I'm on the right track?
So, the forces acting on the bottom link are Fg and bFc (force exerted by middle link on bottom link; pulling up). The ƩF = (0.2 kg)(5 m/s^2) = 1N, so cFb (force exerted by bottom link on middle link; pulling down) = 1N?

And this is what I got for the top link:
ƩF = Fa (applied force of 8.88 N) - Fg (1.96 N) - bFa (force exerted by middle link on top link; pulling down)
bFa = Fa - Fg - ƩF
bFa = (8.88 N) - (1.96 N) - (1 N); = 5.92 N, so aFb = 5.92 N?

 

[haruspex]

Thank you so much! Can you just see if I'm on the right track?
So, the forces acting on the bottom link are Fg and bFc (force exerted by middle link on bottom link; pulling up). The ƩF = (0.2 kg)(5 m/s^2) = 1N, so cFb (force exerted by bottom link on middle link; pulling down) = 1N?

You forgot Fg.

And this is what I got for the top link:
ƩF = Fa (applied force of 8.88 N) - Fg (1.96 N) - bFa (force exerted by middle link on top link; pulling down)
bFa = Fa - Fg - ƩF
bFa = (8.88 N) - (1.96 N) - (1 N); = 5.92 N, so aFb = 5.92 N?

Yes, that looks right.

 

[harujina]

You forgot Fg.

Whoops...

ƩF = (0.2 kg)(5 m/s^2) = 1N
ƩF = bFc - Fg; = Fg + ƩF; = (1.96 N) + (1 N) = 2.96 N so cFb (force exerted by bottom link on middle link; pulling down) = 2.96 N
Is this ok?

 

[haruspex]

Whoops...

ƩF = (0.2 kg)(5 m/s^2) = 1N
ƩF = bFc - Fg; = Fg + ƩF; = (1.96 N) + (1 N) = 2.96 N so cFb (force exerted by bottom link on middle link; pulling down) = 2.96 N
Is this ok?

Yes.