F in Friction Diagram: Explaining Opposite Force to Normal

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Discussion Overview

The discussion revolves around the interpretation of free body diagrams in the context of forces acting on a mechanical system, specifically focusing on the normal force and friction in a diagram involving claspers. Participants explore the implications of omitted forces and the representation of distances between points in the diagrams.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the absence of the force exerted by the claspers in the second free body diagram, suggesting it is a critical oversight.
  • Another participant agrees that the omission of forces that cancel each other out can simplify diagrams, but acknowledges it is a poor choice in this context where normal forces are discussed.
  • Several participants express concern about the educational implications of leaving out normal forces in free body diagrams, emphasizing the importance of accurately representing all forces to derive correct equations.
  • There is a discussion about the distance between points A and B in the diagrams, with some participants questioning its significance and whether it should affect calculations.
  • One participant mentions confusion regarding the calculation of forces, suggesting that the distance may not matter, while another participant argues that it is essential to consider the entire shape rather than isolating parts.
  • Participants debate whether it is acceptable to isolate half of a symmetrical shape for calculations, with differing opinions on the implications of treating the body as rigid versus allowing for rotation at a hinge.

Areas of Agreement / Disagreement

Participants generally agree that the omission of certain forces in the diagrams is problematic, but there is no consensus on the implications of distances between points or the methods for calculating forces in this context. The discussion remains unresolved regarding the best approach to take in analyzing the system.

Contextual Notes

There are limitations in the discussion regarding assumptions about the rigidity of the body and the treatment of forces at the hinge. The participants express varying interpretations of how to approach the calculations based on the diagrams presented.

  • #31


That'll be next Sunday :D We start dynamics. I have a feeling that dynamics is a much more dynamic field than statics (I really, really should be hanged for this). LOL.
 
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  • #32


For this

http://img43.imageshack.us/img43/7305/this1mv.jpg


The Free Body Diagram is that

http://img233.imageshack.us/img233/3918/this2e.jpg

With T being defined as 0.5mg

Notice there are 3 T's acting on point B, with one having an angle!

It seems counter intuitive to me that there's actually MORE WEIGHT applied on the structure than the actual W's weight just because the rope changed direction. Shouldn't, instead of 2T's, it should be just 1T pointing down vertically? I'm not sure how the rope INCREASED the weight, especially if we're told it's weightless.
 
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  • #33


Femme_physics said:
With T being defined as 0.5mg

Notice there are 3 T's acting on point B, with one having an angle!

It seems counter intuitive to me that there's actually MORE WEIGHT applied on the structure than the actual W's weight just because the rope changed direction. Shouldn't, instead of 2T's, it should be just 1T pointing down vertically? I'm not sure how the rope INCREASED the weight, especially if we're told it's weightless.

First off, T is not being defined, but calculated.
Actually these are a couple of free body diagrams rolled together in one.

If you look at the combination of the mass and the bottom pulley D, you'll see that T has to be half of mg to balance the vertical forces.

And yes, there are 3 T's acting on the frame.
If you take a look at the pulley in B and consider only that pulley as a free body diagram, you may be able to imagine that the rope "over" the pulley is pulling it down, which will result in reactive forces at the point where B is pinned into the frame to keep it in place.

Can you tell where the 3 T's are "grabbing" the frame?
That is, what is their working line when you want to do a moment sum?

And yes, I guess the weight mg is increased from the viewpoint of the frame.
Isn't that what pulleys do?
They are usually are applied so the weight is decreased, with the trade off that you have to pull the rope over a longer distance than you otherwise would have.
Applied "in reverse", they increase the weight.
 
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  • #34


First off, T is not being defined, but calculated.

Point taken.

If you look at the combination of the mass and the bottom pulley D, you'll see that T has to be half of mg to balance the vertical forces.
Agreed.

Can you tell where the 3 T's are "grabbing" the frame?
That is, what is their working line when you want to do a moment sum?

In this exercise we're told to ignore the radius of the pulley, so they're all acting on the same joint, B.

And yes, there are 3 T's acting on the frame.
If you take a look at the pulley in B and consider only that pulley as a free body diagram, you may be able to imagine that the rope "over" the pulley is pulling it down, which will result in reactive forces at the point where D is pinned into the frame to keep it in place.

This gets me back to this diagram we've talked about.

http://img855.imageshack.us/img855/5980/tinternal.jpg


You said that the T I've drawn in E is internal force. But what if I would draw the opposite force to it from the pulley? (photoshopped added). Can we now say it's fair analysis of all the exeternal forces? They cancel each other out, after all.

Applied "in reverse", they increase the weight.

So in this case we actually see it applied in reverse? Hmm..interesting. Who would design something like that?
 
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  • #35


Femme_physics said:
In this exercise we're told to ignore the radius of the pulley, so they're all acting on the same joint, B.

That's nice of them, because that circumvents what is actually happening here.
Luckily the result will be the same.

I posted this drawing earlier which may make it clearer.
Note that I have left out the internal tension forces, but I only drew the external forces on the pulley combined with rope and weight.
attachment.php?attachmentid=34436&d=1302946634.gif


Femme_physics said:
This gets me back to this diagram we've talked about.

You said that the T I've drawn in E is internal force. But what if I would draw the opposite force to it from the pulley? (photoshopped added). Can we now say it's fair analysis of all the exeternal forces? They cancel each other out, after all.

I accept your drawing.

Note that they do not cancel each other out completely.
The force at the pulley is transferred to the axis of the pulley, which gives it a different working line.
So in the sum of the horizontal forces they cancel each other out, but in the sum of moments they do not.

Femme_physics said:
So in this case we actually see it applied in reverse? Hmm..interesting. Who would design something like that?

Well, it's not designed in reverse.
If you try to use this contraption, you would lift the weight the mass m by pulling the rope at E. You would pull with a force T, that is half of the actual weight.
However, as a result the weight exerted on the frame is greater than the actual weight.
 
  • #36


Note that I have left out the internal tension forces, but I only drew the external forces on the pulley combined with rope and weight.
Reminds me of the method of joints in trusses, that's pretty much the diagram I'd have to do if I'd look at a certain joint.


I accept your drawing.
Note that they do not cancel each other out completely.
The force at the pulley is transferred to the axis of the pulley, which gives it a different working line.
So in the sum of the horizontal forces they cancel each other out, but in the sum of moments they do not.

I've done sum of all moments on (smachim) A and B, and they perfectly cancel each other out. I can't imagine a point on the structure where they don't cancel each other out even with the sum of all moments.


Well, it's not designed in reverse.
If you try to use this contraption, you would lift the weight the mass m by pulling the rope at E. You would pull with a force T, that is half of the actual weight.
However, as a result the weight exerted on the frame is greater than the actual weight.

I understand now, it did take some time and thought to accept it, but it actually makes sense. Mechanics is beautiful :)
 
  • #37


Femme_physics said:
I've done sum of all moments on (smachim) A and B, and they perfectly cancel each other out. I can't imagine a point on the structure where they don't cancel each other out even with the sum of all moments.

What then is your sum of all moments?

Femme_physics said:
I understand now, it did take some time and thought to accept it, but it actually makes sense. Mechanics is beautiful :)

Yes it is :)
 
  • #39


What then is your sum of all moments?
If I consider point B,

Sigma Mb = 0 ; T x (60+150+250) + T x (60) -T x (60) +Ax x 200 = 0

See, those two T's in the diagram cancel out even with the sum of all moments.

Every so often, in most subjects, a real gem of a textbook is written.
Unfortunately if it is also an older book copies/reprints become like gold dust.

So it is with mechanics - the text extract on the lifting tongs came from this book.

http://www.abebooks.co.uk/servlet/Se...asic+mechanics

The book also covers dynamics and mechanisms in a really accessible, but modern, way.

Hm...well, I'm about to finish statics and start dynamics, so I'm not sure how much it's worth to invest in a book now. You say it also has dynamics, but I want to start studying dynamics with fresh emptiness in my head, actually. If you know what I mean :)
 
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  • #40


Femme_physics said:
If I consider point B,

Sigma Mb = 0 ; T x (60+150+250) + T x (60) -T x (60) +Ax x 200 = 0

See, those two T's in the diagram cancel out even with the sum of all moments.

Yes, this works too, so I'll stop bothering you about it. :)

(Please note that your first term should have a minus sign though. ;()

[edit]And congratulations! :smile: You typed your first formula ever instead of scanning it! :smile: [/edit]
 
  • #41


And congratulations! You typed your first formula ever instead of scanning it!

LOL thank you :)

(Please note that your first term should have a minus sign though. ;()
Argh, yes, I was being careless because I was rushing to prove a point

Thanks :)
 
  • #42


Femme_physics said:
LOL thank you :)

Who needs a scanner if you can't get to it! ;)
 

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