F in Friction Diagram: Explaining Opposite Force to Normal

[Femme_physics]

Whenever there's normal force there must be a force that causes it. Please explain how in this diagram of this object

http://img850.imageshack.us/img850/605/aa1b.jpg

Where first Free Body diagram is this:


http://img534.imageshack.us/img534/9454/aa2zw.jpg


Which I accept as valid, but then when you isolate the second body:


http://img217.imageshack.us/img217/7923/aa3rq.jpg

You see "friction", but you don't see the force applied on the object by the two claspers. Does it make sense?

-FP

 

[I like Serena]

Very sharp!

Yes, there is force exerted by the claspers.
So there would also be a downward force at A and an upward force at B (which would cancel each other out).

 

[Femme_physics]

Aha! It was ignored. I got to tell you, I'm not at all happy about the solution manual *tsk tsk tsk*

Thanks again mei angele custos :)

 

[I like Serena]

Aha! It was ignored. I got to tell you, I'm not at all happy about the solution manual *tsk tsk tsk*

Thanks again mei angele custos :)

To the defense of the solution manual, it is not uncommon to leave out forces like this, that are canceling each other out anyway. It makes the drawing simpler and easier to read.

However, for this particular problem, where they specifically ask about normal forces, it seems a bad oversight. :)

 

[AlephZero]

I agree, leaving the normal forces out is a bit naughty in an educational context.

The main point of draiwng free body diagrams is so you can get the right equations without leaving things out "by accident". Of course once you know how to "do it right" you can often take shortcuts, but IMO you need to know the rules (and be able to demonstrate you know them by using them) before you are allowed to break them.

 

[I like Serena]

I agree, leaving the normal forces out is a bit naughty in an educational context.

The main point of draiwng free body diagrams is so you can get the right equations without leaving things out "by accident". Of course once you know how to "do it right" you can often take shortcuts, but IMO you need to know the rules (and be able to demonstrate you know them by using them) before you are allowed to break them.

I second that!

 

[Femme_physics]

I third that! :)

While we're on the issue of this diagram, allow me to squeeze you experts out of more of your juicy knowledge.

Take a lot on the first diagram of the object, or even its original drawing here:

http://img821.imageshack.us/img821/2672/caseh.jpg


There seems to be a relatively significant distance between A and B. How come in the second body diagram it practically appears to be a joint force, right in the middle, acting on the same plane?

 

[I like Serena]

I third that! :)

While we're on the issue of this diagram, allow me to squeeze you experts out of more of your juicy knowledge.

There seems to be a relatively significant distance between A and B. How come in the second body diagram it practically appears to be a joint force, right in the middle, acting on the same plane?

Very good!

The first and second free body diagrams do not match indeed.
They should have drawn the claspers in the second diagram in the same place as in the first diagram!
*tsk tsk tsk*

 

[Femme_physics]

Yes! Moreover, the original drawing doesn't give the distance between A and B. Shouldn't it matter?

 

[I like Serena]

Yes! Moreover, the original drawing doesn't give the distance between A and B. Shouldn't it matter?

Is there something to calculate then?
It seems to me that all given distances and angles are excess information, that you need to ignore.

However, if you would have to calculate something, the distance between A and B won't matter.
The forces will come out the same (I think).

 

[Femme_physics]

Yes, there is something to calculate, I just got a little confused so I looked at the solution, but then got even more confused and posted it here, heh.

Thanks for all the replies :) I'm glad this is sorted.

 

[Studiot]

Since all the technical stuff is sorted, perhaps you would let us in on a secret.

All your diagrams look as though John Harrison himself might have drawn them, back in 1763.

What sort of textbook are you using?

 

[Femme_physics]

The diagrams in the question are taken from the official mechanical engineers test of Israel's technological workforce training institute. The one I posted was from a 2006 test. The solution manual is of an Ukrainian-Israeli physics teacher, and the only known solution manual for these tests. Although, I'm considering to make my own solution manual though the rate this is going.

I agree they're kinda outdated looking and drab. Hibbeler's doing a much better job at it.

 

[Femme_physics]

Pardon the interruption, but I do believe that when solving a symmetrical shape we get to isolate one half of it. Is that correct? Look what I did trying to solve for Na. It is my humble opinion that if the friction force at Na had any arm, the result would be different. However, the result I turned up with is exactly like the solution manual says. (attached are my calculations. Again, I only used the upper part of the shape)

So this is why this post

However, if you would have to calculate something, the distance between A and B won't matter.
The forces will come out the same (I think).

Is rather conflicting to me.

 

[I like Serena]

Pardon the interruption, but I do believe that when solving a symmetrical shape we get to isolate one half of it. Is that correct? Look what I did trying to solve for Na. It is my humble opinion that if the friction force at Na had any arm, the result would be different. However, the result I turned up with is exactly like the solution manual says. (attached are my calculations. Again, I only used the upper part of the shape)

So this is why this post



Is rather conflicting to me.

No, you always solve the entire shape, although symmetry will often show you beforehand that some forces or moments will for instance cancel out.

However, what you did not take into consideration yet, is that the body you have is not rigid. It has a hinge in the middle allowing free rotation, so you need to treat it as 2 separate free bodies.

 

[Femme_physics]

No, you always solve the entire shape, although symmetry will often show you beforehand that some forces or moments will for instance cancel out.

Well, I tried solving for the entire shape and it's impossible to get to Na that way. However, looking at only half of the object, I get the correct result for Na (as shown in my attachment above).

A 2nd method to solve fo Na would be to isolate one of the diagonal beams. However, again, unless the friction force is acting on the same plane at C, you can't solve for either Na or the friction force. (beam isolation diagram attached from the manual)

However, what you did not take into consideration yet, is that the body you have is not rigid. It has a hinge in the middle allowing free rotation, so you need to treat it as 2 separate free bodies.

But if C is a pin-connected, can't I look at the entire body first with all the "external forces" acting on it before I isolate it? In the 2nd body diagram at my first post, the one I'm using, they didn't write down any force at C, so they're presumably seeing it as a rigid body.

 

[I like Serena]

Well, I tried solving for the entire shape and it's impossible to get to Na that way. However, looking at only half of the object, I get the correct result for Na (as shown in my attachment above).

A 2nd method to solve fo Na would be to isolate one of the diagonal beams. However, again, unless the friction force is acting on the same plane at C, you can't solve for either Na or the friction force. (beam isolation diagram attached from the manual)


But if C is a pin-connected, can't I look at the entire body first with all the "external forces" acting on it before I isolate it? In the 2nd body diagram at my first post, the one I'm using, they didn't write down any force at C, so they're presumably seeing it as a rigid body.

I'm confused.

The last attachment is apparently from you solution manual, and it shows that they have taken one diagonal beam as a free body.
It also contains a force at C.
And if you do the moment sum relative to C, you can solve it for Nb.

Btw, if you take the 2nd body diagram at your first post as a rigid body, it can not "clasp".

 

[Femme_physics]

The last attachment is apparently from you solution manual, and it shows that they have taken one diagonal beam as a free body.
It also contains a force at C.
And if you do the moment sum relative to C, you can solve it for Nb.

Yes, but only if we presume that the friction force has no arm at C! Remember our initial discussion about whether it's important for the friction force to have an arm?

The quote:

However, if you would have to calculate something, the distance between A and B won't matter.
The forces will come out the same (I think).

So in my last attachment of the diagonal beam, that distance would matter!

Btw, if you take the 2nd body diagram at your first post as a rigid body, it can not "clasp".

So the second body diagram is wrong and unrealistic?

 

[I like Serena]

Yes, but only if we presume that the friction force has no arm at C! Remember our initial discussion about whether it's important for the friction force to have an arm?

My bad!
You're ever so right!

So in my last attachment of the diagonal beam, that distance would matter!

Yes, the distance would matter.

So the second body diagram is wrong and unrealistic?

Since we know now that the distance would matter, we need the distance to solve the system. However, since it is not given we will have to make an assumption.

What if the neck of the bottle is very small compared to the contraption?
Then we could neglect the size of the neck of the bottle, that is, treat the distance as zero.

This would make the second diagram more realistic than the first diagram!

 

[Femme_physics]

Great, so our diagram is more realistic than the actual picture drawing lol. I blame them for not giving/defining a proper distance so we have to make assumptions instead. Very tricky. I'd disqualify this question if I was in charge of the tests. Thank you as always Serena for helping me sort through that.

 

[I like Serena]

Great, so our diagram is more realistic than the actual picture drawing lol. I blame them for not giving/defining a proper distance so we have to make assumptions instead. Very tricky. I'd disqualify this question if I was in charge of the tests. Thank you as always Serena for helping me sort through that.

Well, in reality, the contraption would be meant to be used for different size bottles.
So then you would have to do the calculation twice.
Once for a bottle with the smallest neck possible, and once for a bottle with the largest size the contraption can still hold.

If you would do the numbers, you will find that the smaller the neck-size, the smaller the normal force, and with a smaller normal force, comes a smaller friction.

So a neck-size of zero would correspond to the ultimate worst case scenario, where the contraption might not be able to move the bottle.
This is exactly what a user of the contraption would want to know, and what would be a part of the specification of the contraption.

I suspect these deliberations are a bit out of scope for the problem at hand though.

 

[Femme_physics]

Well, in reality, the contraption would be meant to be used for different size bottles.
So then you would have to do the calculation twice.
Once for a bottle with the smallest neck possible, and once for a bottle with the largest size the contraption can still hold.

If you would do the numbers, you will find that the smaller the neck-size, the smaller the normal force, and with a smaller normal force, comes a smaller friction.

So a neck-size of zero would correspond to the ultimate worst case scenario, where the contraption might not be able to move the bottle.
This is exactly what a user of the contraption would want to know, and what would be a part of the specification of the contraption.

I suspect these deliberations are a bit out of scope for the problem at hand though.

Always good to hear your expert's perspective! I think it's important to know the logic behind things, it makes you think, and shows you how interestingly dynamic and rich of information is that field of mechanics (even statics is dynamic, if you catch my drift). It was an interesting read, for sure. Keep it up. :)

 

[Studiot]

The line of action of the pin forces at C must be at right angles to the line of action of F.

Your mechanism is a standard set of lifting tongs.

They have the property that they are self locking. This is a posh way to say that if used for vertical lifting the max weight (W) they can lift is independent of W. It depends only on the coefficient of friction and the geometry of the tongs.
If you would like to PM me an Email address, capable of receiving jpg I will send you an analysis - it runs to 7 pages too much to post here.

 

[Femme_physics]

While we're on the issue of false diagrams. Can anyone explain why there's no force T (rope tension) emerging from point E in this FBD? Compare the FBD of the solution manual to mine. IMO, mine makes a whole lot more sense.

Original

http://img860.imageshack.us/img860/1424/original1c.jpg


Solution manual

http://img580.imageshack.us/img580/2478/solution1j.jpg

Mine

http://img101.imageshack.us/img101/547/mine1on.jpg

 

[I like Serena]

While we're on the issue of false diagrams. Can anyone explain why there's no force T (rope tension) emerging from point E in this FBD? Compare the FBD of the solution manual to mine. IMO, mine makes a whole lot more sense.

These are not free body diagrams.
These are diagrams of the total system with only the external forces drawn.

Actually, the rope tension is not an external force, but the real external force is the gravity force W which is pointing downward.
In the solution manual, they have replaced W by the tension force T (in the same direction) which is naughty.

The force T you drew at E is an internal force, and it's to early to draw that one, because then other internal reactive forces would have to be drawn as well.

[edit]Note that I consider the rope and the wheel at D as separate free bodies.[/edit]

 

[Femme_physics]

*smacks forehead* Right. Internal. My mechanics intuition needed some recalibration, thanks :)

 

[Femme_physics]

Actually, the rope tension is not an external force, but the real external force is the gravity force W which is pointing downward.
In the solution manual, they have replaced W by the tension force T (in the same direction) which is naughty.

I'm not sure why it is "naughty", but since YOU said it's naughty, I will forever henceforth write T as W in such cases!

Not that there's anything so wrong with being naughty... ;)

 

[I like Serena]

I'm not sure why it is "naughty", but since YOU said it's naughty, I will forever henceforth write T as W in such cases!

Not that there's anything so wrong with being naughty... ;)

Ah well, there's actually a 4th free body which hasn't been drawn, and I guess it should have been drawn.
That is the mass that is suspended on the rope.

If we look at the free body diagram of this mass, there are 2 forces:
1. The gravity force W
2. The upward rope tension T that is equal and opposite to W.

The downward rope tension T that has been drawn in the solution-manual-diagram, is actually the reactive force.

And I agree that naughty can be a lot of fun! :P

 

[Femme_physics]

Oh, actually, it's there, I just didn't scan it. Now I did.

In this case it's fair game to write T, yes?

 

[I like Serena]

Oh, actually, it's there, I just didn't scan it. Now I did.

In this case it's fair game to write T, yes?

Yes, now you *have to* write T, although this only becomes relevant when we move to dynamics. :)

 

[Femme_physics]

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.

 

[Femme_physics]

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.

 

[I like Serena]

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.

 

[Femme_physics]

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?

 

[I like Serena]

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.

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.

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.