Directions of Friction Force (Pipe)

[Lisciu]

Homework Statement

The 4000N concrete pipe is being lowered from the truck bed when it is in the position shown:

If the coefficient of static friction at the points of support A and B is 0.4, determine where it begins to slip first: at A or B or both at A and B

Homework Equations

It's not so complicated task.

ΣF_x = 0 = N_a+F_b - Wsinθ
ΣF_y = 0 = N_b+F_a - Wcosθ
ΣM_o = 0 = F_br - F_ar = 0
F_max=μN_max

The Attempt at a Solution

[/B]
The question I have is why the friction at support A have the direction with montion of the pipe and not trying to oppose the motion. We assume that the pipe is moving clockwise direction according to our friction direction at support B. Can someone explain me this ?

 

[haruspex]

3. The Attempt at a Solution

The question I have is why the friction at support A have the direction with montion of the pipe and not trying to oppose the motion. We assume that the pipe is moving clockwise direction according to our friction direction at support B. Can someone explain me this ?

If A and B were fixed and some external torque were acting on the cylinder then when the cylinder slips it will move in the same sense against both supports, so the friction would act the same way around.
But the situation here is different. The points of support, A and B, are about to move further apart. If the cylinder does not slip at A then it will slip clockwise relative to B, establishing that the friction acts anticlockwise on the cylinder there.
Conversely, if it does not slip at B then it will slip anticlockwise relative to A.

 

[Merlin3189]

If you consider moments about the pipe axis, F_A and F_B must be opposite and equal.
The direction you show on your diagram is arbitrary. If you show both as CW for example, one will be positive and the other negative.
Until you calculate a value, you do not know which direction either is acting.
Edit: Haruspex is right. His extra consideration shows their direction at this point.

 

[Lisciu]

But the situation here is different. The points of support, A and B, are about to move further apart. If the cylinder does not slip at A then it will slip clockwise relative to B, establishing that the friction acts anticlockwise on the cylinder there.
Conversely, if it does not slip at B then it will slip anticlockwise relative to A.

I don't understood this. Could you say something more or even give me picture of it? I don't get it what you mean saying: "A and B are about to move further apart". Because the things you saying looks like it can sleeping and B clockwise and counter clockwise at A. I can't imagine the same wheel with diffrent direction of slipping at the same time...It impossible to get slipping to right and other point to left...

If you consider moments about the pipe axis, F_A and F_B must be opposite and equal.
The direction you show on your diagram is arbitrary. If you show both as CW for example, one will be positive and the other negative.
Until you calculate a value, you do not know which direction either is acting.
Edit: Haruspex is right. His extra consideration shows their direction at this point.

Clearly it was what I actually did because I will have that for example F_a and that will mean that I have wrong direction on diagram. But i don't understood the physical aspect of it. Because it can actually rotate for A clockwise and for B counterclockwise for example if we look on those directions of friction... What does mean those direction?

 

[haruspex]

Could you say something more or even give me picture of it? I don't get it what you mean saying: "A and B are about to move further apart".

First, it's important to understand the mechanism on the truck. There is a gate, the upper end of which is at B. The hinge is at the bottom, vertically below A. A telescopic arm angling down to the right from A extends to lower the gate. As it does so, A and B get further apart. That is why the cylinder must slip at one or other, or both, of those two places.
So it may help to consider a simpler version. A horizontal plank length L rests on two supports. The supports are distances X, Y from the ends of the plank respectively. The supports are being pushed further apart. The plank must slip on one or both supports, but it will not slip the same way on both. The right hand support is going to the right, so the frictional force on the plank is to the right. Conversely, the frictional force on the plank from the left hand support is to the left

 

[Lisciu]

Okay I understand what you want to say. But the problem is that for me:

If the support from right hand going to right then the friction is to the left... And when the support from left side going to left the friction is to the right. Why you saying something totally diffrent?

 

[haruspex]

If the support from right hand going to right then the friction is to the left..

It's to the left on the support but it's to the right on the plank.

 

[Lisciu]

Totally don't understand that...and I'm confused...

That's what I see. Looks like I don't understand the concept...

 

[haruspex]

Do you agree that the frictional force on the support is to the left? It opposes the attempted motion of the support relative to the plank, yes?
When friction occurs at an interface, it is always as a pair of equal and opposite forces, action and reaction. It is one way on one body and the other way on the other body.
As far as the plank is concerned, friction from the support is trying to drag it to the right.
To look at it another way, without friction, the plank would move left relative to the support. Since friction opposes relative motion of the surfaces in contact, it acts to the right on the plank.

 

[Lisciu]

Almost I grasp this. But it raise one question to me.

Why we saying that the friction from the support is trying to drag it to the right? Isn't a force that we applied to support is trying to drag it to do right? The friction on the support is working to the left side.

Another example. If you say that the friction it is always as a pair of equal and opposite forces, action and reaction. Then I can say that if we have just one box on surface. There will be friction to the left relative to the box moving to the right. But also the friction to the right relative to the surface. So no friction at all because their are equal...

 

[haruspex]

Why we saying that the friction from the support is trying to drag it to the right? Isn't a force that we applied to support is trying to drag it to do right? The friction on the support is working to the left side.

If you push the support to the right, that is a force applied to the support. It does not act on the plank. If there were no friction between support and plank then there would be no force dragging the plank.

Another example. If you say that the friction it is always as a pair of equal and opposite forces, action and reaction. Then I can say that if we have just one box on surface. There will be friction to the left relative to the box moving to the right. But also the friction to the right relative to the surface. So no friction at all because their are equal...

The two equal and opposite forces act on different bodies. If you only look at one body at a time then you must include these forces. If the static friction is sufficient that there is no slipping then you can choose to consider the two bodies as one for the purposes of a free body diagram. In that case the two frictional forces are internal to that body and do cancel.

 

[Lisciu]

I guess you have a hard time with me. :) I appreciate that. I just back to physics after more then 10 years.

Okay I agree that the force is acting on support only. The second statement is also correct and I understand. The thing bother me is how we determined the direction of friction for the Plank ? If we going to 3rd Law and draw the FBD for support aren't we have only the friction of support going to the left on the ground and to the left acting on the upper edge of support? If so then we going to 3rd law and do FBD for plank and according to 3rd law we take the friction force to the right side? And from FBD we see that there is no other force acting on x-axis so it will looks like we have 0 friction... still something is wrong in my assume.

But what if there is slipping? There is a lot of exercises when is checked the Tmax to consider on which case the body is. But I didn't see the any diffrence then just taking the dynamic coefficient of friction.

Can you give an example of this canceling two frictional force?

 

[Lisciu]

If you push the support to the right, that is a force applied to the support. It does not act on the plank. If there were no friction between support and plank then there would be no force dragging the plank.

The two equal and opposite forces act on different bodies. If you only look at one body at a time then you must include these forces. If the static friction is sufficient that there is no slipping then you can choose to consider the two bodies as one for the purposes of a free body diagram. In that case the two frictional forces are internal to that body and do cancel.

I study our conversation and I decided to clarify that bothers me here. And I will show step by step what the problem for me is.

Step 1 :
Is just recognize two point of contact on the pipe. And as you mentioned the hydraulic mechanism that make those two 2 points further away from each other.

Step 2:
We have normal force from the point of support at A and B normal to the surface. Nothing extraordinary for me.

Step 3:
This where this madness begin. In my workflow I assumed that when the hydraulic mechanism trying to get the pipe down and obviously change the distance between those 2 supports. Then the friction will occur to prevent the rotation of the pipe. If we assume the clockwise direction (we don't know if it's right). Then friction will trying to block the slipping by acting to right side (with angle of course, but not important right now). And the same will be for the point A the pipe will try to rotate at point A and friction will be trying to stop it by acting downard. So actually the problem for me is that even with this plank I can't imagine this situation for the pipe.

Like you said that this pipe try to rotate counterclockwise at A and clockwise at B in the same time.

 

[haruspex]

This where this madness begin. In my workflow I assumed that when the hydraulic mechanism trying to get the pipe down and obviously change the distance between those 2 supports. Then the friction will occur to prevent the rotation of the pipe. If we assume the clockwise direction (we don't know if it's right). Then friction will trying to block the slipping by acting to right side (with angle of course, but not important right now). And the same will be for the point A the pipe will try to rotate at point A and friction will be trying to stop it by acting downard. So actually the problem for me is that even with this plank I can't imagine this situation for the pipe.

Your mistake here is in thinking of the pipe's rotation independently of the movement of the mechanism. Certainly if the mechanism were fixed and something else were twisting the pipe then the two frictions would act in the same sense, clockwise or anticlockwise. In the plank case, it is like taking the supports to be fixed and noting that the friction must act on the plank ro the left at both or to the right at both.

But in this problem you need to consider the two possible rotations in relation to the movement of the mechanism:
1. There is no slip at point A. The slip at B will be clockwise relative to the mechanism; or
2. There is no slip at point B. The slip at A will be anticlockwise relative to the mechanism.

 

[Lisciu]

Heh, okay. But then again I can't imagine this case. How symmetrical thing like pipe can just on slip at B and not at A point. If some portion of the circle will move then it need to move on other side, because without that we need to asume that circle will compress or deform whatever... And to looks like you said ( it's good answer of course) we take one case second case and plot them in one drawing as the can exist in the same time... It's impossible in reality. Where is the heck ?

 

[haruspex]

How symmetrical thing like pipe can just on slip at B and not at A point.

Try it. Balance a cylinder (cup?) on two horizontal parallel fingers, one a bit higher that the other. Move the fingers apart slowly.

 

[Lisciu]

You probably don't believe but it's actually act like I said :D it's rotate anticlockwise. So friction looks like acting clockwise in both points... Still don't get it.

But I do other test easier. I take this cylinder between two fingers in position like you said

Step 1 : Fixed left finger. Moving right finger to right - anticlockwise movement
Step 2 : Fixed right finger. Moving left finger to left - clockwise movement.

This will be then as you mentioned but can't imagine at the same time as I said before rotation on two diffrent direction at the same time. Then the pipe shouldn't move at all.

 

[haruspex]

You probably don't believe but it's actually act like I said :D it's rotate anticlockwise. So friction looks like acting clockwise in both points... Still don't get it.

not sure what you are saying here. Are you saying that the given answer to the original question is that it rotates anticlockwise? (I agree with that.). But that does not mean that friction acts clockwise at both. It will act anticlockwise at B and clockwise at A. Because B is below A, the normal force is greater at B, so the maximum frictional force is greater at B. So that frictional force wins, leading to an anticlockwise rotation.

Step 1 : Fixed left finger. Moving right finger to right - clockwise movement.
Step 2 : Fixed right finger. Moving left finger to left - clockwise movement.

That doesn't sound like the test I suggested. Make sure that one finger is lower than the other, so takes more of the weight. It shouldn't matter which finger you move, just move them slowly apart. My guess is that you accidentally had a different finger lower in the two tests.

 

[Lisciu]

I do so many reps with this test and I conlude that in your way for it i can only observe rotation in antyclockwise in on finger and on other no slipping (in the finger that have more weight on it). But it's still tell that the force of friction are in the same clockwise direction...

 

[haruspex]

I do so many reps with this test and I conlude that in your way for it i can only observe rotation in antyclockwise in on finger and on other no slipping (in the finger that have more weight on it). But it's still tell that the force of friction are in the same clockwise direction...

How do you know which direction the frictional force is from the finger that does not slip?

 

[Lisciu]

Because the wheel can't rotate in two direction at once. I mentioned that before. So friction work opposite to motion.

 

[haruspex]

Because the wheel can't rotate in two direction at once. I mentioned that before. So friction work opposite to motion.

That does not imply the friction is in the same direction.
Let's try a different model, one that involves rotation this time.
A drum stands on a vertical axis. You put your hands against it and push to the right. Your friend, standing on the opposite side, pushes the drum to her left. One of you will win. Do you agree that the two frictional forces being applied are in opposite directions?

 

[Lisciu]

Okay now I Get it :) The main problem I have was to imagine this case. I don't even see it as simple and do some mistakes with assumption.

Thank you haruspex now I see why you're a Gold member!