# Transition from frictionless surface to frictional surface

1. Jun 7, 2014

### Nathanael

Suppose you have an object sliding along a frictionless floor, but it's approaching a line where the floor changes from being frictionless to some constant coefficient of kinetic friction $\mu_k$

When it is partially on the frictionless surface and partially on the frictional surface, would you calculate the frictional force by multiplying $\mu_k$ by the entire weight of the object? Or would you only multiply it by the portion of the weight that is above the frictional surface?

I think you would just multiply $\mu_k$ by the portion of the weight of the object that is above the frictional surface, but my textbook says nothing about this sort of situation.

(My question also applies to the transition between one surface to another when the two surfaces have a different $\mu_k$)

2. Jun 7, 2014

### UltrafastPED

The weight on the frictional surface; the rest of the weight is on the frictionless surface.

3. Jun 7, 2014

### Delta²

I think there is an extreme case where this doesnt hold, if the left portion of the body exerts somehow a net force on the right portion of the body that has a vertical component (parallel to the weight).

4. Jun 8, 2014

### Nathanael

I don't quite understand what you're describing. (Rather, I don't understand how this could occur.)

Can you perhaps think of an example?

5. Jun 8, 2014

### Delta²

Lets say you are attached to the left portion of the body and i am attached to the right portion.

Suppose we can stretch our hands and bodies enough such as to always hold together with our hands lying exactly over the boundary. We can exert a force to each other such that the net force i am applying to you is upwards +F, and (due 3rd law of Newton ) the net force you are applying to me will be downwards -F.

Assuming all that the normal force (from the frictionless surface) on the left portion where you standing will be B1-F, while that on the right portion (from the frictional surface) will be B2+F, where B1.B2 the weight of the left and right portions.

6. Jun 8, 2014

### Nathanael

Oh, I see. So even though there is an amount of weight above the frictional surface, it's possible that the weight is somehow supported by the side over the frictionless surface. (I didn't misinterpret you, right?)

Good point, I didn't think of that unique possibility.

7. Jun 8, 2014

Yes exactly.

8. Jun 8, 2014

### jbriggs444

This scenario does not work due to conservation of angular momentum. The upward force from you amounts to a torque on your partner. The downward force from your partner amounts to a torque on you. Once you trace it all out, there is no interaction you can do to cause an unbalanced force on the object (as long as you stand still and do not start rotating in place at an ever-increasing rate).

Unfortunately, Nathanael's supposition about the distribution of support force still does not hold for all objects. It will pretty clearly work for rugs being dragged across a floor by a horizontal force. The distribution of the support force matches the distribution of the rug's weight. But it fails for tables and chairs. The distribution of support force does not match the distribution of a table's weight. The support force will shift abruptly from one side of the line to the other as each leg crosses the line, not slowly as the table's weight crosses the line.

9. Jun 8, 2014

### Delta²

What conservation of angular momentum you talking about, there are the external normal forces from the surfaces and the weights which are also external forces to the system.

10. Jun 8, 2014

### CWatters

Indeed, the effect might even be enough for the table to tip over.

Diagram may help (It's not intended to be a FBD).

With friction acting below the centre of mass there will be a weight transfer onto the part on the surface that has friction.

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11. Jun 8, 2014

### CWatters

Even for flat bottomed rigid objects on uniform surfaces it's not clear what happens when there is uneven weight distribution. My guess is the coefficient of friction itself changes.

12. Jun 8, 2014

### jbriggs444

Consider a system consisting of you and your partner. Consider an axis of rotation going through your combined center of mass. What forces are present?

There is the force of gravity. But that has no net torque about your center of mass.

There is the interaction between the two of you. But that is internal to the system and cannot change your net angular momentum.

There is the support force of the object holding the two of you up. The point of the exercise was to apply unbalanced forces to the object. It follows that the object is applying unbalanced forces to the two of you. That's a net torque.

Where is the resulting change in your angular momentum?

13. Jun 8, 2014

### AlephZero

That is right as a general principle, but as others have said the hard part is dividing the weight between the two parts of the surface.

If the center of mass of the object is above the surface, when the object is decelerating there is more weight on the front of the object than on the back. It is easier to work out the details if the object touches the ground at a few separate points rather than over an area (e.g. a table, a bike or car, etc).

As an extreme example, think about a tall block of material, or a human sliding on ice. When it hits the friction surface, it might even topple over, rather than continue sliding.

Textbooks don't mention "real world" situations like this, because modelling them "accurately" is complicated, often requires computer simulations rather than algebra and calculus, and is not a very effective way to learn the basic principles.

14. Jun 8, 2014

### dauto

You seem to have forgotten to include the torque of the friction force itself

15. Jun 8, 2014

### jbriggs444

The "friction force itself" would apply between the object and the floor. It does not apply to the system in question -- the two people standing on top and trying to apply an unbalanced external force.

I agree that you are correct in suggesting that there may be a lateral force between the people standing on top of the object and the object itself. If the object is slowing down and the people are standing in place relative to it, there is certainly some such lateral force. Unless their center of gravity is at their feet, that lateral force will produce an external torque on the system composed of the two people.

The key point that I was trying to make was that no internal interaction between the two people on the platform can change their angular momentum. If they are to produce a net external torque, they need to do something other than "I'll push up on your right hand if you push down on my left".

I think that point is made more clearly if one disregards the acceleration of the object. But perhaps I should have stipulated people of negligible height instead.

16. Jun 8, 2014

### UltrafastPED

I was assuming something like a puck; you can imagine what will happen for various situations by considering that object freely sliding on ice, then, suddenly encountering a patch that has sand scattered on it.

The ice skater would probably topple, the table maybe flip over, and the puck (depending on weight, etc) would probably just slow down, then come to a stop.

In the case of the puck it would feel less total force when the edge of the puck encounters the sand, and the total resistive force would increase as the puck makes the transition.

Something similar happens with a toboggan that runs out of snow and hits a patch of grass; if the patch is small, you just feel a jerk. If the patch is bigger, you get thrown forward, and maybe off.

This is my experience from growing up in the frozen north, and spending quality time outdoors during the winter.

17. Jun 8, 2014

### Delta²

I said that the human bodies are attached to the left and right portions respectively which means that they cant move or rotate unless the respective portions do so also. Essentially they become extensions of those portions.

18. Jun 8, 2014

### jbriggs444

If they are extensions of the object, any forces they exert on each other cannot induce any net torque on the object.

19. Jun 8, 2014

### Delta²

Yes the net torque is zero cause we have to equal and opposite forces at the same points where the hands are meeting.

20. Jun 9, 2014

### jbriggs444

Right. But recall the point of the exercise. You want the fellow on the right to be pushing down on the object harder than the fellow on the left. That is, by itself, a non-zero net torque. And that cannot happen.

Try looking at it this way. The down force on the left hand on the fellow on the right amounts to a counter-clockwise torque on him. [Looking at him from the back]. The up force on the right hand on the fellow on the left also amounts to a counter-clockwise torque on him.

You have said that neither fellow starts moving. They stand in place on the object. Let us ignore for the moment any linear acceleration of the object. That means that there must be a counter-balancing torque on each fellow. That torque must come from their feet. The left hand fellow stands more heavily on his left foot than on his right. The right hand fellow stands more heavily on his left foot than on his right.

So while it remains true that the right hand fellow is standing more heavily then the left hand fellow, both their left hand feet are standing more heavily than their right. The net effect is zero.

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