Friction forces and motion resulting from them

[Robittybob1]

Not in the coordinate system you have chosen.

 

 

 

Again, not in the coordinate system you have chosen.

 

This is the problem I mentioned earlier. You can't jump between coordinate systems.

 

 

I don't know what this means. What table?

Do you have just one frictional force or none at all? I can't seem to snap my thinking away from the two frictional forces at the moment. How many do you have?

The table had a list of materials with their coefficients of friction and I can't recall the exact figure (either 0.1 or 0.2) but there was no difference between the static and the dynamic friction coefficient of glass on teflon, I was surprised but I suppose that can happen.

[swansont]

Do you have just one frictional force or none at all? I can't seem to snap my thinking away from the two frictional forces at the moment. How many do you have?

The table had a list of materials with their coefficients of friction and I can't recall the exact figure (either 0.1 or 0.2) but there was no difference between the static and the dynamic friction coefficient of glass on teflon, I was surprised but I suppose that can happen.

 

You break it down into components. There will be a radial component and a tangential one.

 

0.1 or 0.2 isn't very high precision. All that means is the difference is smaller than the precision of the measurement.

[Robittybob1]

 

You break it down into components. There will be a radial component and a tangential one.

 

0.1 or 0.2 isn't very high precision. All that means is the difference is smaller than the precision of the measurement.

Well that is my point; if there two friction forces that combine into one force why not just keep on analysing it as if the two forces keep acting separately?

What I am needing to do is to understand the maths of positive work being done by friction.

Formulas for work done by friction - here please.

Edited May 31, 2015 by Robittybob1

[swansont]

W = F(dot)d

 

Forces orthogonal to displacement do zero work.

[Robittybob1]

W = F(dot)d

 

Forces orthogonal to displacement do zero work.

But that goes in the face of one of the questions at the beginning of the thread in that we had concluded the orthogonal forces combine vectorially and add together, so how would you get the force to always be orthogonal?

Push down on a car stuck in the mud isn't doing any useful work, if the others are trying to push it out, OK it works there, but when the object can move freely in the xy plane forces in the same plane at right angles to each other will add to produce a stronger force. I am wondering if you can have a force in the xy plane that is orthogonal to displacement in the xy plane.

[i know that is not quite answering what you said, but just a thought.]

In the rotating turntable the distance in the radial direction is r(final) - r(initial). The object slides that distance so that amount of work is done in that action of dragging the mass across the surface. In the other direction (tangential) it is the arc length back to where is slips off the rim.

 

Now if someone is paying for work done, the force of friction on the diagonal is the same as the pushing the mass in any individual direction on the surface, yet the combined distances (x and y components added) are less than the resultant! I'm a bit lost with that one.

[J.C.MacSwell]

Well that is my point; if there two friction forces that combine into one force why not just keep on analysing it as if the two forces keep acting separately?

What I am needing to do is to understand the maths of positive work being done by friction.

Formulas for work done by friction - here please.

Not sure if this is being missed, but once in the dynamic friction range in any direction it is in that range in any other direction, until it comes to a stop. (there are transitional effects, but not significant here)

[swansont]

But that goes in the face of one of the questions at the beginning of the thread in that we had concluded the orthogonal forces combine vectorially and add together, so how would you get the force to always be orthogonal?

Did you understand this:

 

You break it down into components. There will be a radial component and a tangential one.

[Robittybob1]

Not sure if this is being missed, but once in the dynamic friction range in any direction it is in that range in any other direction, until it comes to a stop. (there are transitional effects, but not significant here)

What I am coming to understand is that the resultant direction the object takes determines the alignment of the dynamic friction force.

This dynamic friction force vector could be broken down into the radial and tangential components, but the components are less than the dynamic friction force needed to be overcome to make the object slide in any one component direction.

 

So does that mean any centripetal force being provided by the dynamic friction (the radial component of it) declines as the object takes on a more transverse trajectory? * Does this failing centripetal force contribute to the apparent increase in the radial speed that the object obtains as it reaches the edge? *

 

* * This could be a very important factor that I haven't heard being discussed before.

Edited June 1, 2015 by Robittybob1

[Robittybob1]

One aspect of the idea is bothering me. What would happen in the situation of an extremely large disk, would the object be carried around for more than one turn of the spiral once it has dynamic friction?

[swansont]

It could. It depends on the coefficient of friction. The wheel could be turning quickly before the static friction fails, and the dynamic friction could be only slightly smaller.

[Robittybob1]

It could. It depends on the coefficient of friction. The wheel could be turning quickly before the static friction fails, and the dynamic friction could be only slightly smaller.

That could be true if the mass starts off very close to the center of rotation. Thanks.

[studiot]

Robittybob,

 

I note you have spent several pages considering what happens in two directions.

 

This, more advanced question has puzzled many and can be approached not from the direction you have taken but from the other direction.

Start with a body moving in a straight line against friction and then consider what happens as it enters a slight bend and a radial force appears.

This is a simpler approach.

 

You might like to use all that others here have you to consider this question that appeared elsewhere.

 

 

If a car is accelerating on a straight level road with a force on the driving tires equal to uN (max available without slipping) and then enters a slight curve, now friction acts centripetally also, but since there's none left and friction is a vector force, the car cannot safely negotiate the curve without slipping outward, no matter how slight the curve or speed may be. Is this a correct statement?? Or does friction not obey vector law, that is, a force of up to uN is available in both perpendicular and tangential directions?

Edited June 3, 2015 by studiot

[Robittybob1]

Robittybob,

 

I note you have spent several pages considering what happens in two directions.

 

This, more advanced question has puzzled many and can be approached not from the direction you have taken but from the other direction.

Start with a body moving in a straight line against friction and then consider what happens as it enters a slight bend and a radial force appears.

This is a simpler approach.

 

You might like to use all that others here have you to consider this question that appeared elsewhere.

 

With rubber friction is supposed to be different. Commercial - Firestone Tires - Where the rubber meets the road!

Drive down memory lane for a minute!

 

There is the motor sport of drifting where what you describe is the purpose and the art.

Edited June 3, 2015 by Robittybob1

[studiot]

The numbers might be different for pneumatic tyres, but the principles remain the same.

 

The physicist that I had a most enlightening discussion about this with is also an advanced/skid driving instructor.

 

You can treat this as a learning opportunitity or you can be a smart___ and remain no wiser.

[Robittybob1]

The numbers might be different for pneumatic tyres, but the principles remain the same.

 

The physicist that I had a most enlightening discussion about this with is also an advanced/skid driving instructor.

 

You can treat this as a learning opportunitity or you can be a smart___ and remain no wiser.

Were you going to share the link to the discussion?

[Robittybob1]

"Centrifugal force is a fictitious force that is only appropriate to create pseudo equilibrium.

It cannot and does not move anything outwards."

 

When the radial component of the dynamic friction is less than the required centripetal force to enable the object to travel in a circle, the object no longer travels in a circle but a spiral. I am trying to see if the results of friction can be explained without using the word "centrifugal force".

[studiot]

 

Studiot post#1

1) Friction always acts to oppose motion or the tendency to move.

 

Coulomb's first law of friction.

 

It is important to realiise that a body can only move in one net direction.

You can resolve that net tendency to move in as many ways as you like, but there is only one net direction.

 

the net Friction always acts in that direction and is equal to the static or dynamic coefficient times the reaction depending upon whether the object is moving or not.

 

You cannot have the static coefficient in effect in one direction and the dynamic in another.

[studiot]

Since there has been some talk of objects being held in place by friction in one direction, whilst being moved along by another force in some other direction here is a simple development of what actually happens.

 

Start by placing a moderately heavy block on a rough table as in Fig1.

 

Apply and maintain a fixed push at A, insufficient to move the block as in Fig 2

This can be achieved qith a spring loaded push rod.

The action of the friction force exactly opposes A.

 

Now, whilst maintaining A, apply an increasing second force B at right angles to A as in Fig 3.

The fig shows B equal to A and the tendency to move angeled midway between A and B.

Note that the opposing friction force is larger than A or B, but still enough to hold the block still.

 

Continue to increase B and the tendency to move rotates round, becoming more nearly parallel to B as in fig 4

 

Finally the combined push of A and B is enough to move the block along aome line close to B as in Fig 5.

Note that the block now moves away from A, (and therefore at right angles to B) although we know that friction is enough to hold the block against A alone.

That is the block does not move parallel to B. In fact, so long as A is non zero, the movement is never parallel to B.

Measurement of B would show that B alone is also not large enough to overcome static friction.

 

This experiment shows that you cannot have a force moving an object against dynamic friction directly along its line of action whilst another force holds that object from moving off that line by static friction.

 

 

 

 

 

[Robittybob1]

Does your diagram show a moving surface? With a spinning turntable with a mass held onto it by friction alone. At a certain rotational speed it does slip outward but since the surface further out is rotating at a higher tangential speed the path it takes is the resultant of being slowed by friction in the radial direction but accelerated in the tangential direction which in turn accelerates it in the radial direction further overcoming the slowing due to friction.

I am certain of this. Are you saying that is impossible? There is only one resultant direction so it moves in a curve across the turntable surface. With respect to the ground I am uncertain as to its path, but it isn't slipping on the ground, and even Swansont and I thought it might be possible the mass could do a full "circle" during the slip phase.

That will be my next level of investigation (the motion WRT the ground).

[studiot]

 

Does your diagram show a moving surface?

 

 

Are you serious?

 

Is that your only comment on the work I out in to help?

 

Yes the table could be mounted on a railway truck travelling in an evacuated straight level tunnel at a steady 25 m/s.

Do you understand what the frictional force must be between the table top and the block in those circumstances?

Edited June 4, 2015 by studiot

[Robittybob1]

 

 

Are you serious?

 

Is that your only comment on the work I out in to help?

 

Yes the table could be mounted on a railway truck travelling in an evacuated straight level tunnel at a steady 25 m/s.

Do you understand what the frictional force must be between the table top and the block in those circumstances?

OK it matters not how fast the truck goes in that circumstance. I should have said "accelerating surface" rather than just "moving surface", for the surface of turntable is accelerating even when the angular velocity stays the same.

 

I am just trying to get you to declare whether you thought an accelerating surface gives you a different result than what your diagram shows. There is nothing wrong with the physics of your diagrams but it doesn't cover the situation of circular motion.

[studiot]

 

I am just trying to get you to declare whether you thought an accelerating surface gives you a different result than what your diagram shows. There is nothing wrong with the physics of your diagrams but it doesn't cover the situation of circular motion.

 

 

One thing I have learned over many years is that when I want to understand something is to start with something simple (and preferably well known) and build up my picture, along with my understanding.

 

That is what I am trying to (help you ) do here.

 

So the next stage is not to jump straight to rotating systems but allow my train to accelerate along the tunnel.

 

This should establish the principle that there is now another force acting on the block, that should be taken into account.

 

A force is (by definition) a line object - A force acts along A line, not two lines or ten lines, one line.

 

What we want to determine is: What line?

 

My train is a simpler system because that line does not change.

Introducing rotation is more complicated because that line is constantly changing.

 

This is no different from the normal reaction constantly changing in direction as the objects slips down the dome in the current Norton dome thread (I don't know if you have seen that one, but I have stepped out it of because of the hostile and juvenile reactions I received to my thoughts).

 

Anything that is continuously changing direction must be subject to a rotation.

Edited June 5, 2015 by studiot

[Robittybob1]

If the train's acceleration is along force B the magnitude of B would increase (change) by factor (a=f/m) proportional to the acceleration.

With force A pushing at 90 degrees to force B and inline with the motion of the train (accelerating in the direction of B) at some stage the mass will fall off the train at an angle to the direction of motion.

Edited June 5, 2015 by Robittybob1