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Because the car's body is not fixed and the wheel will accelerate with it.bob012345 said:Then how can the wheel accelerate if the two blue forces are equal and opposite and both act on the wheel?
Because the car's body is not fixed and the wheel will accelerate with it.bob012345 said:Then how can the wheel accelerate if the two blue forces are equal and opposite and both act on the wheel?
No, torque is not a force, it is a force multiplied by a distance. The units are Nm not N.rudransh verma said:So ##f_r-Torque=ma## is wrong or the eqn ##f_r-F_a=ma##. Torque is the force which generates static friction and this friction accelerates the car forward.
My point is that if you draw a free-body diagram of the wheel alone, which is perfectly valid, it better have a net force on it if it is undergoing acceleration. You cannot say the net force on the wheel is zero but the car carries it along with it. Every part of the car has a net force on it if it accelerates with the car. Even the driver.jack action said:Because the car's body is not fixed and the wheel will accelerate with it.
No,rudransh verma said:So ##f_r-Torque=ma## is wrong or the eqn ##f_r-F_a=ma##. Torque is the force which generates static friction and this friction accelerates the car forward.
Not if it's massless.bob012345 said:it better have a net force on it if it is undergoing acceleration.
Static friction is a force. Forces do not need to be generated by other forces.rudransh verma said:If there is static friction where is the force that generates it? Please answer this one question.
If the OP's question would be exactly the same, except for the car that started at an initial velocity until it stops (thus braking), the answer would be exactly the same.bob012345 said:I just think it is a misleading diagram in the context of this thread.
Good point. Frankly, I've entered a bit of a brain fog regarding the free-body diagrams that I have to work out on this problem beyond the basic idea of the static friction being the driving force.jack action said:If the OP's question would be exactly the same, except for the car that started at an initial velocity until it stops (thus braking), the answer would be exactly the same.
But the free body diagram - representing the car as a block - would show the friction force in the other direction, i.e opposite of motion, i.e. the "expected way" of a block sliding on the ground (even though there is no sliding involved).
How can you explain the change of friction force direction without explaining how a wheel works and how the direction of the applied torque at the wheel plays a major role in this?
Ok, I resolved my brain fog issue over the free body diagram of the wheel alone. What bugged me in the diagram is that it looked like the blue Axle force to the right was exactly equal to the static friction force to the left which could not be true under acceleration. The total forces left and right are slightly different and total to a net force of the exact amount necessary to accelerate the wheel of mass ##m## by acceleration ##a##. Likewise the driving torque at the axle is greater than the counter torque due to all other torques by exactly the amount consistent with an angular acceleration ##\alpha= \large \frac{a}{R}##.bob012345 said:Good point. Frankly, I've entered a bit of a brain fog regarding the free-body diagrams that I have to work out on this problem beyond the basic idea of the static friction being the driving force.![]()
If the green arrows are intended to denote rolling resistance then the nature of that effect could be more accurately denoted by a shift in the upward red arrow (normal force) so that it is no longer directly under the axle. The contact patch experiences more force as it compresses into the ground than as it rebounds from it. So the center of force is forward of the center of the contact patch. That is one source of rolling resistance.bob012345 said:
No doubt the actual forces and torques are complex but with a no-slip condition whatever the retarding torques are there must be a co-existing retarding force since ##a## and ##\alpha## are consistent unless the wheels are either slipping or sliding.jbriggs444 said:If the green arrows are intended to denote rolling resistance then the nature of that effect could be more accurately denoted by a shift in the upward red arrow (normal force) so that it is no longer directly under the axle. The contact patch experiences more force as it compresses into the ground than as it rebounds from it. So the center of force is forward of the center of the contact patch. That is one source of rolling resistance.
Another source of rolling resistance is frictional torque at the axle.
Either way, when superimposed on everything else, the net effect is more directly equivalent to a retarding torque rather than a retarding force.
But a retarding force pair that involves a forward torque on the wheel? That's just wrong.bob012345 said:No doubt the actual forces and torques are complex but with a no-slip condition whatever the retarding torques are there must be a co-existing retarding force since ##a## and ##\alpha## are consistent.
Not sure what you mean. Please define force pair and forward torque.jbriggs444 said:But a retarding force pair that involves a forward torque on the wheel? That's just wrong.
Look at your two green arrows. You have one at the axle acting leftward and one at the contact patch acting rightward. That is a counter-clockwise torque acting to make the wheel rotate more rapidly.bob012345 said:Not sure what you mean.
I did not make that diagram nor originally post it but I reposted it to argue I thought it was wrong. Here is another free-body diagram I found of a wheel slowing down by rolling resistance. It looks like it would have a torque that would speed the wheel up but we know it does not so there is something about the way friction works at the point of contact that I'm just not getting. It is easy to see how ##F_{rr}## will act to slow down the wheel but I don't see how the torque ##RF_{rr}## acts in the correct direction. I wonder if it has something to do with the point of contact actually being the center of rotation? And where is the static friction in this diagram? The brain fog continues...jbriggs444 said:Look at your two green arrows. You have one at the axle acting leftward and one at the contact patch acting rightward. That is a counter-clockwise torque acting to make the wheel rotate more rapidly.
The true effect is a clockwise torque acting to make the wheel rotate more slowly.
If the wheel is rolling at speed then there is a forward torque from the axle and a rearward torque from rolling resistance. Rolling resistance is a torque, not a force.hutchphd said:If the wheel is already rolling at speed then $$F_{rr}=0 ~~exactly$$ Remember $$F_{rr} \leq \mu N $$ and the angular acceleration is 0
The diagram I showed assumes the wheel is slowing down. I think the diagram is again wrong like the other one. There appears to be a lot of incorrect free-body diagrams available.hutchphd said:If the wheel is already rolling at speed then $$F_{rr}=0 ~~exactly$$ Remember $$F_{rr} \leq \mu N $$ and the angular acceleration is 0
It must be both if the wheel is not slipping. What I think resolves this is that the rolling resistance is a distributed force acting over a region of contact (a patch) in such a way that it opposes the forward motion and create a torque which slows the wheel.jbriggs444 said:If the wheel is rolling at speed then there is a forward torque from the axle and a rearward torque from rolling resistance. Rolling resistance is a torque, not a force.
The Post #85 diagram shows rolling resistance as a simple horizontal force acting at the base of the wheel, vertically below the wheel's centre. This is an inaccurate simplfication and (as has already been noted) gives a torque which would tend to accelerate the wheel!bob012345 said:I did not make that diagram nor originally post it but I reposted it to argue I thought it was wrong. Here is another free-body diagram I found of a wheel slowing down by rolling resistance. It looks like it would have a torque that would speed the wheel up but we know it does not so there is something about the way friction works at the point of contact that I'm just not getting. It is easy to see how ##F_{rr}## will act to slow down the wheel but I don't see how the torque ##RF_{rr}## acts in the correct direction. I wonder if it has something to do with the point of contact actually being the center of rotation? And where is the static friction in this diagram? The brain fog continues...
https://www.physicsforums.com/attachments/296631
https://archive.thepocketlab.com/educators/lesson/rolling-resistance-physics-lab
bob012345 said:Ok, I resolved my brain fog issue over the free body diagram of the wheel alone. What bugged me in the diagram is that it looked like the blue Axle force to the right was exactly equal to the static friction force to the left which could not be true under acceleration. The total forces left and right are slightly different and total to a net force of the exact amount necessary to accelerate the wheel of mass ##m## by acceleration ##a##. Likewise the driving torque at the axle is greater than the counter torque due to all other torques by exactly the amount consistent with an angular acceleration ##\alpha= \large \frac{a}{R}##.View attachment 296601
If you have access to PF, you presumably also have access to some great physics-friendly drawing tools available on the net, e.g. GeoGebra ##-## https://geogebra.org/m/NvzumAMV, LaTeX/TIKZ ##-## https://texample.net/tikz/examples/all/, Inkscape ##-## https://inkscape.org/gallery/?tags=physics.bob012345 said:It would be nice if there was a simple drawing tool here at PF.
All preamble aside, what if one wishes to use the term "rolling friction"? Personally I would be inclined not to use it ever. Is there a reason to put it in the elementary curriculum (as yet another frictional force??) Just askin'kuruman said:A picture is worth 103 words.