Again?
Does the static coeff of friction “mean” anything in relation to this problem if nothing is moving and isnt going to start moving ?
Now, if it was zero, or lower than dynamic, I could see that complicating things.
My finger is pushing exactly normal to the board, and the board is also pushing normal, if by “push” you mean the normal force. The frictional force is a different force from that, and points perpendicularly to the normal force.
Quoth Baracus:
I commend you for the nice, conceptually-simple experiment one could use to measure this. It doesn’t seem implausible at all to me that the two scales would register different readings: Consider the extreme case where the eraser is glued to one slab but freely sliding against the other, for instance. But, of course, “plausible that they might have different readings” does not imply that they would, in fact, have different readings, in any given situation.
Ah, I understand the philosophical conundrum now. I would still argue that the friction forces are in proportion to their static friction coefficients and the sum total to the force of gravity upon the eraser. However, this is, if you excuse the term, an academic question, as there is no practical way to measure friction forces that exceed that required to break the static condition; that is, a frictional force component is strictly a reaction to an applied force on the object, and so the net friction components in given direction can never exceed the applied force. If the maximum friction force that could be developed exceeds the applied force there is no way to calculate the distribution. You can satisfy this for yourself geometrically by making the "static mu"s the same value but giving the eraser and board an angle to normal and watch how the friction forces between the two contact change proportional to the cosines of the angles; although the forces go up as a trigonometric relationship, they can’t (obviously) exceed the applied force by gravity.
Now, the tribology of the micromechanics of contact (i.e. how deformation of the contact surfaces changes the frictional effects) is another issue entirely, but I gather that is not what you’re interested in, as it delves not only into continuum mechanics but also metallurgy/materials science, solid state boundary interactions, and some more esoteric bits of micromechanics of thermofluids. We could speak of that, but not intelligibly, as (in my experience) such discussions are only useful in the context of very specific, applied problems with experimental data to fit half-baked theories to. This is why Mathworks developed the MATLAB Curve Fitting Toolbox–to turn craptastic data scatter into genuine, no-bullshit logarithmic forth-order polynomial fits.
Stranger
Coefficient of friction has no importance in statics. In a static situation it is impossible to determine the force of friction from the coefficient of friction and normal force. The force of friction is always the last to be calculated and that is done so by balancing all of the other forces. It is assumed that the coefficient of friction is sufficiently large to generate the necessary friction force.
For the explanation below, x is horizontal with 0 at the center of gravity, and positive into the board, y is vertical with 0 being the middle of the eraser and positive towards the top, and z is perpendicular to those.
Basically you have four unknowns with only three equations to use to solve. You don’t know the friction force from your finger (F[sub]f[/sub]), the friction force from the blackboard (F[sub]b[/sub]), the normal force (F[sub]N[/sub]), and the distance of F[sub]N[/sub] from y=0. The three equations you have are forces in the x dimension, forces in the y dimension, and the moment about the z axis.
F[sub]g[/sub]=Force due to gravity
F[sub]P[/sub]= Pushing
Y axis is: F[sub]f[/sub]+F[sub]b[/sub]=F[sub]g[/sub]
X axis is: F[sub]N[/sub]=F[sub]P[/sub]
Z moment is 0=-F[sub]g[/sub]*x-F[sub]N[/sub]*y+F[sub]B[/sub]*2x
Chronos- This can not be solved without modulus of elasticity information for the eraser. Assuming that you push on the middle of the eraser, the friction forces will equal each other if the eraser is symmetrical about the x-axis. In that case, the deformation is symmetrical and therefore the force is symmetrical. Or so I think.
The unsymmetrical case is a little more complicated and I haven’t quite noodled it out yet. My first guess is that if you have an unsymmetrical modulus regime then you will end up with different friction forces.
I was going to add something along these lines. In order to satisfy my curiosity, I just did the experiment with a kleenex box and the wall, and I could vary the angle at which I was pushing with my finger such that I was pushing a little up or even a little down without the object sliding, since the friction between the object and the wall was great enough to make it stay put for a whole range of force vectors from my finger.
However, after realizing this, I also noticed that all but one of these solutions violate a stated condition of the original problem: the force from your finger is specified as being horizontal. So by the original problem definition, there is no upward or downward force from your finger, and the entire upward force that opposes gravity must be provided by the friction between the eraser and the board. If the problem is stated in a way that allows you to be pressing against the eraser but doesn’t specify the exact direction as “horizontal”, then there are many solutions.
Well, the normal force is exactly normal to the board by definition, but it’s not completely clear what Chronos means. But he is clear that there are tangential forces on the finger.
In your OP, is F[sub]N[/sub] the total force, or the normal force?
Assuming it is the normal force only, then there is also a friction force F[sub]f[/sub] between your finger and the eraser, which is unspecified in your problem, and not determinable from the OP. I held an eraser to the wall with my finger this morning, and could easily vary F[sub]f[/sub] all over the place, by many times the weight of the eraser.
Try this:
Case A: F[sub]f[/sub] = W (weight of the eraser). The frictional force from the whiteboard be zero.
Case B: F[sub]f[/sub] = 0. The frictional force from the whiteboard be W.
Case C: F[sub]f[/sub] = -W. The frictional force from the whiteboard be 2W.
All these cases have the same F[sub]N[/sub], and are consistent with your OP.
Not quite ZenBeam’s point. (ETA: although ZB can speak for himself if he wishes, I see he slipped in another post whilst I was composing this one.)
Consider for a moment: Suppose you go whole-hog and actually build some apparatus to test what exactly happens. Rock-solid base, carefully machined smooth and vertical parts, some kind of instrumented tower in place of your arm, specialty materials, sensors out the wazoo, experimental physics at its finest.
OK, then, test procedure. You want your “arm” to apply a force that’s absolutely horizontal, so you place the eraser into the apparatus, carefully wheel your instrumented tower “arm” into place and actuate the force. But how do you know the force is absolutely horizontal? Neither the tower “arm” nor the eraser is infinitely rigid, so small shifts could make you force direction go off-kilter. Easy enough; it’s an instrumented tower. You read the vertical and horizontal components of the force from transducers, and make small adjustments to the location of the tower base until the vertical component of the force is zero.
Now, on to actually measuring the friction. Since everything is nice and square, the friction component of the force between the “arm” and the eraser is the vertical component of the same force. So what does your transducer read for this vertical force?
It reads zero, because that’s the way you set up the experiment.
If you want it to read anything other than zero, you have to redefine what “my finger is pushing exactly normal to the board” means. Force is a vector, and your arm doesn’t care that you’re decomposing that vector in a certain way. Your arm just applies the vectoral force in whatever direction you tell it to.
What am I, chopped liver? Is this thing on?
Again, the coefficient of friction doesn’t enter into it at all.
To repeat my analysis (did I post it in Sanskrit or something?):
If you go with the simplest case, an eraser that contacts the finger and the board at exactly one point each, with the center of mass exactly halfway between the finger and the board, then : The normal forces on the finger and board must be equal, or the eraser moves sideways. Friction at the board plus friction at the finger must equal gravity on the center of mass, or the eraser falls.
Now, looking at torques around the center of mass: The two normal forces and gravity force are all pointed in line with the center of mass, so no torques from them.
Therefore, the torque from the finger friction must be the same as the torque from the board friction. Since they’re both perpendicular to the line from the center of mass, and the same distance from the center of mass, that means the two forces must be equal (or the eraser rotates).
As I said, if the center of mass is offset towards the board or finger, the friction force on that side increases and the other decreases (torques balance; shorter lever means higher force). Also as I said, if we assume an extended but rigid eraser and board, with the friction and normal forces spread evenly over the board/eraser interface, keeping the finger a point, then the friction force on the board side is now larger (and the finger less), because again torques have to balance.
Now, this analysis fails if the required frictional forces anywhere are larger than the maximum static friction force. This is the only case where the coefficient of friction matters. If we suddenly replace the board with a teflon-coated one, then the friction force on the finger goes up (and the eraser tries to rotate around the finger because torques no longer balance. This leads to a greater normal force at the top of the board because the eraser can’t rotate through the board, and there’s a complicated rebalancing of forces). Of course if the maximum static friction on the finger isn’t enough to support the eraser either, then it slips downwards.
Yes, and there is also apparently an echo 
Hint, post 2 (and yet another one by me) maybe ?
The static COF only enters into it to determine if its a stable situation, or one that that possibly becomes unstable as forces or the exact geometry are modified.
That’s true for the case where the eraser contacts the finger and the board at exactly one point each, because the no-torque requirement automatically sets the force anges. However, the conclusions don’t hold for surface contact.
In particular, even if the normal force from the finger is aligned with the center of the erser, there’s no reason the force from the board will be, unless you construct the problem to enforce that result.
All of which doesn’t affect your point that the coefficient of friction doesn’t enter into the soluion of the problem.
No, even if the two friction forces aren’t equal, which is very likely, the eraser still won’t rotate. Because that torque will make the top of the eraser push against the board and the board will push back with a force of equal magnitude and opposing direction, generating a counter torque.
Take a rod, drill a hole though the center and mount it on a stand like an upside-down see-saw. (Make sure the pivot joint is frictionless.) Now replace your finger with the end of the rod; slide the stand closer to the eraser until the rod is pushing against the eraser with enough pressure to hold it in place.
Now, because the eraser exerts a vertical (frictional) force on the rod tip, the rod will pivot downward; to keep the rod horizontal, you will need to pull down on the other end of the rod. Use a spring-scale to pull on the other end of the rod (vertically) just enough to keep it horizontal. The reading on your spring-scale is the “finger” friction. The difference between that and the weight of the erase is the board friction.
You’re right that the coefficients of friction don’t determine the forces, but I think all the talk about torque is just clouding the matter. In Case A of my last post, the force on the eraser from the whiteboard is greater at the top and less at the bottom. Vice-versa for Case B. So the torque on the eraser isn’t determined either.
When you go to your “simplest case”, the torque must be zero, so you’ve added an additional constraint not present in the OP, and the problem becomes solvable.
You don’t know that the rod will pivot down. If there’s enough force from the rod that the static friction from the whiteboard alone will support the eraser, it won’t move (assuming a rigid eraser).
That’s essentially impossible. Unless the finger (or rod) is fictionless there will be some distribution of frictional force between the finger and the board.
Try this: Push some flat block-like object against a wall or board with your finger as Chronos described. Now, with a finger on your other hand, gently push down on your pushing finger until something starts to slide. Did your finger slide against the block or did the block slide against the board? I bet the block slides against board. If so, that shows that the greater frictional force is actually on the finger side.
No it isn’t. The rod is pressed against the eraser. For the rod to pivot downward, the eraser must move downward. For the eraser to move downward, it must overcome the friction between the eraser and the whiteboard.
Well, if the system isn’t moving then the total torque on the eraser (around each and every point) has to be zero. That’s kind of the definition of torque.
Now, I didn’t spell out the assumption, which I thought was part of Chronos’ original statement of the problem, that the finger is held steady, without trying to pull up or down on the eraser (except as necessary to hold the finger in place by counteracting the frictional force supporting the eraser). You’re right that if the finger is pulling up or down by some unknown amount then we can’t solve the system, but there’s not much point in trying to solve a system where we don’t know all the inputs. So I thought the ‘additional constraint’ was implied in the OP. Note that the problem is still solvable in the less-simple cases where the eraser extends along the board.
I agree here.
This made me realize that the problem is subtler. There are three friction forces at work. One between hand and eraser and two between eraser and board, one due to the normal force applied by the hand and another due to the normal force generated by this torque.
Good thing this wasnt problem #1 on the physics GED.
Everything is static, so the upward force of the board on the eraser (Fb) plus the upward force of the finger on the eraser (Ff) must equal the weight of the eraser (W): Fb + Ff = W. If this is not true then we have violated Newton’s laws.
If, as you said, “the static friction from the whiteboard alone will support the eraser” then Fb = W and, ergo, Ff = 0. That completely answers Chronos’ question: Fb = W; Ff = 0; Ff/Fb = 0. You are saying: “If the answer is A, the answer is A.”
But in real life, there IS some finite distribution of force; Ff != 0. The eraser will slip down slightly and the rod will pivot slightly, pull away from the eraser (the rod has a rounded tip), and allow the eraser to fall to the ground.
You can measure the force needed to stabilize the rod and that gives you a direct measurement of Ff.
I’m kind of lost regarding this discussion of torques, and I’m engineer that designs and analyzes mechanisms (or at least I used to be before being assimilated by the Middle Management Collective). There are definitely friction forces on the eraser, which can be seen by drawing a free-body diagram, and for a non-elastic case in which the force are applied with their vectors through the c.g. or parallel to one another the forces all cancel out. For the friction forces not being equal you simply adjust the location of the normal force application to create an opposing couple to make the sum of moments go to zero, but this is really more of a mathematical formalism than a practical reality. The eraser isn’t going to come spinning off the board because you don’t have a theoretical force balance, the forces will simply redistribute their effective point of application to whatever equilibrium demands. If you hold the point of application for one of the forces as being fixed (i.e. the reaction force from the board going through the c.g.) then you can figure out where the point of application of the applied force should be, but if you let both sides vary the problem is statically indeterminate, as you have two unknowns and one equation (the moment).
Stranger