That could be correct, but it relies on a deeper understanding of how friction is developed on each surface (which I admit is beyond my level expertise).
Thought experiment here: imagine you’re holding an eraser against a board as Chronos describes. I come along and place a weight on top of it. Not a large enough weight to move the eraser, but enough to significantly change the total weight. Since the weight increases, the counteracting frictional forces must also increase, right? How does that actually occur such that the frictional forces on either side increase only in proportion to their own frictional coefficient?
Point being, I *think *that the forces developed are going to depend at least in part on local elasticity and perhaps the time history of loading–things that are not captured in the simplified frictional coefficient. Which means this is a case where the simple friction model breaks down, and trying to “solve” the problem with that model just gives a problem with too many variables and not enough equations.
Again, not my area of expertise, but I’ll be surprised if a more complex modelling of friction developemnt isn’t required.
There are two frictional forces, one on each surface.
Consider the case of two rigid blocks, a smaller one atop a larger one, and the larger block sitting on a rigid plane.
Consider what happens if the smaller block is placed offset from the center of the larger block. If the forces are uniformly distributed over the contact areas, then that means the normal force between the large block and the plane is uniformly distributed, and it’s centroid is at the center point of the large block.
And, because the smaller block is placed offset from the center of the larger block, the force from the smaller block on the larger block is *not *in the center of the larger block.
These two forces (along with the weight of the large block) create a moment, and the large block goes spinning off into the night.
I concede that in the case of a rectangular object that can’t rotate a net torque due to friciton is not an issue. Might be interesting to consider a round/cylindrical object though.
I’m at work, so I don’t have a lot of time to devote to this, but the Force Diagram is going to look exactly the same (in the assumption that your box is a point particle), even if you have disagreements with other parts. (If you don’t think so – what other forces are there?)
As far as I can see the only difference is using the coefficient of static friction vs. the coefficient of sliding friction.
The reason static vs. sliding friction is relevant is that with sliding friction, the frictional force is always equal to the normal force times the coefficient of friction, whereas with static, it’s less than or equal to the normal times the coefficient, and in the problem here presented, we’re in the “less than” regime. If I push harder on the box in your link, I’ll increase the frictional forces, and slow down the sliding box (perhaps slowing it down very abruptly, if I’m pushing hard enough). With this problem, though, if I push harder on the eraser, I don’t increase the frictional forces (at least, not their total), because the sum of the friction forces will never be greater than the weight.
If we’re treating the eraser as a rigid body, the forces are as you outline in your post. The static friction forces never exceed the sliding limit force and will be in proportion to their coefficients, the sum to equal the gravitational force on the eraser regardless of how hard you push. The “extra” forces exerted on the eraser by you and the corresponding reaction from the blackboard are reacted out of the total eraser-blackboard-you system by the floor to which your feet are grounding and the wall to which the blackboard is mounted. Push hard enough and you’ll either exceed the sliding limit friction on your shoes or (less likely, unless you are the Six Million Dollar Man, who clearly isn’t subject to the laws of physics) you’ll push over the wall.
If we treat the eraser as a deformable body, then things are a little different. The overall free-body diagram from a border drawn around the outline of the eraser is still the same, including the friction forces, but now you can have all sorts of deformation going on inside the eraser which may change its configuration and the distribution of pressure on the contact surfaces, i.e. it changes shape and applied forces are no longer normal. Except for a very simple body, the only way to get an accurate measure of deformation and resulting force distribution is through some numerical simulation method like Finite Element Analysis.
I’m not certain if this answers the question because reading back through the thread (which has grown considerably since I started to write a response) it’s not entirely clear to me what the question is. Is it an issue of how pressure is distributed, or what the net forces on the eraser are, or what?
If this is true, it’s the answer to the question (basically, how is the total frictional force apportioned between the two surfaces). But are the coefficients of friction actually relevant in the case where the the object is not almost slipping? That’s the part I’m uncertain of.
And I don’t really care about deformation, here (which obviously requires an approach much more sophisticated than the elementary level implied by coefficients of friction). I should perhaps have specified the object to be something rigid like a brick or a block of steel, instead, in the first place.
why two static frictional forces? The problem doesn’t change if your finger is frictionless. the friction between the board and the eraser balances the force of gravity. The end. Or if it is unsatisfactory to have a frictionless finger, you can divide the vertical static frictional force up any way you want as long as the sum equals the weight. If you are determined to solve both static frictional forces, you need to know the coefficient of friction between each surface. As you said, it will be different for each. It is always so. the cof has to be measured independently.
now that I think I understand the question. I had posted before reading the entire thread-oops.
not a friction expert, but it seems to me that the ratio of forces should be equal to the ratio of cofs. When one measures a coeff of friction, isn’t one measuring the friction between two surfaces? That is, the cof between steel and wood will be different than the cof between glass and wood. I don’t see how one can describe the cof for just wood. So there is a cof for finger/eraser and another for eraser/board. Since these must be indepentently determined, doesn’t that give you your answer?
Would there necessarily have to be a huge frictional component from the active pushing? What if you had an extremely tiny point push in with that force?
Or perhaps I should say, is the frictional force from the active push affected by the area over which it’s applied?
If you pushed with an immeasureably small point from one side onto a wall with sufficient friction of its own, the friction exerted by the point object would be negligible, and the friction exerted by the wall would be the sum of gravity and the inward push but in the opposite direction, and wouldn’t exert a torque because it’s distributed over the entire side of the rigid rectangular object.
On the other hand, I wonder what would happen if you pushed with an immeasureably small point from a point other than the center of the rigid object. My gut feeling both from physics and practical experience tells me that there would be a torque which might cause the object to slip away from you (imagine pushing very hard less than 10% from the edge, even straight forward. Of course, given a frictional enough wall it would still be possible since the side of the wall receiving the torque will push back harder. )
Your problem doesn’t have a unique solution. Consider if your finger pushes slightly up or down (or left or right). The eraser still doesn’t move. The breakdown of how much downward force is in the finger, and how much is in the board can vary over wide range, limited by the static coefficients of friction.
If you want, you could specify that your finger is pushing exactly normal to the (perfectly vertical) board. Then all the downward force of the eraser must be opposed by the board. Similarly, if you specify that the board is pushing exactly horizontally, then your finger must supply the upward force. (You can’t specify that both push exactly horizontally, since the forces won’t balance.)
As it is stated, your problem is underspecified to give you a unique breakdown of what is supplying the upward force to keep the eraser stationary.
I am not convinced that it is either. Another thought experiment:
Take your eraser and clamp it between it between two vertical slabs of material of equal mass but different coefficients of friction. Suspend the whole shebang from a pair of scales; one for each slab. The reading on each scale should be the weight of the slab plus the downward force of friction from the eraser. According to the proportional line of thinking, the slab with the greater coefficient of friction should weigh more because it has more friction pulling down on it. Frankly, I don’t see how this could be if everything is static.
There we go; that’s the answer. After mulling it over for a while, I had arrived at the same conclusion.
To phrase it another way, in the original problem, the magnitude and angle of the force Chronos is applying with his hand must be independently set at the beginning. That allows all three equations of equilibrium to be solved uniquely.
You need a better model for the static friction coefficient to answer this (a function rather than the peak value simplification). This will require either using some “theory of friction” or some experimentation.