The theoretical maximum frictional force would be different, but the frictional force would only be was was needed to exactly equal the force trying to make it slide.
Provided the F-applied (n this case Fg) is less than f-max, then f=Fg
Damn edit time limits…
Thus, provided Fg (the force trying to move the eraser) is less than the sum of f-max-board and f-max-finger, the eraser will not move.
Further, provided both f-max-board and f-max-finger are greater than Fg/2 then f-finger = f-board.
If Fg/2 > f-max-board, then f-board = f-max-board leaving f-finger = Fg - f-max-board. (That is f-finger > f-board). Switch the board/finger subscripts for the opposite case.
Since this is not a marginal case f-finger = f-board = Fg/2
I did the math and think I found the answer. I’ll post it here in case anybody is interested, which is doubtful, and it’s not easy to understand f without a diagram anyway. I’ll just write the quadratic because the solution is immediate:
C*f[sup]2[/sup] + (B+AC)*f + AB = 0
where
A = “half the weight of the eraser”
B = “magnitude of horizontal push-force”
C = (sqrt(2)/2) * sqrt(w² + h²) / w
where w is the width of the eraser and h is the height.
Encouragingly this can be easily proven to give real values for A > 0, B > 0, C > 0. I’d like to look at it further but it’s really late here.
Oh yeah, f is one of the two friction forces acting along the board (the other is equal to weight/2 - f) .
I would be very interested in seeing the derivation of that, given that I’ve now convinced myself that the frictional forces can take any value at all.
Usually when you talk about an applied force there is no implicit preclusion of a friction force resulting from that applied force. In addition to that, I can’t think of any instance where someone combined a normal and friction force into one net force.
The answer to this was implied by other posters, but I don’t think that your statement was ever directly addressed.
In any event, I disagree with your statement.
Even if the force from your finger is directly horizontal, an upward frictional force is created between your finger and the eraser. This is the whole point of Chronos’ original problem statement.
Changing the angle of force exerted by your finger on the eraser needlessly complicates things.
True, if the problem statement is set up schematically with a horizontal force arrow, but in this case, we have a real world problem, in which the horizontal force is exerted by an actual object. In this case, the horizontal force is exerted by a finger. Fingers do not act on a point; instead there is an area of contact between the finger and the eraser, and thus a frictional force is implied. Again, I thought that this was the whole point of the problem statement.
ETA: Whoops, just noticed that you wrote “preclusion” there, Treis. Are we actually agreeing here?
Yes I think so. Saying that there is no friction force because the problem states that the applied force is horizontal isn’t a satisfying answer.
Are you convinced of this only in the case where the finger could be pushing up or down with an unkown amount of force?
Because in the case where the finger is pushing horizontally and only exerts a vertical force to counteract friction, the force is determined (assuming a rigid eraser, and the finger is a point and the blackboard is smooth and even, etc.). It’s just Statics 101 from that point: draw a diagram, put in the gravity and other external forces, add forces to the supports to balance gravity, then check torques at all points and add the necessary forces to the supports to balance the torques.
Why doesn’t this approach work? Do you see any problems with my previous analyses? (I mean, I did pretty good in undergraduate physics/engineering classes back in the day; I think I still know what I’m doing).
But I would be interested in Pedro’s calculations, cause I really don’t see how you get an equation with forces squared.
You can’t? Forces are vectors, and are decomposed into orthogonal components for convenience.
Consider a block (which has weight) sitting on a wedge, where the top surface of the wedge is not horizontal. What is the force from the block on the wedge?
There’s nothing wrong with describing the force from the block on the wedge as two orthogonal component: one normal to the wedge surface and one (frictional) component parallel to it. But neither is there anything wrong with describing the force as a strictly vertical force, which is the combination of the normal and frictional.
And, I should note, there’s some parallel between this scenario and the original eraser problem, in that you know the frictional force between the block and the wedge only because you know exactly the direction in which the body force of the block is applied.
I’m not sure I understand what you mean by that. Here is the free-body diagram I used. The answer is for f_BE in the diagram.
Any component of force not normal to the wall supplied by the finger is due to friction.
Look at Cases A, B,and C from post 46. All of those only have vertical forces due to friction, but have different vertical forces. How can you rule any of them out? That’s the issue. It’s not whether you can find a solution, but whether you can show that that solution is the only solution.
I never said that you couldn’t do so. I said that I’ve never seen it done.
And now you have. Don’t say I never did anything for ya. 
(There’s also a standard journal bearing frictional analysis technique that relies on combining normal and frictional force into a total force and calculating a moment from that, but that’s sort of off-topic.)
Which leads me to a further thought. Let me re-adjust Chronos’s original problem statement:
Suppose Chronos has a *massless *eraser made out of some magnetic material. He lays a dry-erase board on the ground and sets up a powerful magnet at the edge of the board. When the eraser is placed on the board, it’s attracted to the magnet with a certain force.
Now Chronos does a single-hand handstand atop the eraser, which of course applies a force equal to his weight in a vertical direction, and holds very still. However, Chronos forgot to check one very important thing: he doesn’t know if the ground is flat. The ground is, in fact, at some *unknown *angle to the vertical.
The question is now: what are the frictional forces between the eraser and the whiteboard, and the eraser and Chronos’s hand?
The answer to this problem depends on the parameter Chronos forgot to check, namely the angle between the board surface and the vertical. And this problem is exactly the same as the one in Chronos’s OP.
The point is that if you a relying on this to solve the problem it would push the question into the realm of trick questions. First by relying on a non-standard wording definition and second by breaking the relationship between the way objects act in the problem and how they would in reality.
Because I’ve assumed that the finger is held in place, so it’s not applying any vertical force except that necessary to counteract any force from the sponge – this may be different from what Chronos meant in the OP, but it seemed a reasonable assumption (I mean, I assumed a physicist wouldn’t ask an underspecified problem). Or is the answer you’re looking for: Because in solving a static forces problem, you find the solution with the minimum amount of force from the supports necessary to counteract external forces (and of course balance all resulting torques).
Again, with my assumption a better model than a finger (which could decide to push up or down), imagine a rod that’s clamped in place (with a spring-loaded tip that pushes horizontally on the sponge and transmits any veritical force to the clamps). Now clearly, if we set up such a system, pinning a sponge to the wall, there’s going to be a single resulting distribution of forces, right? I mean, the force on the rod isn’t going to randomly jump up and down and back and forth: there’s a single real answer, right? Since nature can solve the problem and get a single answer, why can’t we?
Sure, there’s a different answer if we unclamp the rod and start pushing down on it, so it starts pulling down on the sponge instead of supporting the sponge, but in that case we’ve changed the external force applied, so of course there’s a different answer.
Wait, what? If by “non-standard wording definition” you’re talking about the direction of the applied force, then you’re missing the point. The applied force can be in any direction you want. If you don’t specify what direction you apply the force in, then you can’t solve the problem, simple as that.
And I have no idea what you mean by “breaking the relationship between the way objects act in the problem and how they would in reality.”
But the OP did specify what direction the force is applied in. It described a force vector F[sub]N[/sub] which is explicitly specified as “horizontal”.
In order to allow for the finger imparting some vertical force on the eraser, you have to either:
- interpret “horizontal” as “horizontalish”
or - assume that when the OP described the force the finger was imparting on the eraser, it was intentionally seperating this force into two vectors: one that is horizontal, and is explicitly mentioned, and one that is vertical, and is not mentioned.
To me, both of those interpretations would imply that the problem was badly worded. I prefer to assume the problem was worded just fine and meant exactly what it said.
For the record, while I did not intend this to be a trick question, it certainly seems to have worked out that way. The correct answer is that I can make the friction force between my hand and the eraser anything I want, and that then determines the friction force between the board and eraser.
What do you mean “except that necessary to counteract any force from the sponge”? If you mean that the force of finger on eraser is equal and opposite to the force of eraser on finger, well, that’s just Newton’s Third Law, and is true no matter what the forces are, or how they’re arranged.
Not really. If you push in the normal direction to the (ideal, rigid, flat) eraser surface, the friction force between hand and eraser is constant.