Circular argument. What’s the necessary force in Case A? W. What’s the necessary force in Case B? 0.
If you pin it to the wall, there is indeed some single physical force, measurable in principle. If you remove it, then pin it to the wall again, there’s again a single force, but it may be different than the first one. The amount will depend on details of how you go about pinning the sponge to the wall (as well as properties of the sponge, wall, and rod).
I agree that you can’t solve the problem as stated. However, twisting the words of the problem into something you can isn’t a satisfying solution. Take this question for example:
A person is holding an eraser against a frictionless blackboard by pushing horizontally on the eraser. If the coefficient of friction of friction is .8 and the mass is m, what is the minimum force the person needs to apply.
According to your interpretation the problem is unsolvable. However, every other physics student and teacher I am familiar with would interpret it as solvable.
I think he means that he is assuming that your “finger force” is not adding any vertical component other than the friction force that prevents the eraser from falling - that is:
Otherwise you could be “pushing up” on the eraser and in fact have a downward Fb - this could be any value up to F[sub]N[/sub]*u[sub]b(static)[/sub]
That is how I interpreted the question as well originally.
Here are some thought experiments:
Hold the eraser in your left hand against the whiteboard, pressed against it and applying a force upward equal to the weight of the eraser (so there’s no frictional force from the whiteboard). Press your finger from the other hand against the eraser exactly normal to the wall, with enough force to hold the eraser to the wall. Release the eraser from your left hand.
Repeat the experiment, but pushing up on the eraser with two or three times that force (the friction with the whiteboard is holding it from moving).
Repeat the experiment, but pushing down with a force equal to the weight of the eraser.
Do these all have the same forces? Not likely.
Also, my finger has a lot more give (shear modulus, and note the large range of typical values in the table) than a whiteboard. In the first case (depending on the eraser), the eraser could still end up with almost all of the force holding it up coming from the whiteboard, and very little from the finger.
I’m not sure how you came to the conclusion in your last sentence. When I do your experiment #1, and then push my finger parallel to the board (while trying to keep it pointed straight into the board) it usually happens that the eraser slides across the board rather than my finger across the eraser. This suggests that the finger-eraser friction is greater than the eraser-board friction.
Why is everybody going on about the direction of applied force? :rolleyes:
The OP is clear:
Bolding mine. This is the answer, in case you missed it before: http://img26.imageshack.us/img26/7505/diagramgq.png.
The only ambiguity is whether the hand is freely moving in the vertical direction or not. The first case is trivial (the friction against the board is equal to the weight of the eraser), the second is answered above. Note that taking torque into account is fundamental to attack the second problem.
It would be easy to generalize to arbitrary directions for the pushing force. Obviously this direction must be specified for a complete description of the problem.
According to that diagram, the finger is pushing against the eraser with force (f_EH + F_p), which is not horizontal.
(ETA: unless f_EH has a magnitude of zero)
No, F_p is pushing force, f_EH is eraser-hand friction (due to gravity).
Save the rolleyes. It’s valid to describe the horizontal component of force as “pressing horizontally on it with a force FN” Further, the OP goes on to say:
So he’s not assuming there are only horizontal forces.
You haven’t shown that it’s a unique solution. In particular, show why Cases A, B, and C from post 46 are not valid solutions. As I said before, 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 wasn’t talking about friction, but about how the force develops, depending on details of how the eraser came to be held to the board. You’re holding it in place with your left hand, supporting its weight. Your finger is pinning it to the board with a purely horizontal force, and the board is pushing back with a purely horizontal force. You then remove your left hand. The eraser will drop a small amount to develop the frictional force with the finger and whiteboard to support it. For a stiff eraser, stiff whiteboard, and soft finger, the force will develop faster at the whiteboard-eraser junction than at the eraser-finger junction, because the skin of the finger tip has more give.
Maybe I’m the unusual one here, but I was equating “pushing” with “imparting force.” The if the hand is imparting upward force due to friction, it’s “pushing” upward, is it not?
Right, actually that was my stupid mistake. Friction will be down and pushing up (by the arm holding the hand stationary). But otherwise that takes us back to the trivial case I mentioned before. I will correct it before this gets confusing again.
No, wait, it’s right, because I am only representing external forces to the system comprised by the eraser.
This is the case where the eraser is stationary but the hand is freely moving in the vertical direction: http://img403.imageshack.us/img403/8528/diagramqg2.png. A more apt description would be a massless horizontal rod doing the pushing, supported vertically by a frictionless roller bearing.
To be perfectly clear, here: I was the OP. I did not get the problem from anywhere; I made it up myself, so any problem with the wording is my fault. The wording of the OP is in fact confusing, and it is that very confusion resulting from the poor wording that caused me to not get the answer right away. Now that other folks have pointed it out, I have realized that the way I asked the question was confusing, and that the frictional force can have any value at all.
Some of the more amusing spam I’ve seen in a while.
(link broken)
