# This should be an easy physics problem... **URL:** \ **Category:** Factual Questions\ **Created:** [October 14, 2009, 6:49pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819 "2009-10-14T18:49:26Z")\ **Posts on this page:** 15\ **Page:** 6
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 16, 2009, 9:54pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/101 "2009-10-16T21:54:32Z")
> [@Quercus](#): > > 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 Circular argument. What’s the necessary force in Case A? W. What’s the necessary force in Case B? 0. > [@](#): > > 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? 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). ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [October 16, 2009, 10:05pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/102 "2009-10-16T22:05:04Z")
> [@zut](#): > > 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. 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. ---
**Author:** ![Delayed\_Reflex](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/delayed_reflex/32/3172_2.png) [@Delayed\_Reflex](https://boards.straightdope.com/u/Delayed_Reflex)\ **Post date:** [October 16, 2009, 10:10pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/103 "2009-10-16T22:10:36Z")
> [@Chronos](#): > > 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. 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: 1. Ff + Fb = m[sub]eraser[/sub]g 2. Fb is in the same direction as Ff (up) and is \>=0 3. Ff is minimized 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. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 16, 2009, 11:41pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/104 "2009-10-16T23:41:11Z")
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](http://en.wikipedia.org/wiki/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. ---
**Author:** ![EdwardLost](https://avatars.discourse-cdn.com/v4/letter/e/bbe5ce/32.png) [@EdwardLost](https://boards.straightdope.com/u/EdwardLost)\ **Post date:** [October 17, 2009, 12:21am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/105 "2009-10-17T00:21:39Z")
> [@ZenBeam](#): > > 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](http://en.wikipedia.org/wiki/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. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 17, 2009, 12:34am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/106 "2009-10-17T00:34:41Z")
Why is everybody going on about the direction of applied force? :rolleyes: The OP is clear: > [@](#): > > Let’s say I’m holding an eraser stationary up against a whiteboard by pressing **horizontally** on it with a force FN. FN is very large, such that the eraser is nowhere remotely close to sliding. The question is simply, what are all of the forces acting on the eraser? Bolding mine. This is the answer, in case you missed it before: [ImageShack - Best place for all of your image hosting and image sharing needs](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. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [October 17, 2009, 1:22am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/107 "2009-10-17T01:22:46Z")
> [@Pedro](#): > > Why is everybody going on about the direction of applied force? :rolleyes: > … > Bolding mine. This is the answer, in case you missed it before: [ImageShack - Best place for all of your image hosting and image sharing needs](http://img26.imageshack.us/img26/7505/diagramgq.png). 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) ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 17, 2009, 1:29am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/108 "2009-10-17T01:29:07Z")
> [@ntucker](#): > > 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). ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 17, 2009, 1:34am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/109 "2009-10-17T01:34:40Z")
> [@Pedro](#): > > Why is everybody going on about the direction of applied force? :rolleyes: > > The OP is clear: 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: > [@](#): > > Then, in the vertical direction, there’s the gravitational force of the Earth on the eraser, pulling down, and since we have two normal forces, there are two static frictional forces (one from my hand and one from the board), both pointing up to oppose gravity. So he’s not assuming there are only horizontal forces. > [@](#): > > This is the answer, in case you missed it before: 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. > [@EdwardLost](#): > > 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. 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. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [October 17, 2009, 1:35am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/110 "2009-10-17T01:35:41Z")
> [@Pedro](#): > > No, F\_p is pushing force, f\_EH is eraser-hand friction (due to gravity). 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? ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 17, 2009, 1:49am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/111 "2009-10-17T01:49:29Z")
> [@ntucker](#): > > 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. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 17, 2009, 2:01am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/112 "2009-10-17T02:01:16Z")
No, wait, it’s right, because I am only representing external forces to the system comprised by the eraser. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 17, 2009, 2:29am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/113 "2009-10-17T02:29:14Z")
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](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. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [October 17, 2009, 4:49am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/114 "2009-10-17T04:49:02Z")
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. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [October 17, 2009, 6:51am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/115 "2009-10-17T06:51:53Z")
> [@On another topic](https://boards.straightdope.com/t/514127/3): > > clearly, the sum of the two [new york asian escorts](http://www.example.com) frictional forces must be equal in magnitude to the weight, since, again, a = 0. Some of the more amusing spam I’ve seen in a while. (link broken) [Previous page](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819.md?page=5)