# 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:** 20\ **Page:** 4
**Author:** ![EdwardLost](https://avatars.discourse-cdn.com/v4/letter/e/bbe5ce/32.png) [@EdwardLost](https://boards.straightdope.com/u/EdwardLost)\ **Post date:** [October 15, 2009, 7:14pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/61 "2009-10-15T19:14:16Z")
> [@Stranger\_On\_A\_Train](#): > > 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 I agree with you: the discussion of torques is no help. We know they are balanced, and they don’t help us answer the question of what downward (frictional) force the erase applies to the finger. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 8:27pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/62 "2009-10-15T20:27:56Z")
The torque accounts for the fact that (as **Quercus** correctly pointed out) the friction force on the hand side of the eraser (one vector) is equal to the sum of the friction forces on the board side of the eraser (two vectors), _without_ implying both surfaces have the same friction properties (I’m avoiding the term “coefficient” on purpose so no one gets hung up on that). ---
**Author:** ![Tom\_Tildrum](https://avatars.discourse-cdn.com/v4/letter/t/e95f7d/32.png) [@Tom\_Tildrum](https://boards.straightdope.com/u/Tom_Tildrum)\ **Post date:** [October 15, 2009, 8:28pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/63 "2009-10-15T20:28:32Z")
> [@Chronos](#): > > 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. But the only way for the object to be not “almost slipping” is for something to be deforming, right? Once you’ve got it stable but almost slipping, if you apply more force, your finger deforms to bring more finger surface into contact with the eraser. Or imagine instead a steel C-clamp suspended from above, with the open side facing down. You put a steel block into the clamp and tighten until it’s not slipping. How do you tighten it past that point? If everything’s rigid, I don’t see that you can. ---
**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 15, 2009, 8:33pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/64 "2009-10-15T20:33:08Z")
OK, I think I get it now. The problem is incompletely specified. If I lean upward or downward into the eraser, then I’m varying the frictional force between hand and eraser (and therefore also the one between board and eraser), and the normal force is still perpendicular, as it must be by definition. The cause for confusion is the use of the imprecise concept “pushing force”: I can, in an imprecise sense of the word, push the eraser up or down the board (or even sideways), but the force that’s acting non-perpendicularly must be a frictional force. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 8:52pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/65 "2009-10-15T20:52:50Z")
I don’t think it’s incompletely specified and I don’t see what’s confusing about the term pushing force. I still think the problem is that the model used for friction ([Coulomb friction](http://en.wikipedia.org/wiki/Friction#Coulomb_friction)) is too weak. Come up with something better that replaces the inequality with a proper equation and you’ll have your answer. Problem is this is quite difficult to do. ---
**Author:** ![adhay](https://avatars.discourse-cdn.com/v4/letter/a/ecd19e/32.png) [@adhay](https://boards.straightdope.com/u/adhay)\ **Post date:** [October 15, 2009, 10:08pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/66 "2009-10-15T22:08:02Z")
IANAP or anything else for that matter but I do have an interest in physics. I’d like suggest a thought experiment. Fingerforce is applied thru the end of a long and guided (no torque) metal bar, 1 inch square. The metal can have any qualities you like but to start, at the pressures we are talking about, it is incompressible and coefficients of friction are low but may vary between participants. For the Erasermass, we start with a similar metal bar, 100 inches long. It’s shape will change in the course of the experiment but it’s mass remains constant. The wall is a sun-sized cube of metal monitored by gyroscopes. Displacement=Wallforce. For a start, assume that Fingerforce is twice that necessary to support Erasermass against Wallforce. Reduce Fingerforce, what happens when it breaks? Varying the shape of Erasermass to a 1X10 plate, what happens? In either case, it seems a lot depends on what’s assumed about the coefficient(s) of static friction. Forget compressibility. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [October 15, 2009, 10:10pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/67 "2009-10-15T22:10:59Z")
> [@Chronos](#): > > OK, I think I get it now. The problem is incompletely specified. If I lean upward or downward into the eraser, then I’m varying the frictional force between hand and eraser (and therefore also the one between board and eraser), and the normal force is still perpendicular, as it must be by definition. The cause for confusion is the use of the imprecise concept “pushing force”: I can, in an imprecise sense of the word, push the eraser up or down the board (or even sideways), but the force that’s acting non-perpendicularly must be a frictional force. Yes. You yourself are determining the frictional force by your choice of “pushing direction.” ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 15, 2009, 10:11pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/68 "2009-10-15T22:11:37Z")
> [@EdwardLost](#): > > 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. Absolutely. > [@](#): > > 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. See the middle paragraph of post 37. > [@](#): > > 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. In real life, the eraser and rod (or finger) both have some give, so the rod doesn’t immediately pull away from the eraser. You’re setting up the rod so that it’s pushing against the eraser. That force isn’t going to drop to zero discontinuously. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [October 15, 2009, 10:18pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/69 "2009-10-15T22:18:56Z")
> [@EdwardLost](#): > > I agree with you: the discussion of torques is no help. We know they are balanced, and they don’t help us answer the question of what downward (frictional) force the erase applies to the finger. First of all, there is no way to measure a force except by allowing it to do some measure of work. Even a load cell or pressure plate gives force by measuring some kind of deflection or displacement, so unless the (rigid) eraser is allowed to move if but slightly, there is not only any way to directly measure the force but we can’t even say that it would exceed the value at which the eraser would start to slip. Second, the force developed as a reaction to the an applied force simply can’t exceed the applied force. The friction force can’t exceed that applied by gravitational attraction. This discussion of torques doesn’t add to that applied force unless the finger is pushing up or down. As I understand it, the finger force is determined to be normal. If we locate the position of the finger force at a particular point and assess the friction forces contributed by the eraser-finger and eraser-wall interfaces as being equal in proportion to their respective static friction coefficients and summing to the applied gravitational force, the locus of pressure–the centoid of force application–from the wall will move to whatever location is required to make a complementary couple which balances out whatever imbalance exists in the friction forces so the sum of all moments is zero. If we don’t have a defined point of application–if the hand is pushing along the entire surface of the eraser–then the problem is statically indeterminate. **Pedro** , I don’t see any reason, especially given the simplifications of the problem (rigidity, purely normal forces, no consideration for surface effects), why Coulomb friction doesn’t apply; indeed, this is the very scenario that satisfies the conditions for Coulomb friction. Deviations from the Coulomb assumptions generally occur when you have non-linearities at the interface (viscous lubrication, electrostatic effects, vacuum adhesion). However, unless we’ve defined the force on one side as focused at a particular point, and accept that the developed forces will never exceed the applied gravity load (and apportion the forces equal to their μ[sub]s[/sub]) we have a situation of two unknowns (the loci of the normal force application) and one equation (the moment about the eraser c.g. or some other arbitrary point in the system) which is not soluble. Stranger ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 10:26pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/70 "2009-10-15T22:26:49Z")
> [@zut](#): > > Yes. You yourself are determining the frictional force by your choice of “pushing direction.” What? Why not assume a perfectly horizontal push? ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 10:34pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/71 "2009-10-15T22:34:18Z")
> [@Stranger\_On\_A\_Train](#): > > **Pedro** , I don’t see any reason, especially given the simplifications of the problem (rigidity, purely normal forces, no consideration for surface effects), why Coulomb friction doesn’t apply; indeed, this is the very scenario that satisfies the conditions for Coulomb friction. Deviations from the Coulomb assumptions generally occur when you have non-linearities at the interface (viscous lubrication, electrostatic effects, vacuum adhesion). However, unless we’ve defined the force on one side as focused at a particular point, and accept that the developed forces will never exceed the applied gravity load (and apportion the forces equal to their μ[sub]s[/sub]) we have a situation of two unknowns (the loci of the normal force application) and one equation (the moment about the eraser c.g. or some other arbitrary point in the system) which is not soluble. I’m not saying it doesn’t apply, I’m just saying it’s not good enough to yield a definite answer, e.g. Sum(Friction\_board) = 12.84%\*Weight and Sum(Friction\_finger) = 87.16%\*Weight, assuming a perfectly horizontal push against a vertical board. ---
**Author:** ![spinky](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@spinky](https://boards.straightdope.com/u/spinky)\ **Post date:** [October 15, 2009, 10:34pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/72 "2009-10-15T22:34:33Z")
Apologies for not following every branch of the conversation here, but I want to clarify something. Based on this statement from the OP: > [@Chronos](#): > > I’m holding an eraser stationary up against a whiteboard by pressing horizontally on it with a force F[sub]N[/sub]. If you interpret that as saying that the force from your finger is strictly horizontal, then by definition, there is no upward force whatsoever provided by your finger, and thus the entire weight of the eraser is being supported by friction between the eraser and the wall. Does anyone disagree with that? There’s still a question of whether this is a reasonable interpretation of the OP, but if that’s the way you interpret it, then the answer seems clear. ---
**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 15, 2009, 10:37pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/73 "2009-10-15T22:37:27Z")
> [@](#): > > What? Why not assume a perfectly horizontal push? What we’re thinking of as the push is the sum of the normal and frictional forces exerted by the finger. You can assume a perfectly horizontal push, but in that case, you’re asserting from the outset that the frictional force from the finger is zero, which solves the problem. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 10:40pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/74 "2009-10-15T22:40:50Z")
> [@Pedro](#): > > I’m not saying it doesn’t apply, I’m just saying it’s not good enough to yield a definite answer, e.g. Sum(Friction\_board) = 12.84%\*Weight and Sum(Friction\_finger) = 87.16%\*Weight, assuming a perfectly horizontal push against a vertical board. Sorry, I screwed up there, it shouldn’t say sum because as I argued above the sum has to be 50% each side, but you get the point I think. That’s the friction due solely to the normal force along the axis of the “pushing force”. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 15, 2009, 10:53pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/75 "2009-10-15T22:53:32Z")
> [@Chronos](#): > > What we’re thinking of as the push is the sum of the normal and frictional forces exerted by the finger. You can assume a perfectly horizontal push, but in that case, you’re asserting from the outset that the frictional force from the finger is zero, which solves the problem. Why? 😕 Assume a frictionless board (but not finger) and an horizontal push. Are you saying the eraser will fall to the ground in that case? ---
**Author:** ![treis](https://avatars.discourse-cdn.com/v4/letter/t/bc79bd/32.png) [@treis](https://boards.straightdope.com/u/treis)\ **Post date:** [October 15, 2009, 10:56pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/76 "2009-10-15T22:56:28Z")
> [@Chronos](#): > > What we’re thinking of as the push is the sum of the normal and frictional forces exerted by the finger. You can assume a perfectly horizontal push, but in that case, you’re asserting from the outset that the frictional force from the finger is zero, which solves the problem. I would call this an unsatisfying answer bordering on a trick question. ---
**Author:** ![EdwardLost](https://avatars.discourse-cdn.com/v4/letter/e/bbe5ce/32.png) [@EdwardLost](https://boards.straightdope.com/u/EdwardLost)\ **Post date:** [October 16, 2009, 12:19am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/77 "2009-10-16T00:19:35Z")
We could angle the finger around and complicate the situation, but I think the spirit of the problem is: For a force with a given normal component, Fn, what will be the values of the two vertical forces Ff [finger] and Fb [board]? (Of course Ff + Fb = W.) There is a single answer possible – that is, for a particular finger, a particular board, a particular eraser and a particular Fn, Ff and Fb will have particular values. I don’t think it’s possible to calculate Ff(Fn) and Fb(Fn) from first principles, but it would be possible to _measure_ them. To perform the measurement you need an apparatus that separates the horizontal “thruster” from the vertical “lifter”. That was what I was trying to get at by my see-saw rod concept. When you find the counterweight necessary to hold the rod (finger) horizontal you have measured Ff. As mentioned by others, I don’t think the mu’s (static friction coefficients) will help much, except in the particular case where Fn is reduced to the point where the eraser is just about to slip. Then Ff/Fb will equal muf/mub. My guess is that at larger Fn, Ff and Fb will both approach 1/2W. ---
**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, 12:44am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/78 "2009-10-16T00:44:40Z")
I’ve done the experiment. I’ve held an actual eraser up to an actual whiteboard with my actual finger. Ff can be several W up, several W down, several W left, several W to the right, and anywhere in between. There isn’t a unique solution to the problem. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [October 16, 2009, 12:48am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/79 "2009-10-16T00:48:36Z")
> [@Pedro](#): > > Why? 😕 > > Assume a frictionless board (but not finger) and an horizontal push. Are you saying the eraser will fall to the ground in that case? Yes, although the original intention of the problem was not to have a frictionless board. If that’s confusing, then consider this: in order to remain stationary, the weight of the eraser must be resisted by another force. Where would this force come from? > [@treis](#): > > I would call this an unsatisfying answer bordering on a trick question. Disagree. Or, rather, the original question wasn’t _intended_ to be a trick. It’s just that it’s not intuitively obvious that the \*direction \*of the “total” finger force is a parameter that can be arbitrarily set by choice at the beginning of the problem. If it helps, append “What am I missing?” to the OP. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [October 16, 2009, 1:27am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/80 "2009-10-16T01:27:04Z")
> [@zut](#): > > Yes, although the original intention of the problem was not to have a frictionless board. > > If that’s confusing, then consider this: in order to remain stationary, the weight of the eraser must be resisted by another force. Where would this force come from? Ok, I think I get what you mean now. You are suggesting the finger would follow the eraser on its downward trajectory? But that violates the spirit, if not the letter of the original question. It’s a cop out and the more interesting problem of a completely rigid finger in the vertical direction is still open. [Previous page](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819.md?page=3) [Next page](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819.md?page=5)