# 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:** 2
**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 14, 2009, 8:49pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/21 "2009-10-14T20:49:30Z")
> [@groman](#): > > So why is the ratio of forces not the ratio of coefficients of static friction? You seem to dismiss that but considering that more slippery the board or your fingers are, more force you have to exert on the eraser to keep it from sliding, it conceptually makes sense. 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. > [@BrandonR](#): > > Okay, I drew the situation and did a FBD. In the horizontal direction we have the force F you’re pushing the eraser with, and a resulting normal force from the board. In the horizontal direction we have the weight, which is countered by the frictional force. Total of four forces, all adding up to static equilibrium. Am I missing something? There are two frictional forces, one on each surface. ---
**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 14, 2009, 8:58pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/22 "2009-10-14T20:58:44Z")
> [@L. G. Butts](#): > > This is why I have to assume that all bodies are completely rigid. In this case the forces are uniformly distributed over the contact areas. I don’t believe that’s true. 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. ---
**Author:** ![Baracus](https://avatars.discourse-cdn.com/v4/letter/b/2bfe46/32.png) [@Baracus](https://boards.straightdope.com/u/Baracus)\ **Post date:** [October 14, 2009, 8:59pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/23 "2009-10-14T20:59:20Z")
> [@ntucker](#): > > I doubt this is true. There may be a torque on the eraser from the upward frictional forces being unequal, but in that case, the way in which the eraser was being pressed against the board would provide opposing torque, right? > > Imagine that the finger-eraser connection is basically frictionless – he’s holding it up by pressing an ice cube against it and it’s nice and smooth on its back. The upward force is provided entirely by the friction between the eraser and the board, which causes a torque, but rotation is prevented by the horizontal force from the ice cube and a corresponding horizontal force between the board and the bottom corner of the eraser, which work together to provide a counter-torque. 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. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [October 14, 2009, 9:28pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/24 "2009-10-14T21:28:04Z")
> [@](#): > > Different system: 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. ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [October 14, 2009, 9:44pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/25 "2009-10-14T21:44:23Z")
> [@zut](#): > > I don’t believe that’s true. > > 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. Yikes, you’re right. That seems counter-intuitive to me, but you are absolutely right. ---
**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 14, 2009, 9:44pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/26 "2009-10-14T21:44:32Z")
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. ---
**Author:** ![Happy\_Fun\_Ball](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/happy_fun_ball/32/17664_2.png) [@Happy\_Fun\_Ball](https://boards.straightdope.com/u/Happy_Fun_Ball)\ **Post date:** [October 14, 2009, 10:03pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/27 "2009-10-14T22:03:14Z")
I still think this is solvable, but I haven’t had a statics/dynamics class in ~15 years. I can’t spend any more time on this though until after work. ---
**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 14, 2009, 10:34pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/28 "2009-10-14T22:34:36Z")
> [@CalMeacham](#): > > 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?) It’s not what other forces there _are_, it’s what other forces there _aren’t_. From your link: > [@](#): > > A person holds a 10 kg box against a smooth (i.e. **frictionless** ) wall (as it slides down) by applying a perfectly horizontal force of 300 N. Not so applicable when the discussion is about the effect of friction. ---
**Author:** ![CalMeacham](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/calmeacham/32/35_2.png) [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)\ **Post date:** [October 14, 2009, 11:17pm UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/29 "2009-10-14T23:17:27Z")
Sorry – I thought that was the problem with friction. ---
**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, 1:11am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/30 "2009-10-15T01:11:50Z")
> [@Chronos](#): > > As many of you probably know, I’m a physicist. But I’ve recently come up with a problem that seems like it should be easy, but has several professors and grad students stumped. > > Let’s say I’m holding an eraser stationary up against a whiteboard by pressing horizontally on it with a force F[sub]N[/sub]. F[sub]N[/sub] 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? 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? Stranger ---
**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, 1:32am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/31 "2009-10-15T01:32:08Z")
> [@](#): > > The static friction forces never exceed the sliding limit force and will be in proportion to their coefficients, 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. ---
**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, 1:33am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/32 "2009-10-15T01:33:15Z")
Or, to rephrase the question: There are two frictional forces on the object. What is the magnitude of each? ---
**Author:** ![rbroome](https://avatars.discourse-cdn.com/v4/letter/r/838e76/32.png) [@rbroome](https://boards.straightdope.com/u/rbroome)\ **Post date:** [October 15, 2009, 1:34am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/33 "2009-10-15T01:34:23Z")
> [@Chronos](#): > > As many of you probably know, I’m a physicist. But I’ve recently come up with a problem that seems like it should be easy, but has several professors and grad students stumped. > > Let’s say I’m holding an eraser stationary up against a whiteboard by pressing horizontally on it with a force F[sub]N[/sub]. F[sub]N[/sub] 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? > > Well, obviously the horizontal forces are a normal force from me on the eraser, and a normal force from the board on the eraser, and since a = 0, these cancel out and are equal in magnitude by Newton’s Second Law. 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. > > Now, clearly, the sum of the two frictional forces must be equal in magnitude to the weight, since, again, a = 0. But what’s the breakdown? How much force is in each? The surfaces are different for the two frictional forces, so we can’t invoke symmetry. We could try using the coefficients of static friction, but those are only supposed to be relevant in the marginal case, where the object is about to start sliding, and we’re far from marginal here, since I’m pressing so hard. But obviously, there must be some way to model this and get an answer. How? 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. I guess I am not seeing what the problem is. ---
**Author:** ![rbroome](https://avatars.discourse-cdn.com/v4/letter/r/838e76/32.png) [@rbroome](https://boards.straightdope.com/u/rbroome)\ **Post date:** [October 15, 2009, 1:51am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/34 "2009-10-15T01:51:06Z")
> [@Chronos](#): > > Or, to rephrase the question: > > There are two frictional forces on the object. What is the magnitude of each? 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? ---
**Author:** ![Ludovic](https://avatars.discourse-cdn.com/v4/letter/l/7ab992/32.png) [@Ludovic](https://boards.straightdope.com/u/Ludovic)\ **Post date:** [October 15, 2009, 1:57am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/35 "2009-10-15T01:57:03Z")
> [@Chronos](#): > > Or, to rephrase the question: > > There are two frictional forces on the object. What is the magnitude of each? 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. ) ---
**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, 1:58am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/36 "2009-10-15T01:58:07Z")
> [@rbroome](#): > > not a friction expert, but it seems to me that the ratio of forces should be equal to the ratio of cofs. I’m not a friction expert either, but I don’t see why that should necessarily be so, other than in a rough way. (further musing on post 21) ---
**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, 2:19am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/37 "2009-10-15T02:19:29Z")
> [@Chronos](#): > > Or, to rephrase the question: > > There are two frictional forces on the object. What is the magnitude of each? 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. ---
**Author:** ![Baracus](https://avatars.discourse-cdn.com/v4/letter/b/2bfe46/32.png) [@Baracus](https://boards.straightdope.com/u/Baracus)\ **Post date:** [October 15, 2009, 2:28am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/38 "2009-10-15T02:28:48Z")
> [@zut](#): > > I’m not a friction expert either, but I don’t see why that should necessarily be so, other than in a rough way. (further musing on post 21) 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. ---
**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, 2:46am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/39 "2009-10-15T02:46:27Z")
> [@ZenBeam](#): > > 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.) 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. ---
**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, 2:52am UTC](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819/40 "2009-10-15T02:52:16Z")
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. [Previous page](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819.md?page=1) [Next page](https://boards.straightdope.com/t/this-should-be-an-easy-physics-problem/513819.md?page=3)