This should be an easy physics problem...

Factual Questions
1

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?

2

WAG

If it aint moving, I think the coefficients of friction are irrelevant or cancel or sumptin.

Note, I’ve given this about 10 seconds of thought.

Where is my Nobel ? :slight_smile:

3

Why not? Like billfish678 said, as long as it isn’t moving the coefficient of friction doesn’t matter. If you assumed unequal upward forces, then you would have a torque on the eraser and all hell would break loose.

4

Static friction is going to be there when the objects are not moving, so friction is still an issue.

As to the answer… no clue. I can’t even imagine how you’d experimentally measure the force on one side of the eraser separately from the other side.

5

Let’s start with an obvious point: The eraser is being pulled down by the Earth and the Earth is being pulled up by the eraser. A lot of people miss that, but a Physicist certainly wouldn’t. A typing error perhaps. Those forces balance and so when you start getting to the frictional forces, they also balance. (I.e., there’s equal frictional forces up and down.) Where do you want to go from here?

6

Good question. I’m just an engineer, but you may recall that I used to teach physics a few years back.

Anyway, my intuition is that there is no easy way to determine the answer without more information. Anyway, here are my thoughts:

One way to start looking at the problem is to consider the extreme cases. For example, suppose that the whiteboard was a frictionless surface. In this case, all of the frictional force would come from the friction arising between your hand and the eraser.

Conversely, if your hand was a frictionless surface, all of the frictional force would have to come from the friction arising between the whiteboard and the eraser.

(If both the whiteboard and your hand were frictionless surfaces, the eraser would simply fall down.)

So as the normal force from your hand on the eraser decreased and the eraser came near sliding, I agree with you that in this marginal case, the relative magnitude of the two frictional forces would depend on the relative coefficients of static friction.

However, even if a great deal of normal force is applied, I still think the relative coefficients of static friction would come into play. After all, if we apply a relatively high, constant normal force, and make the surface of the whiteboard more and more smooth, approaching that of a frictionless surface, the relative magnitude of the two frictional forces will change in a smooth function. The frictional force between the eraser and whiteboard will not just drop abruptly to zero. So I think that the coefficients of static friction do matter, even if a great deal of normal force is applied.

Of course, friction is a tricky subject, and its my understanding that the equation for frictional force is just an approximation of the real world.

7

There’s no typing error, just some conceptual errors. On your part. :slight_smile:

You have to consider the forces on a given object. In this case, we are considering the eraser. The forces on the Earth are irrelevant.

While it is true that the force of gravity that the Earth exerts on the eraser is equivalent in magnitude to the force of gravity that the eraser exerts on the Earth, this fact is irrelevant when we consider what forces are acting on the eraser. These two forces only “balance” if you consider the Earth-eraser system together as a whole. They do not “balance” on any one object.

Also, in the case presented by the OP, there are no frictional forces acting in the downward direction. The frictional forces act only upward to oppose the downward gravitational force exerted by the Earth on the eraser.

8

Eraser is squishy, board isn’t perfectly stiff either. You’ll end up asymmetrically pushing your rectangular eraser into a somewhat rectangular depression in the dryboard. There’ll be non-horizontal forces acting on the eraser from all around the edges of the depression.
I expect those forces will be enough to make an eraser pushed against a dryboard much more stable than say a flat chunk of stainless pushed against a thick slab of polished stainless.

9

Why is this hard? Can’t you use the fact that the sum of the torques will also be zero (as Baracus implied)? On the board side you have distributed normal and tangential forces (that I would assume are uniform) and on the finger side I would treat this as a point source (both normal and tangential). It should be relatively easy to figure out the distributions.

I’m off to do some back of the envelope calculations…

10

How about this page from the MIT RELATE Physics Teaching Wiki?

https://wikis.mit.edu/confluence/pages/viewpage.action?pageId=31655490
You can click on the tabs at the top and anything with a blue arrow beside it. In particular, go to Approach and then to Diagrammatic Representation and Mathematical Representation.

Disclaimer: I’m involved in putting this wiki together, although I had nothing to do with this particular “page”.

11

Yeah, one of the other grad students thought of using torques, but the problem there is that an extended surface is in contact with the board at least, and possibly also the hand. I know what the total normal force of the board on the eraser is, but I don’t know what the pressure is at each individual point on the surface, which I would need in order to calculate torque.

And if the squishiness of erasers and whiteboards is a problem, feel free to replace it with a block of steel or whatever. The only reason I picked an eraser is, first, that it was the most convenient object available as a visual aid when I asked the other grad students, and second, because it makes it clear that the surfaces are asymmetric.

CalMeacham, that page is clearly a related problem, but it has the key difference that in all those problems, the object is sliding, so you have kinetic friction, not static.

12

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.

13

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?

The frictional force is equal to W up til the point sliding is about to begin, at which point its mu*N. It all seems to balance out and those are all the forces on the eraser that I can see…

14

No. You don’t know the locations of the centroids of the normal forces. If the frictional forces on each side of the eraser are different, the locations of the reaction force centroid will shift to provide an offsetting torque.

Different system:

15

Thought experiment time. Construct an incline split down the middle with rubber on one side and chalkboard on the other. Place half the erase on the rubber side and half on the chalboard side. Adjust the incline such the both halves do not slide down the incline. Both halves are experiencing the same normal force to the surface and the fricitonal force is the same in both cases, i.e. equal to the gravitational component parallel to the incline. If we then glue the two halves together and straddle the rubber-chalk divide nothing changes, right? They both contribute half of the total frictional force keeping the thing from sliding down the incline despite having different coefficients of friction.

I think there are two cases:

Case 1: The normal force x coefficient of friction for each surface is at least half of the weight of the object. In this case, both frictional forces will be equal as with my thought experiment above.

Case 2: The normal force x coefficient of friction of one of the surfaces is less than half the weight of the object. In this case, there will be a net torque causing the horizontal forces to no longer be evenly balanced causing the fricitonal forces to be mumble, mumble, mumble…and there you have it!

16

CalMeacham’s wiki assumes the box is a point particle, this to me is the biggest problem with it, not necessarily that it has motion and a frictionless wall. With a point particle, the friction forces are always symmetric.

I can solve this problem (and would be happy to post a solution after work today) if I make the following assumptions:

Board is perfectly rigid.
Eraser is perfectly rigid.
Finger applies force to an infinitesimal point.

The first two assumption lead to the condition where the normal and static friction forces on the board side are distributed uniformly over the area. With this, it becomes trivial to solve by noting that the net forces and net torques must equal zero.

I’ll post a solution, complete with a FBD and the reduction of the distributed force system to a statically equivalent simple system after work if nobody has solved it before that.

17

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.

18

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.

19

This is right as far as I can see. This leads to a nonuniform distribution of forces on the board side. I don’t think this is possible with perfectly rigid objects.

20

I’m a little confused. You started by implying this was an idealized-basic-physics-land object and you were looking for an idealized-basic-physics-land answer, but now seem to object to the idealized general answer by saying that you don’t know enough detail about the real physical object.
I mean, we can do a super-idealized model where the eraser only contacts the board and the finger at one point each. If the center of mass is halfway between the finger and board, then balanced torque means that the friction forces at the board and finger have to be equal. If the eraser isn’t symmetric from board to finger, then the friction forces will be stronger on the side (finger or board) closest to the center of mass and weaker on the other (simple algebra should give quantitative answers here, which I’m too busy to do).
Now we can go to a slightly less idealized model, where the finger is a single point, but the eraser contacts the board along a vertical range. We’ll also assume that the eraser is completely rigid, and that it and the board are uniform, which means that the normal force and friction force are equally distributed along the vertical contact range. We’ll assume the center of mass is still at exactly the level of the finger, and the eraser is symmetric top-to-bottom.

In this case, some of the friction forces at the board, above and below the finger, aren’t aligned perfectly perpendicular to a line from the center of mass. Therefore, they’re contributing a little less torque about the center of mass, therefore, the friction forces at the board have to be overall a little higher. So if the center of mass is halfway between the finger and board, the friction forces will be higher at the board then at the finger. Just as before, if the COM isn’t halfway, the friction forces rise on the side closer to the COM. It’s going to take some of that there fancy calculus to get a quantitative answer, and I’m definitely not doing that, but it’s certainly possible in principle.

If the finger contacts the eraser above or below the COM, things get more complicated, because now the normal forces aren’t in line with the COM, and so they start contributing to torque around the COM. I’ll leave that for bored grad students to analyze.
I suppose next you could make a model with a deformable eraser, calculating stresses, deformations, resulting changes in stress and second-order deformations, etc. etc., but that’s for a really bored grad student trying to learn a fancy stress modelling computer system.