1. Rusty all round

I came across this problem in a subject I'm writing some short tutorials for. I thought I'd toss it in here, and compare the answers and approaches with my own:<hr>A garage door is mounted on an overhead rail as shown. The wheels have rusted so that they do not roll, but slide along the track. The coefficient of sliding friction is 0.5. The distance between the wheels is 4m; each is 1m from the edge of the door. The door is symmetrical and weighs 160N. It is pushed to the left at constant speed by a horizontal force P, 2m below the wheels. Find the vertical component of force exerted on each wheel by the track.<hr>And believe me, the wheels weren't the only rusty component when it came to solving this one! <img src=/S/grin.gif border=0 alt=grin width=15 height=15>

Alan

2. Re: Rusty all round

<P ID="edit" class=small>(Edited by sdckapr on 11-Aug-04 06:23. Reread and realized the wheels are part ot the door and changed answer accordingly)</P>Perhaps, I am missing something, but:
<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>since the vertical/horizontal components are independent, you have the force downward on the wheels which go downward on the support. The force pushing it is irrelevant as is the friction component. SInce you have 2 wheels (I assume the same) attached to the door (they are thus part of the 160N total weight), the force on each downward is 80N.</font color=yellow></span hi>

Steve

3. Re: Rusty all round

Hi Steve

Your analysis goes as far as <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>satisfying the condition that the door is in a state of "static" translational equilibrium. But there is another condition to be satisfied.</font color=yellow></span hi> I suspect that your solution might be one of the common ones I get for this problem.

Alan

4. Re: Rusty all round

I am not sure where you are going with static translational equilibrium. I find it hard enough to keep track of the concepts let alone some of the terminology after decades of non-use. (Though I admit to doing some with my daughter last year). Her "terms" for some of the concepts were different than I recall.

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>But as far as I remember vertical is vertical and horizontal is horizontal and there is no interaction. If the door is not moving up/down just left/right the vertical component does not change and the total is equal to the weight and it is equally distributed between the 2 wheels.

It should not matter whether it is sliding or not. Friction force is along the surface (horizontal in this case) though it is proportional to the vertical "normal force" its direction is only horizontal in this example.</font color=yellow></span hi>

Steve

5. Re: Rusty all round

What I was hinting at with the terminology (non-standard terminology I'm sure) was in the "translational" inclusion. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>There are other types of motion (and equilibrium) too - other than translational.</font color=yellow></span hi>

Alan

6. Re: Rusty all round

I'm afraid I don't know what you are hinting at. Based on all I recall, I can't think of any answer than what I already provided.

Steve

7. Re: Rusty all round

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Most of these (I think) are covered by the "coefficient of friction" of the material. Even if you view the railing as a series of peaks, valleys, and obstructions, you are looking at the average overall - so you are not moving it vertically.

There might be something if we speak of rotating the door slightly so there is more force on one wheel over the other, but as indicated the force is horizontal, with no indication of rotation (if so we would need more info on the angle.</font color=yellow></span hi>

Steve

8. Re: Rusty all round

Hey Timbo, I really like these ideas you're tossing round. I was actually asked to write these tutorials because the requestor seemed to like the approach I'd taken to other topics, trying to get away from the standard formulaic/ mathematical treatment and encourage prediction and visualization - much like you've done.

I should have stated that the force P is always applied horizontally, so can't agree with all of your conclusions. But most significant is the first part of observation 2. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If I move the force P up the door, it becomes easier to move the door</font color=yellow></span hi>. And I think your conclusion is good, based on your visualizations.

Alan

9. Re: Rusty all round

Steve,

You've just about hit on it in your second paragraph, and you're quite right - <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>the door doesn't rotate</font color=yellow></span hi>. This, together with Tim's point 2. is what makes this problem a bit more complex than <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>just balancing the forces</font color=yellow></span hi>.

Alan

10. Re: Rusty all round

Given that I don't even remember knowing about this stuff and, thus, approaching this from an 'expectation' point of view, are the following observations valid?

1. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If I place a grain of rice in front of the wheels and then push the door with an increasing force, initially, the grain of rice will stop the door from moving until there is enough force to force the wheel up and over the grain of rice (thus showing vertical force). If that is the case, then as the rice grain only increases the friction factor by a bit this implies that this is a vertical force component with or without the rice grain. Further implying that the vertical force is a function of the friction.</font color=yellow></span hi>
2. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If I move the force P up the door, it becomes easier to move the door thus implying more vertical force is generated.</font color=yellow></span hi>
3. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If I move wheels apart, then i expect that i will need to exert more force to move the door and so the distance between the wheels comes into play (somehow)</font color=yellow></span hi>
4. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If the friction was zero, then i could move the door with a very little force, implying that all off my force would be used horizontally and none vertically.</font color=yellow></span hi>
5. <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>If the friction was infinite, then i could not move the door regardless of how hard I pushed it, implying that all off my force would be used vertical and none horizontal.</font color=yellow></span hi>

So, <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>this means that the answer is some function of friction, force, distance between wheels and horizontal application of force. I'm thinking that the angle of the application of the to the wheels comes into play and thus the tangent (or some other trig function) should be used. I expect that the answer will be the weight of the doors plus the amount required to overcome the friction.</font color=yellow></span hi>

11. Re: Rusty all round

I hear what you are saying about horizontal and vertical force. I suppose that this exposes that I'm not sure how friction works. Does the force P 'lift' the door enough to overcome (or maybe 'bypass' is a better word) the friction or is there a threshold force below which the door doesn't move and above which it does?

Also, what about the situation where the right wheel is stuck fast to the track, then as P is increased the door would rotate clockwise (say P^ is the point where the door starts to rotate). Wouldn't moving the wheel closer to the point of application of the force reduce P^?

Finally, what about the situation where the beam can support 160N and not an ounce more. If moving the door increases the total vertical force then it'll come crashing down. I don't think that that will actually happen in practice.

I've just re-read the question and <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>I think that the total vertical force is 160N but it isn't spread uniformly over the two wheels. The closer wheel will support more and the wheel further away less. I leave it to someone else to apportion the blame / force. </font color=yellow></span hi>

12. Re: Rusty all round

Well done Timbo! Your hidden text is the key to the "more complex than first looks" aspect of the problem. But the next bit of trickery is how to solve that, given only the data provided.

Alan

13. Re: Rusty all round

<hr>Does the force P 'lift' the door enough to overcome (or maybe 'bypass' is a better word) the friction or is there a threshold force below which the door doesn't move and above which it does?<hr>

Yes, a "threshold force". You have "2 forces" which you apply. the force needed to overcome friction and the net force that accelerates the object. If the force you apply is less than or equal to the friction force the object does not move.

Any amount over that is force used to calculate the acceleration of the object.

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Based on what I recall (which seems to be "incomplete"), the friction force on each wheel is 0.5 * "the normal force". The normal force is the force opposite the force of gravity which prevents it from falling: it is the force exerted by the railing on the door. By my calcs the Fn on each wheel is 80N, so the friction force on each is 40N. Thus it should take 80N (40+40) of horizontal force to move the door. If (eg) 100N are applied then there is 20 N net force and the door is accelerating at:
9.8(m/s^2) / 160 N * 20N = 1.2 m/s^2 [Note: 160N/(9.8 m/s^2) is the doors mass].

I don't see how the placement of the horizontal force would be an issue, unless there is some rotation: if you press on the bottom, you put all the weight on the near wheel, if you press on the top you lift it off the near wheel, though this doesn't change how I calculate so, it seems (again) my education was not complete.</font color=yellow></span hi>

Steve

14. Re: Rusty all round

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>I guess I either miss the trick or my education is just incomplete, I can't see how even with different forces on each one (which I only see via rotation), but it doesn't make difference in my calcs, it still seems to be the dame total force down and the same force to move (unless you rotate).

If you rotate, then you are pushing upwards and "lifting the door" decreasing the normal force and decreasing friction so less force is needed to move.</font color=yellow></span hi>

Steve

15. Re: Rusty all round

Steve,

I think you're just about there. A key fact is that <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>the door isn't rotating</font color=yellow></span hi>.
A further hint: <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>The problem is not actually symmetrical, since P is being applied to only one side</font color=yellow></span hi>.

Alan

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