# Homework Help: How to Keep a Sign from Collapsing

1. Jan 19, 2015

### postfan

1. The problem statement, all variables and given/known data
. The sign shown below consists of two uniform legs attached by a frictionless hinge. The coefficient of friction
between the ground and the legs is µ. Which of the following gives the maximum value of θ such that the sign will
not collapse?

(A) sin θ = 2µ
(B) sin θ/2 = µ/2
(C) tan θ/2 = µ
(D) tan θ = 2µ
(E) tan θ/2 = 2µ

Page 2, Diagram 6

2. Relevant equations

3. The attempt at a solution
I drew a FBD and got a force of gravity for each leg with force mg down (relative to the surface on the ground) , normal force of mgcos((180-theta)/2) up (relative to the surface of the leg), and friction of mu*N towards the interior of the sign. The question is where do I go from here?

2. Jan 19, 2015

### phinds

Since the two forces you have computed don't oppose each other, you aren't going to get far that way. How about you find the force that the friction force IS opposing?

3. Jan 19, 2015

### postfan

Friction opposes the horizontal component of the normal force which is 1/2*mg*sin(180-theta), right?

4. Jan 19, 2015

### phinds

Well, what do you think?

5. Jan 19, 2015

### postfan

Well I am pretty sure that that's it, but I want to make sure.

6. Jan 19, 2015

### phinds

I didn't look at your figures but you have the right idea. The friction is a horizontal force so obviously the only force that can directly oppose it is a horizontal force.

7. Jan 19, 2015

### postfan

So is the next same just to equate the horizontal component of the normal force with the force of friction?

8. Jan 19, 2015

### phinds

You seem very hesitant to embrace the obvious. I suggest that you spend some time studying force vectors in various kinds of problems, not just ones where the forces that matter happen to be horizontal.

9. Jan 19, 2015

### postfan

I'm hesitant because I already equated it with each other and got mu=cos(90-theta/2) which is not one of the answer choices.

10. Jan 19, 2015

### phinds

That could be due to a calculation error, but can't be due to the forces not cancelling each other when the end is right at the point of moving.

My point is that once you've done enough of these problems you'll have an instinctive feel for what forces have to cancel each other and an answer that seems wrong will just lead you to recheck your calculations, not your logic.

11. Jan 19, 2015

### BvU

I always learned that normal forces are just that: normal forces. Normal (perpendicular) to the horizontal plane in this case. So I don't think there is such a thing as a horizontal component of the normal force in this exercise.

12. Jan 19, 2015

### postfan

Ok so in that case m*g*sin(theta) = mu*m*g*cos(90-theta/2) ?
If so anyway to simplify cos(90-theta/2)?

13. Jan 19, 2015

### phinds

That's correct. But there IS a horizontal component to the ladder's force on the floor. There are two resultant forces from the ladder's force on the floor, the normal force and the horizontal force.

14. Jan 19, 2015

### postfan

OK I'm confused.

15. Jan 19, 2015

### phinds

To such an utterly vague statement, I can only say, well that's unfortunate. I'm not. Do you have a question? Do you understand force vectors at all? If you don't then that's where to start.

16. Jan 19, 2015

### postfan

You two are telling me conflicting things.

17. Jan 19, 2015

### phinds

OK, I certainly don't mean to be. What two things are you talking about? Do you want me to guess or would you like to tell me?

18. Jan 19, 2015

### postfan

You are telling me to use the horizontal component of the normal force but BvU is saying to just use the normal force without decomposing it.

19. Jan 19, 2015

### phinds

No, I most emphatically am NOT telling you to use a component of the normal force. Please re-read what I said in posts #6 and #13

20. Jan 19, 2015

### postfan

I was using the horizontal component of the normal force.
And you said ...

21. Jan 19, 2015

### phinds

Ah ... that's where the confusion set it. Sorry. Since there IS no horizontal component to the normal force, I just read this as meaning that you were using the horizontal component for the force that the ladder exerts on the floor, which is the right thing to do. Since, as I said, I did not look at your calculations, I did not see that that apparently was not what you were doing.

I AGAIN urge you to read what I said in post #13.

I'm done with this conversation. Since you clearly do not understand force vectors, I can only say once again that what you need to do is learn about force vectors in general before trying to solve a problem involving force vectors.

22. Jan 20, 2015

### BvU

One place where there is a horizontal force in action is at the floor level. There is one other place.
Look under the relevant equations to see what you have available as conditions for equilibrium (i.e. non-collapse). It may well be that the resultant there is zero, but it helps in the math for each of the legs.

23. Jan 20, 2015

### BvU

I'm sorry to see that Phinds is worn out a little. My impression is that's partially caused by the medium we are working with: it's easy to trigger misunderstandings with such indirect communication. Can't be helped at the moment. I am completely convinced of Phinds good will and expertise. I am also convinced of postfans sincerity, curiosity and courage in tackling a challenging set of AAPT contest questions, so I'll go along a bit more.

Going back to square 1 (post #1?) with this exercise (usually going under the tags ladder - wall, e.g. here -- check out post #4 for some relevant equations !!), I think a lot of hassle would have been avoided with the picture (good for 1k words, they say) you drew ! (Can you still show it ?)

Things start well in post #1with $2mg$ down acting at the center of mass of the whole contraption -- awkward place, but probably cancels anyway. Easier to deal with if we use symmetry and look at one leg only: $mg$ halfway the ladder. (You can see I'm too lazy to carry the 2 around, so I use m for the mass of one leg only).
[edit:} Can't do that: too confusing.

Things start well in post #1 with $mg\$ down acting at the center of mass of the whole contraption -- awkward place, but probably cancels anyway.
Easier to deal with if we use symmetry and look at one leg only: ${1\over 2}mg\$ halfway the ladder. Straight down.​

Then a second force "normal force $mg\; \cos\left ( (180^\circ - \theta)/2 \right )$ up (relative to the surface of the leg) " -- here you can see how useful a picture would have been ! As well as a bit of basic trig: $\cos (\pi/2 - \alpha) = \sin\alpha$. Apparently we are projecting $mg$ (That is the actual normal force) on a line perpendicular to the leg. For the right leg would point a bit up and to the right. Project that onto the floor and presto: there is a horizontal component where there was no horizontal component before. No good.
Would be wiser here to not lump the two legs together. One normal force is ${1\over 2}mg$ pointing straight up at the point the right leg is on the floor. You can guess the second normal force is ${1\over 2}mg$ pointing straight up at the point the left leg is on the floor. Together they nicely add up to $mg$ pointing straight up and this is compensating gravity: the stand won't sink into the floor, nor will it float up. Both normal forces are equal, which is good too: the thing won't play dog and lift one leg :) or start rotating in a vertical plane any other way.
Third force in post #1: $\mu N$ inwards. Let me not be corny and go on about lumping together. There is $\mu {1\over 2}mg$ pointing to the left at the point the right leg is on the floor, etc. Left and right add up to 0. Everybody happy -- if that were all. And it isn't, as the bloke who looked at each leg separately has noticed already: the legs lean against each other and a fourth force comes in the picture. Two actually: left on right and vice versa.

Now we can have a sidebar about the direction of these forces -- acting at the hinge, which is frictionless, thank goodness.
Is it purely horizontal, or is it in the direction the tip would want to go (i.e. perpendicular to the leg), so that the legs don't just lean against one another, but also ON one another. I'm in favor of the latter (thinking of one heavy leg and one light leg). I hope someone else reads this far and is knowledgeable enough to settle this.
The good news is it doesn't matter, because of the symmetry: both normal forces at the bottom will remain at ${1\over 2}mg$ . And at the top we only need the horizontal component -- to compensate $\mu {1\over 2}mg$ at the bottom.
As they say: one picture... The above isn't far short of a thousand words (and cost me a lot of time).

Back to work: four forces per leg, dragged together in such a way that the force balance is in proper order. One more condition for equilibrium to fulfil. My bet is the ${1\over 2}mg\$ divides out :) and the L too.

24. Jan 20, 2015

### postfan

Thanks for your confidence BvU!! I've read and reread your post several times (plus drawn a few FBDs!) and this is the part I still don't understand.
How is mg "the actual normal force" and not mg*sin(theta/2)?

25. Jan 21, 2015

### BvU

Would the normal force be 0 when theta is zero ?