1. Apr 19, 2005

### ninjagowoowoo

I have a few problems where I just need a push in the right direction:

(1) A tightly stretched "high wire" is 46.5 m long. It sags 3.49 m when a 59.7 kg tightrope walker stands at its center. What is the tension in the wire? I have no idea what to do on this problem...

(2)A 36.6 kg round table is supported by three legs placed equal distances apart on the edge. What minimum mass, placed on the table's edge, will cause the table to overturn? Neglect the mass of the legs. Wouldn't I need to know some other information like radius or something?

(3)A cube of side l = 120 cm rests on a rough floor. It is subjected to a steady horizontal pull, F, exerted a distance h = 83.0 cm above the floor. As F is increased, the block will either begin to slide, or begin to tip over. What is the maximum coefficient of static friction for which the block begins to slide rather than tip? Wouldn't I ned to know F to figure this out? Keep in mind my answer has to be a numerical value, not a formula.

Any tips on any problem would be greatly appreciated. Thank you so much for helping...

2. Apr 19, 2005

### James R

(1) The shape of the wire is an isosceles triangle with "height" 3.49 m. You can work out the angles at the two sides (on either side of the tightrope walker), and then resolve forces on the walker (two tension forces + gravity).

3. Apr 19, 2005

### ninjagowoowoo

Thanks, any ideas on the other questions?

Last edited: Apr 20, 2005
4. Apr 20, 2005

### squib

I'm stuck on those last two too...

5. Apr 20, 2005

### ninjagowoowoo

did you get numbers 5 or 6? My brain just wont work today. The last two questions regarding the damn ladder?

6. Apr 20, 2005

### squib

Yeah... I set x and y axis as they are assigned in the prob, then just found torques. Torques are not perpendicular to the things they are acting on, so you must convert. Most the conversions were cos(theta) i think... although I think one was sin(90-theta). That help?

7. Apr 20, 2005

### ninjagowoowoo

Haha, I am seriously having a rediculous amount of difficulty with this. Did you use the center of the ladder as your pivot point? Then for torque you're using rFsin(theta) right? Is that what you meant by converting? Is F just the weight of the ladder?... I think I'm just having a difficult time identifying the forces acting on the ladder that contribute to torque.

8. Apr 20, 2005

### whozum

If you want some help, a problem, or some pictures wouldnt hurt.

Torque = rFsin(theta) is the torque equation

cos(x) = sin(90-x) is a trig identity

9. Apr 20, 2005

### ninjagowoowoo

The problems I'm most concerned with are in the first post, I just got sidetracked.

The problems are pretty self explanitory, but if you want some pics i'd be happy to upload some.

Last edited: Apr 20, 2005
10. Apr 20, 2005

### whozum

I dont see any ladders in the first three problems.

11. Apr 20, 2005

### ninjagowoowoo

I just changed my last post, sorry

12. Apr 20, 2005

### whozum

You've got me on #2 and #3.

13. Apr 20, 2005

### ninjagowoowoo

yeah they're crazy... I don't know what my professor is thinking.

14. Apr 20, 2005

### whozum

Ok, the ladder will pivot about its ground contact. The center of mass would just be at the center, or 3.87m from the pivot point. You can find the torque in the counterclockwise direction from there, but keep in mind that the force is not just mg. Now since the torques sum to zero, the wall is providing the same torque but in the clockwise direction.

The forces in the x and y directions both equal 0. The force towards the ground (-y) is F = mgsin(71.4), what can you say about the normal force (in x and y) applied on the ladder?

A similar argument for the wall works.

15. Apr 20, 2005

### squib

I picked the point against the wall as the rotation point

16. Apr 20, 2005

### whozum

squib: Either one works. Now find the torque in the clockwise direction from that point, can you find the force, and radius?

17. Apr 20, 2005

### ninjagowoowoo

Thanks a lot. I really appreciate all the help, now if only we can find someone that can do the second two probs. I think I'm going to repost them.

18. Apr 20, 2005

### squib

On the cube problem I seem to come up with u = (a*h)/(g*(l/2))

My problem is that I can't figure out what a is...

this could also be written u = (F*h)/(mg*(l/2)), F being the constant force...

19. Apr 20, 2005

### ninjagowoowoo

yeah, I can't find a way to express u without using F or A. Bah.

20. Apr 20, 2005

### ninjagowoowoo

do you think maximizing F or A would give us an answer? Eg: putting all the constants in, taking the derivative of the whole thing and setting it equal to zero? Then we could use this force to put back into your equations. Does that sound plausible? I think I'll try it.

bah ... darned equation has no max (duh). That doesnt work

Last edited: Apr 20, 2005