# Holding up a Cone

1. Sep 14, 2010

### mmmboh

[PLAIN]http://img688.imageshack.us/img688/3941/fingernorm.jpg [Broken]

Because the cone is being held up, I said mg=2fy, where f is the friction force, so mg=2uNcos(x) (where x=theta), and I got N=mg/(2ucos(x)).

The cone also isn't moving from side to side, so I said Nx=fx, so fsin(x)= uNsin(x)=Ncos(x), so u=cot(x), then I replace u in my first equation to get N=mg*tan(x)/(2cos(x))...
I have a feeling this is wrong though. Help please.

Last edited by a moderator: May 4, 2017
2. Sep 14, 2010

### Mindscrape

I think in this case you are safer, and better off, by not rotating your coordinate axis and taking advantage of symmetry.

3. Sep 15, 2010

### mmmboh

I didn't think I did rotate it. But besides that, what is wrong with my solution?

4. Sep 15, 2010

### mmmboh

Is the correct equation 2Nsinx+mg=2uNcosx ? And so N=mg/(2ucosx-2sinx). And so the denominator can't be zero, so u is greater than tanx? This seems right.

Last edited: Sep 15, 2010
5. Sep 16, 2010

### Mindscrape

Ah okay, I'm sorry, I guess I didn't follow everything you did to begin with.

The sum of all forces in y should be

2NµcosØ-2NsinØ-mg=0

In the x directions, we can see from symmetry that everything will cancel out. I can write this one out if you want too, but symmetry makes it trivial.

Yes, the static friction must be greater than tanØ. We know that if Ø=90º then it's impossible to hold the cone.

Edit: Corrected the signs.

Last edited: Sep 16, 2010
6. Sep 16, 2010

### novop

That's not right... gravity and the normal force are in the negative vertical direction, the only positive vertical contribution is from friction.

Sum of forces in y should be:

2uNcosØ - 2NsinØ - mg = 0
N(2ucosØ-2sinØ)=mg

7. Sep 16, 2010

### Mindscrape

Yeah, that's right. For some reason my head was cloudy this morning.

N=mg/(2µcosØ-2sinØ)