Static equilibrium with a suspended sphere on a wall

[gills]

Homework Statement

Homework Equations

\SigmaFi = 0
\Sigma\tau = 0

Nw = force of the wall
Ff = Frictional force
\mu = frictional coefficient
T = Tension

The Attempt at a Solution

\SigmaFx = Nw - Tsin(30) = 0

\SigmaFy = -mg + Tcos(30) + Ff = 0

\Sigma\tau = (R/2)mg - (3/2)R*Ff = 0

That's what i have, but i have a feeling that I'm missing something. Any help would be great. Especially if it's Doc Al!

Thanks
Tom

 

[Doc Al]

\SigmaFx = Nw - Tsin(30) = 0

\SigmaFy = -mg + Tcos(30) + Ff = 0

Right idea, but check your sines and cosines.

\Sigma\tau = (R/2)mg - (3/2)R*Ff = 0

Good!

The only thing you're missing is the relationship between the force of friction and \mu. What is it?

 

[gills]

Right idea, but check your sines and cosines.


Good!

The only thing you're missing is the relationship between the force of friction and \mu. What is it?

Doc,

I think the trig is correct. Unless I'm smoking something, but the opposite of that 30 degree angle is the x component of T, right? And the adjacent is the y component? Unless you want me to use Tcos60 for the xcomponent and Tsin60 for the y?

Anyway --->

Ff = \muNw --->

Nw = Tsin30 --> from the x-components, so -->

Ff = \mu*Tsin30

 

[Doc Al]

I think the trig is correct. Unless I'm smoking something, but the opposite of that 30 degree angle is the x component of T, right? And the adjacent is the y component?

Yep, you're right. (D'oh!)

Anyway --->

Ff = \muNw --->

Nw = Tsin30 --> from the x-components, so -->

Ff = \mu*Tsin30

Right. So solve those equations together and you'll find the value for \mu.

 

[gills]

Yep, you're right. (D'oh!)



Right. So solve those equations together and you'll find the value for \mu.

worked out on paper, i got 0.866 which is the right answer!

You know what confuses me the most is that you can solve for an actual number in a question with all variables. What is the best way to tell that variables will actually cancel out and you'll end up with computable trig functions and a constant? My brain is still trying to get used to it.

 

[Doc Al]

What is the best way to tell that variables will actually cancel out and you'll end up with computable trig functions and a constant?

If the number of unknowns matches the number of independent equations--that's a good sign.

But the real secret is to have the confidence to set up the equations and attempt to solve them without knowing in advance that it will work out. That's how you learn. Some students (not you!) are afraid to make a move unless they see how to get the final answer in one step; with that attitude they'll never improve their problem-solving skills.

After you've solved a zillion problems, you'll have an intuition about what's solvable and what's not. But don't wait for that!