Finding the Maximum Angle for Equilibrium in a Rotational Statics Problem

[µ³]

http://img103.imageshack.us/img103/6237/yffigure11407dw.jpg
"One end of a uniform meter stick is placed against a vertical wall View Figure . The other end is held by a lightweight cord that makes an angle \theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400 ."

I can't for the life of me get the equation for this.
Part A:
What is the maximum value the angle \theta can have if the stick is to remain in equilibrium?
So far I've tried using the last end as a pivot point.
so that would give me
Balancing torque:
\mu T cos(\theta) = mg(1-x)
and balancing total forces
\mu T cos(\theta) + T sin(\theta) = mg
combining the two
produces
T sin(\theta) = mgx
dividing the third by the first gives me
tan(\theta) = \mu \frac{x}{1-x}
and there is no way I can get a "max theta". Naturally I can't get Part B and C but I figured if you guys can help me with A it will be easy to get the rest.

Thanks in advanced.

 

[Hootenanny]

What is the mass of the meter rule? Is it light?

 

[µ³]

What is the mass of the meter rule? Is it light?

negligible I presume.

 

[Gokul43201]

Can you draw the forces in that figure?

So far I've tried using the last end as a pivot point.

This is the end of the ruler not touching the wall, right?

so that would give me
Balancing torque:
\mu T cos(\theta) = mg(1-x)

I see a mistake here. What torque does the string exert about the point of attachment of the string? And are there no other forces on the ruler that exert a torque about this point?

Again, never do a force problem without a free body diagram.

 

[µ³]

Can you draw the forces in that figure?
This is the end of the ruler not touching the wall, right?

I see a mistake here. What torque does the string exert about the point of attachment of the string? And are there no other forces on the ruler that exert a torque about this point?

Again, never do a force problem without a free body diagram.

Yeah that's the end of the ruler not touching the wall. I thought about the string itself exhibiting torque but since I chose it as a pivot point it wouldn't make much sense if it did. Can you suggest an alternate pivot point? I also considered using the end touching the wall as pivot point and that gave me the equation
T sin(\theta) = mgx
which was already derived from the first two. I also considered using the ruler's center of mass a pivot point and that didn't give me a sensical answer either.
http://img445.imageshack.us/img445/4207/yffigure11407dw28yw.jpg
Pretty sure that's right.

 

[Hootenanny]

Perhaps taking moments about the centre of the rule since we don't know its mass? I haven't worked through this problem so don't take my word for it.

 

[µ³]

Perhaps taking moments about the centre of the rule since we don't know its mass? I haven't worked through this problem so don't take my word for it.

1/2 \mu T cos(\theta) + mg(x-1/2) = 1/2 T sin(\theta)
\mu T cos(\theta) + T sin(\theta) = mg

so...
\mu T cos(\theta) + 2mgx -mg= T sin(\theta)
\mu T cos(\theta) + 2mgx -\mu T cos(\theta) - Tsin(\theta)= T sin(\theta)
an equation I already had
mgx = T sin(\theta)
and reapplying the second equation
you get
\mu T cos(\theta) + T sin(\theta) = mg
\mu T cos(\theta) + mgx = mg
\mu T cos(\theta) = mg(1-x)
and we're back where we started. So so far, I've chosen the left end (ruler touching wall), the right end (string pulling on ruler), and midpoint and gotten the same answer for all of them. Argrhgrh what am I doing wrong?