Help with torque and equilibrium

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The discussion focuses on the equilibrium of a uniform meter stick supported against a wall, with a lightweight cord at a 15-degree angle. The coefficient of static friction between the stick and the wall is 0.370. The user seeks to determine the minimum distance (x) from the wall for the stick to remain in equilibrium while a block of equal weight is suspended from it. Key equations derived include T*sin(θ)*L = M*g*0.5L + M*g*x and 2*M*g = T*sin(θ) + u*T*cos(θ), addressing both translational and rotational equilibrium.

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Thanks for the many helpful posts/answers I've found through google! I ran into one I can't figure out, any help is much appreciated!...

One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle @ with the stick. The coefficient of static friction between the end of the meter stick and the wall is .370.

Let the angle betwewen the cord and the stick be @=15 degrees. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

Here is a quick illustration I made... www.theshaggy.com/meter_stick.jpg
 
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I tried setting

sin@*T = m*g*x + m*g*(L/2) ...making tension... T = m*g*(x+L/2)/sin@

so the F_x = u*cos@*T ///u=coefficient of friction

and F_y = m*g*(x+L/2) - T*sin@

the I'd set them equal to each other, which ended up wrong. I think I'm on the right track but can't figure out where I am going wrong.
 
Equillibrium

Please make the F_y equation clear.Dimensions do not match

The two equations might be as follows:
T*sin@*L=M*g*0.5L+M*g*x

and

2*M*g=T*sin@+u*T*cos@
These two equations are considering the rotational and translational equillibrium of the body
 

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