# Help with torque and equilibrium

1. Jun 8, 2006

### mrtheshaggy

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

2. Jun 8, 2006

### mrtheshaggy

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 eachother, which ended up wrong. I think I'm on the right track but can't figure out where I am going wrong.

3. Jun 8, 2006

### brp1387

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