Stress on a beam supported by mass-less cable, friction at play, no angle given

[brutalmouse]

Homework Statement

One end of a uniform 3 m beam, that weighs 15 N, is supported by a mass-less cable. The other end rests against the wall. The coefficient of static friction between the wall and the beam is μs=.5

Figure: http://i.imgur.com/Ad9R2.jpg?1

Determine the minimum distance x from the wall that an additional weight of 15 N can be hung without causing the beam to slip.

Homework Equations

Ʃ Torque
Ʃ Forcesx=0
Ʃ Forcesy=0

force in x are the ones from the wall which we'll call our normal force, and the x portion of tension from cable

force in y are the friction lended by the wall, which is equivalent to μ*normalforce in x, and also y direction of tension

The Attempt at a Solution

I don't know how to sum torques here. I really don't know what to do without that θ

 

[haruspex]

force in x are the ones from the wall which we'll call our normal force, and the x portion of tension from cable
force in y are the friction lended by the wall, which is equivalent to μ*normalforce in x, and also y direction of tension

So turn the above into equations. Please use algebraic names, not the given values.
For the torque, you have to choose a convenient point to take moments about.
I would choose the point where the cable attaches to the beam, but you could take the other end, or the midpoint. It should come out the same in the end.

Btw, your post says the beam is 3m, but the figure shows 6m.

 

[brutalmouse]

Thanks for the reply. In drastic need of help..

ƩFx =Nx-Tx=0 so Nx=Tx
ƩFy =(μs*Nx)+Ty=0
Ʃτ = What do I put here? I'm lost.

 

[haruspex]

ƩFx =Nx-Tx=0 so Nx=Tx

If the angle to the vertical is θ and the tension T, what's Tx.
(You can figure out tan θ from the diagram.)

ƩFy =(μs*Nx)+Ty=0

You've left out the masses.
For the torque, what point do you want to take moments about, what forces have moments about that point, and what moment does each exert?

 

[brutalmouse]

So let's pretend in the question she didnt make the mistake of saying 3m. Then tanθ=5/6 so theta = 39.8°

Tx =Mgcos(39.8°)?


Lets set it at the far right. Tension from the cable is involved due to the Mg.

Pretty dusty on this May I get a little more actual guidance?

 

[haruspex]

So let's pretend in the question she didnt make the mistake of saying 3m. Then tanθ=5/6 so theta = 39.8°

I wouldn't bother calculating angles at this stage. Chances are we'll only be interested in trig functions of theta, and maybe only in the tangent. So just leave it as t = tan(theta) = 5/6.

Tx =Mgcos(39.8°)?

Not so fast. Let's get the three basic equations sorted first. You previously wrote Nx=Tx, and that is correct, so now we have Nx = T cos(θ), right?
What forces have components in the y direction?
What equation can you then write down for the balance of forces in the y direction?
Which forces have a turning moment about the cable end of the beam?