Problem dealing with Equilibrium & Torque

[joshncsu]

Homework Statement

A 40 kg, 5.0-m-long beam is supported by, but not attached to, the two posts. A 20 kg boy starts walking along the beam. How close can he get to the right end of the beam without it tipping?

The left post under the very left end of the beam, and the first post is 3.0 m to the right of it. I've attached an image to assist in visualizing.

Homework Equations

To be in equilibrium:
$$\sum \vec{\tau} = 0$$
$$\sum \vec{F} = 0$$

Torque:
$$\vec{\tau} = r \cdot \vec{F_\perp}$$

Weight:
$$\vec{w} = m \cdot \vec{g}$$

The Attempt at a Solution

Since I know the Torque must be 0 to keep the beam from rotating, I get:

$$40 \cdot 9.8 + 20 \cdot 9.8 = 3 \cdot \vec{F}_\text{right beam}$$
3 is the distance from the pivot point (the left beam) to the right beam and F₂ is force of the right post.

F_net must also be 0, so I get the following:

$$\vec{F}_\text{left beam} + \vec{F}_\text{right beam} = 40 \cdot 9.8 + 20 \cdot 9.8$$

I'm stuck because I've got 2 functions and 3 unknown variables (F_{left beam}, F_{right beam}, x (distance from the pivot point to the boy)).

I assume there is a way to find these but I'm not sure where to go from here or if I've made a mistake.

 

[Doc Al]

Hints:
Where does the weight of the beam act?
When the beam is just about to tip, about which support is it pivoting? At that point, what's the force at the other support?

 

[joshncsu]

Thank you for the hints! Your 2nd hint is what led me to figure it out, your first hint made me realize that I messed up showing my work in Latex (sorry if that confused you).

Anyway, I got 4 m when solving for x, which looks right!

 

[Doc Al]

Excellent.

 

[Doc Al]

How do I figure out what the force on the other support is?

If the board is just about to tip, what must be the force on the other support?

(Realize that you are replying to a thread that is several years old.)