http://www.smartphysics.com/images/c...angingbeam.png

What else could be done in order to be able to hold a heavier sign?
1.while still keeping it horizontal, attach the wire to the end of the beam
2.keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam
3.attach the sign on the beam closer to the wall
4.shorten the length of the wire attaching the box to the beam

I don't quite understand

I think it's safe to assume that the weak point in the whole contraption is the wire. Therefore, we're trying to find solutions that will result in less tension in the wire.

This whole structure is in equilibrium (meaning it's not accelerating in any way). Therefore, all forces and torques must add to zero. For purposes of this problem, the only thing we really care about is the torque created by the weight of the sign and the beam (trying to turn the beam counter-clockwise). We know that it is counteracted by the torque from the wire holding it up.

Since torque = force*perpendicular distance, we know that if we increase the perpendicular distance for the wire (while maintaining the same angle), that automatically reduces the force (i.e. tension) on the wire for the same amount of torque. Therefore, option 1 would indeed reduce tension in the wire and, therefore, allow for a heavier sign.

We also know that the force (tension) in the wire is directed along it's length. However, only the component of the force which is perpendicular to the beam contributes to the torque counteracting the sign. If the wire connects to the beam at a 90-degree angle, we know that ALL of the tension in the wire is directly counteracting the torque from the sign. However, if the wire connects at a shallow angle, only a small portion of the total force is directed perpendicular to the beam. Since that portion is the same in either case, the total tension in the wire must be enormous when it's at a shallow angle! Therefore, the closer to perpendicular you get, the less tension is in the wire and the heavier a sign the it can hold up. So option 2 is valid as well.

If you move the sign closer to the wall, you've reduced it's perpendicular distance to the axis of rotation. Therefore, you've reduced the torque applied by the sign, so the wire doesn't need to supply as much counteracting torque, which means the tension in the wire will be less. Therefore option 3 is valid also.

Finally, changing the length of the wire from which the sign hangs has no effect on the perpendicular distance to the wall (the axis of rotation). Therefore, option 4 would NOT change the amount of torque the wire has to counteract, and therefore does NOT help to increase the weight capacity of the setup.

Does that make more sense now?