Equilibrium and Torque Problems...PLEASE HELP!?

Question

1. Three boys are trying to balance on a seesaw which consists of a fulcrum rock acting as a pivot at the center and a light board 3.6m long. Two boys are already on either end. One has a mass of 50kg, and the other a mass of 35kg. Where should the third boy, whose mass is 25 kg, place himself so as to balance the seesaw?

2. Find the tension in two wires supporting a 39kg traffic light with angles 51 and 39 degrees (measured from the ceiling to the wire).

3. A meter stick with a mass of 235 g is supported horizontally by two vertical strings one at the 0cm mark and the other at the 90cm mark. What is the tension in both strings?

4. Three forces are applied to a tree sapling to stabilize it. The the angle between F1 and F2 is 110 degrees. If F1=282N and F2=355N, find F3 in magnitude and degrees. In other words, find the degrees between F1 and F3 for the degrees part of the answer.

Answer 1

Equilibrium just means that the sum of the forces and the sum of the torques equal zero.

Just as a refresher:
torque = r x F = r*F*sin(x)
force = m * a

Number one is a torque problem. If you draw a diagram, placing the 50 kg boy on the left side (1.8 m from center, therefore r = 1.8), and the 35 kg boy on the right side (1.8 m from center, therefore r = 1.8), you should recognize that they will each be torquing the seesaw in different directions.

By inspection, with the 50 kg boy on the left side, the seesaw should rotate counterclockwise. Thus, the third boy should stand on the right hand side to counter the net torque.

If the seesaw is balanced, then the boys are standing perpendicularly to the board and so torque = r*F (notice the sine function is eliminated since x = 90 degrees).

Now just sum the forces and solve:
50*1.8 = 35*1.8 + 25*r
r = 1.08 m towards the 35 kg boy

(Notice I put the 50 kg boy on the left hand side of the equation and the other two on the right hand side of the equation. You could also put them all on the same side (equal to zero) by adding a negative sign to denote the difference in directions of torque.)

I think the rest are just force problems, although I admit I didn't look over them too carefully. Still, I hope this is enough to get you started on torque equilibrium problems.