Torque of a square question?

Question

A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, 1 and 2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of 2 is 3 times that of 1. Find the distances a and b that locate the axis. Note that a and b are not drawn to scale.

I'm not sure where to begin with this question. Can someone give me some tips? I'd greatly appreciate it.

Thanks.

Here is an image

http://www.webassign.net/CJ/p9-10.gif

Answer 1

Evaluate the individual torques T. Since they are oppositely directed they must be equal.
In A, T1 = F1*b, T2 = F2*a = 3F1*a
Then b/a = 3. T\At the moment that's all one can say about a and b. The locus of points satisfying b/a=3 is a straight line from the top edge, 1/3 m from the right, to the bottom right corner.
In B, T1 = F1*(1-a) (where 1 is the side of the square), and T2 = F2*b = 3F1*b
Then (1-a)/b = 3. The locus of points satisfying (1-a)/b=3 is a straight line from the right edge, 1/3 m from the bottom, to the bottom left corner.
Now we have two intersecting lines, and the intersection point satisfies both equations. So the locations can be found graphically or algebraically. The algebraic solution is
b/a = 3
(1-a)/b = 3 = 1/b - a/b = 1/b - 1/3
1/b = 3.333
b = 0.3
a = 0.1
Checking,
(1-a)/b = 0.9/0.3 = 3 (checks OK)