Mass in Equilibrium with Variable Length Cable

[srg]

Homework Statement

Consider the diagram below. A 1000 kg block being suspended in equilibrium by cables c and a. Cable c is on a pulley which can vary the length of the cable. Write the constraints of equilibrium. Find the tension in each cable based on varying lengths of cable c between 1.2 m and 2.2 m.
http://srg.sdf.org/images/PF/MassVarLength.png

Homework Equations

Law of Cosines: c^2=a^2+b^2-2ab\cos{C}
Constraints of Equilibrium: \sum{F_x}=\sum{F_y}=0

The Attempt at a Solution

I drew a free body diagram:
http://srg.sdf.org/images/PF/VarLengthDiagram.png

I used the law of cosines to begin writing the constraints of equilibrium.
\sum{F_x}=0 \rightarrow T_c(\frac{b^2+c^2-a^2}{2bc})-T_b(\frac{a^2+b^2-c^2}{2ab})=0
\sum{F_y}=0 \rightarrow T_c\sin{\arccos{\frac{b^2+c^2-a^2}{2bc}}}+T_b\sin{\arccos{\frac{a^2+b^2-c^2}{2ab}}}-9810=0

Since everything is known except for T_c and Tb, they can be solved for. However, this system of equations doesn't seem easily solved for. I suspect I went wrong when I took the sin(arccos()), but I needed the sine of the angle when all I had was the law of cosines.

This course has been incredibly frustrating, and I'm only 5 weeks in. I'm beginning to feel I'm not really cut out for this.

 

[OldEngr63]

As I read the problem statement, it sounds like the length AB is fixed at 1.0 m. Assuming this is correct, the geometry will change as the length BC changes.

In your attempt at a solution, you seem to have used the same expression for the for the cosine of the angle on both sides. Redraw the figure, in an unsymmetric form, and define the angles on both sides separately. See if this does not get you started.