To solve for the force of tension in both rope 1 and 2, both forces have to broken down into their x and y components and then solve the resulting system of equations.

• The system is stationary
• Gravity works in the negative y direction, with $g=9.81m/s^{2}$
• There are two forces of tension, on for rope 1, and one for rope 2
• Mass of the object?
• Angles of the ropes?

• Either force of tension of the ropes

• The lengths of the two ropes is irrelevant, only the angle matters to solve for the two forces
• The net force is zero since the system is stationary

• I have a drawing but I'm having a hard time adding pictures

We assume the net force is zero, so $$F_{net} = \sum F = \sum F_{x} + \sum F_{y} = 0.$$ To find the forces acting the the x direction, we have to decompose the tension of both ropes. For rope 1, $$T_{1,x} = T_{1}\sin{\alpha}$$ For rope 2, $$T_{2,x} = T_{2}\sin{\beta}.$$ One important note is since the force of tension of the two ropes point in opposite directions, their signs must be different. As convention, the force pointing to the left will be negative in these calculations. Now we know the net force in the x direction $$\sum F_{x} = T_{2}\sin{\beta} - T_{1}\sin{\alpha}=0.$$ Next, we will use the same technique to decompose the forces in the y direction. For rope one, $$T_{1,y} = T_{1}\cos{\alpha}$$ For rope 2, $$T_{2,y} = T_{2}\cos{\beta}.$$ When finding the net force in the y direction, we cannot forget our assumption that gravity also works in the dimension in the opposite direction as our tension forces. The net force in the y direction is, $$\sum F_{y} = T_{1}\cos{\alpha}+T_{2}\cos{\beta}-Mg = 0.$$