We need to show that the tension in the rope, T, is = to 2Mgcos(theta)
I've taken Moments about A but keep getting T = (1/2)Mgcos(theta).
I've ignored the moments at C as they are in equilibrium already I'm assuming. Meaning (1.5a)Rc = (1.5a)2Mgcos(theta). Including them in my calculation doesn't make a difference.
Any help would be appreciated.
If you drop a perpendicular from any point on $AB$ to the horizontal line, you will see that the angle between that line and a line perpendicular to $AB$ is $\theta$ because both are complementary to the same angle.
The weight of the beam operates at the center of mass of the beam which is $a$ units from the origin. This gives us that the tension of the rope has to counter-balance the weight of the beam acting at $a$ and the weight of the object at $1.5a$. This gives:
$$2aT=1.5a(2M)g\cos{\theta}+Mga\cos{\theta}\implies 2aT=4Mg\cos{\theta}\implies T=2Mg\cos{\theta}$$