Equilibrium equations

You have one horizontal beam. Its left point is attached to a free pivot at the vertical wall. Its right point is suspended by a rope from the ceiling. Nothing is moved. So $\sum F=0$ and $\sum M=0$. To analyze this, we write $F_a-F_b=0$, where $F_a$ is the force from the rope and $F_b$ is the force of the weight of the beam. At the same time, $Μ_a-M_b=0$, where $M_a$ is the moment from $F_a$ and $M_b$ is the moment from $M_b$. Am I right?

Now we have the same system, but the beam's right point is dropped a bit, so it's not horizontal, but it has an angle. What are the equilibrium equations? Are they exactly the same?

• There will be an extra term in the reaction from the pivot. Just resolve all the forces along your choice of axes, and include the term for hinge reaction. – GRrocks Nov 8 '15 at 12:00
• can you tell me the equations please? I don't get what you mean. What will be the difference exactly? – ergon Nov 8 '15 at 23:46
• Homework and 'check-my-work' type questions are off-topic for Physics.SE. Please see this post about how to ask homework problems. Additionally, responding to your own post in that way is not a constructive way to convince people to help you. – DilithiumMatrix Nov 10 '15 at 0:13
• The modification you describe doesn't change the underlying principles at all. The only difference in the calculation is that you should describe the equilibrium in both the x and y directions, instead of just in the y-direction like you describe initially. – DilithiumMatrix Nov 10 '15 at 0:14
• did I say it changes the underlying principles? I said it MAY change the equations. Can you tell me please the exact equations? Consider x,y as vertical and horizontal axes. thanks! – ergon Nov 10 '15 at 20:43

The Equations for the $\sum F=0$ will be the same as the forces will maintain the same orientation. But the equations of the $\sum M=0$ will not be the same as the forces will be at an angle to the point of reference. You may have to use a little bit of sines and cosines to find the perpendicular distances to the forces from the point of reference.
About the force balance: A force is missing. $F_a$ is the tension in the rope, $F_b$ is the weight, but what about the hinge at the wall? This point also helps to hold up the beam, so your force balance must include this: $$\sum F=F_a+F_b+F_c=0$$