Final position of a hanging beam I have a problem where I want to find the final position of a hanging beam supported on its two ends by ropes. The ropes are of different length, so the beam makes an angle with the horizontal plane, and the angle of the ropes relative to the vertical plane are different.
The only information I have is:

*

*The weight (48 kg) and location of the center of gravity of the beam

*The lengths of the beam, ropes, and distances between ropes

I want to find the final angles $\alpha, \theta, \phi$. I know that logically this problem will have some form of iteration as we need to give the beam an initial condition and then calculate its final position, but I've been getting caught up in the calculations and I've been unable to get a final answer. Could anyone help?

 A: This is a four-bar linkage, and it has 1DOF. We can use the angle $\alpha$ as the driving DOF. All other angles can be derived from this one.
Here is a re-worked diagram with point labels and all information needed to defined the kinematics of the system

Write two expressions constraining the top horizontal span in place.
$$ \begin{aligned} S_{{\rm DH}}&=\ell_{{\rm AD}}\sin\varphi+\ell_{{\rm AB}}\cos\alpha+\ell_{{\rm BH}}\sin\theta\\0&=-\ell_{{\rm AD}}\cos\varphi+\ell_{{\rm AB}}\sin\alpha+\ell_{{\rm BH}}\cos\theta \end{aligned}  \tag{1,2}$$
Next write 6 expressions for the positions of A, B and the CG, point C, in terms of the kinematics
$$ \begin{aligned}
x_{A}&=\ell_{{\rm AD}}\sin\varphi\\y_{A}&=-\ell_{{\rm AD}}\cos\varphi\\x_{B}&=S_{{\rm DH}}-\ell_{{\rm BH}}\sin\theta\\y_{B}&=-\ell_{{\rm BH}}\cos\theta\\x_{C}&=\ell_{{\rm AD}}\sin\varphi+\tfrac{1}{2}\ell_{{\rm AB}}\cos\alpha-h\sin\alpha\\y_{C}&=-\ell_{{\rm AD}}\cos\varphi+\tfrac{1}{2}\ell_{{\rm AB}}\sin\alpha+h\cos\alpha \end{aligned} $$

Finally write 3 expressions for the force/torque balance of the system
$$ \begin{aligned} 0&=-T_{{\rm AD}}\sin\varphi+T_{{\rm BH}}\sin\theta\\W&=T_{{\rm AD}}\cos\varphi+T_{{\rm BH}}\cos\theta\\0&=-x_{C}W+x_{B}T_{{\rm BH}}\cos\theta-y_{B}T_{{\rm BH}}\sin\theta \end{aligned} \tag{3,4,5}$$
The above 5 equations 1,2,3,4,5 are to be used for the solution of the following 5 unknowns. Two tension values $T_{\rm AD}$ and $T_{\rm BH}$, and three angles $\alpha$, $\varphi$ and $\theta$.
Of course this is highly non-liner problem with a rather complicated analytical solution. Careful elimination of variables in a step by step process can produce a solution.
In these problems sometimes trig functions are transformed into rational polynomials using the tan-half-angle substitution $t = \tan\left( \tfrac{\theta}{2} \right)$ which leads to $$ \small \begin{aligned} \sin \theta & = \frac{2 t}{1+t^2} \\ \cos \theta & = \frac{1-t^2}{1+t^2} \end{aligned}$$

It just dawned to me that there is a geometric solution.

The torque balance means that the point G where the two support ropes extensions meet, has to be on the line of action of gravity. This is also the instantaneous pivot point for the bar.
This is best solved using a CAD package where the lengths of segments can be fixed and constraints can be used to create the construction above. With some skill the solution can be found with GeoGebra also.

I decided to look a bit closer to this using Linkage

I tracked the CG position as the bar swayed back and forth and found indeed that the above suggested solution would be stable since the CG path is at a minimum height.
Although the CG is above the pivot, hinting at instability, when the bar sways, point G moves side to side more than the bar, controlling the inverted pendulum. This is akin to balancing a standing broom on your hand, and moving the hand ahead of the broom swaying, in order to catch its fall.
Really interesting situation here.
