Angle of mass hanging from two points connected at two points I'm trying to calculate the angle (to the floor) a mass would hang if it were connected from either end to two points above it on the ceiling.
Let's call the distance between the points on the ceiling D1 and the distance between the points on the mass D2. The length of the first line L1 and the second L2. For the purpose of this example let's assume the center of gravity of the mass is directly in the middle and the lines are rigid.

My initial thought was to find the lowest place the center point of a line of size D2 could exist if its ends were coincident with circles of size L1 and L2 drawn from their respective positions on the ceiling (the ends of line D1). Then you could calculate the angle of that line relative to the ground if the positions of its endpoints were known.
Not really sure how to approach this issue. Any advice would be greatly appreciated. Thanks!
 A: The real problem here is to find a closed form to the sought angle. Solving the problem with the help of a symbolic processor, the closed form found is unmanageable. So we show here, a way to obtain a solution which can be closed but unmanageable.
Calling
$$
\cases{
R(\theta) = \left(
\begin{array}{cc}
 \cos (\theta ) & -\sin (\theta ) \\
 \sin (\theta ) & \cos (\theta ) \\
\end{array}
\right)\\
\hat e_x = (-1,0)\\
\hat e_y = (0,1)\\
A=(0,0)\\
B = A + l_1R(\theta_1)\cdot \hat e_x\\
C = B + d_2R(\cos(\theta_1+\theta_2)\cdot \hat e_x\\
D = C + l_2R(\cos(\theta_1+\theta_2+\theta_3)\cdot \hat e_x
}
$$
we have
$$
U = \frac 12 m g (B+C)\cdot \hat e_y 
$$
representing the mass potential energy. So we can formulate the lagrangian
$$
L = U+\mu_1((D-A)\cdot\hat e_x+d_1) +\mu_2(D-A)\cdot \hat e_y
$$
The minimum energy configuration can be found as a stationary point to  $L$. Those stationary points are the solutions for
$$
\nabla L = 
\left\{
\begin{array}{lcl}
 -\frac{1}{2} d_2 (g m+2 \mu_2) \cos (\theta_1+\theta_2)-\mu_1 (d_2 \sin(\theta_1+\theta_2)+l_1 \sin (\theta_1))-l_1 \cos (\theta_1) (g m+\mu_2)-l_2(\mu_1 \sin (\theta_1+\theta_2+\theta_3)+\mu_2 \cos (\theta_1+\theta_2+\theta_3)) & = & 0\\
 -\frac{1}{2} d_2 (g m+2 \mu_2) \cos (\theta_1+\theta_2)-\mu_1 (d_2 \sin(\theta_1+\theta_2)+l_2 \sin (\theta_1+\theta_2+\theta_3))-l_2\mu_2\cos(\theta_1+\theta_2+\theta_3) & = & 0 \\
 -l_2 (\mu_1 \sin (\theta_1+\theta_2+\theta_3)+\mu_2 \cos (\theta_1+\theta_2+\theta_3)) & = & 0\\
 d_1+d_2 \cos (\theta_1+\theta_2)+l_1 \cos (\theta_1)+l_2 \cos(\theta_1+\theta_2+\theta_3) & = & 0\\
 -d_2 \sin (\theta_1+\theta_2)-l_1 \sin (\theta_1)-l_2 \sin (\theta_1+\theta_2+\theta_3)& = & 0
\end{array}
\right.
$$
Solving for
$$
l_1 = 3,\ l_2 = 2,\ d_1 = 5,\ d_2 = 3,\ g = 10,\ m = 1
$$
we get
$$
\theta_1 = 2.04536,\ \theta_2 = 1.36987,\ \theta_3 = 0.917741,\ \mu_1 = 2.24493,\ \mu_2 = -5.6301
$$

A: It's bland trigonometry, and I should not do your homework/take-home for you, denying you the pleasure of devising a more elegant way of organizing the problem. Here is a hint on how the angle sought, θ, can be solved for. I drew the diagram so it exhibits the crucial features of the problem more naturally.
Call C the center (of mass) of the bar D₂, and consider the quadrilateral ABGE of fixed sides D₁, L₂, D₂, L₁, before applying any physical constraints.    The internal angles $\phi_1, \phi_2,\omega_2,\omega_1$ sum to 2π,
$$
\phi_1+\phi_2+\omega_1+ \omega_2= 2\pi, \tag{1}
$$
and you see the diagonal AG may be shrunk or expanded freely preserving the sides: it is the only degree of freedom of the problem, to be determined by "physics" (static equilibrium).   Work with angles.

Projecting the three sides AE, EG, and GB onto AB, you have
$$
D_1=L_1 \cos \phi_1+ L_2 \cos \phi_2+ D_2\cos\theta, \tag{2}
$$
and onto the vertical axis,
$$
L_1\sin\phi_1=L_2\sin\phi_2 +D_2\sin \theta . \tag{3}
$$
From the triangle AE[unlabelled lowermost left corner],
$$
\theta + \phi_1+ \omega_1=\pi, \tag{4}
$$
so that, from (1),
$$
-\theta + \phi_2+ \omega_2=\pi. \tag{5}
$$
So, (4,5,2,3) constrain all four angles of the quadrilateral to the hitherto free angle θ, to be determined by one additional static equilibrium condition ("physics").
The equilibrium condition requires that, as suggested by the faint pencil lines, for forces of strength $f_1, f_2$ along $L_1,L_2$, respectively: the horizontal forces the suspension strings apply to E and G cancel, (the vertical forces add to cancel the weight of the rod applied to the center of gravity C, and do not constrain the system); and, in addition, the torque around C vanishes, so the rod does not rotate, and equilibrates at θ. That is,
$$
f_1 \cos\phi_1 = f_2 \cos\phi_2, \qquad  f_1\sin\omega_1= f_2\sin\omega_2    \implies  
$$
$$
\cos \phi_1 \sin\omega_2=\cos \phi_2 \sin\omega_1.  \tag{6}
$$
This is to say that (2,3,4,5,6) suffice to determine θ.
There should be a tasteful arrangement to produce a solution.

*

*Your solution has an evident $L_2\to 0$ limit, as G and B merge to morph the quadrilateral into a triangle.

