Hanging in balance

Consider the rod $AD$ . For this rod to be in equilibrium, the tensions in the two rods attached to it and the weight of the rod must have concurrent lines of action (so that the moment of the forces about $C$ is zero).

If the rod $AD$ has length $2L$ , then $AB = L\cos\beta$ , while $BC = 3L\sin\beta$ (there are three right-angled triangles that can be used to calculate $BC$ . Thus $\tan\alpha \; = \; \frac{BC}{AB} \; = \; 3\tan\beta.$ making the answer $\boxed{3}$ .

Here was my chain of reasoning:

1) The entire set of four rods has a weight of $4mg$ , which is evenly distributed between the two anchor points (see above).

2) The upward $2mg$ force on the first rod is balanced by the weight ( $mg$ ) at its midpoint, and by a downward force with magnitude $(mg)$ supplied by the second rod.

3) With respect to the anchor point, the torque (clockwise) on the first rod due to the vertical forces is equal to:

$\tau_{v1} = mg \frac{l_1}{2} \cos(\alpha) + mg l_1 \cos(\alpha) = \frac{3}{2} mg l_1 \cos(\alpha)$

4) This torque must be balanced by a rightward horizontal force supplied by the second rod:

$\frac{3}{2} mg l_1 \cos(\alpha) = F_x l_1 \sin(\alpha) \\ F_x = \frac{3}{2} mg \, \cot(\alpha)$

This force is balanced by a horizontal reaction force at the anchor point.

5) By Newton's Third Law, the forces supplied by the second rod to the first rod must be reciprocated (with opposite sign) by the first rod. See the diagram. It is therefore apparent that the vertical forces on the second rod are balanced.

6) In order for the horizontal forces on the second rod to balance, another horizontal force of value $F_x$ (rightward) must be supplied by the second rod's mirrored equivalent (not shown).

7) With respect to the contact point with the first rod, the torque (clockwise) on the second rod due to the vertical forces is equal to:

$\tau_{v2} = mg \frac{l_2}{2} \cos(\beta) - F_x l_2 \sin(\beta)$

8) Substituting in the known value of $F_x$ and applying the torque equilibrium condition for the second rod gives:

$mg \frac{l_2}{2} \cos(\beta) - \frac{3}{2} mg \, \cot(\alpha) l_2 \sin(\beta) = 0 \\ mg \frac{l_2}{2} \cos(\beta) = \frac{3}{2} mg \, \cot(\alpha) l_2 \sin(\beta) \\ \cos(\beta) = 3 \, \cot(\alpha) \sin(\beta) \\ \frac{1}{\tan(\beta)} = \frac{3}{\tan(\alpha)} \\ \frac{\tan(\alpha)}{\tan(\beta)} = \boxed{3}.$

Assuming 2L to be length of each rod, since horizontal distance between the 2 fixed ends to be constant,cosα +cosβ=constant. dβ/dα = -sin(α)/sin(β). To have minimum potential energy of the system the height of the centre of mass of the system should be minimum=H=Lsin(α)+(2Lsin(α)+Lsin(β)). Making dH=0 , dβ/dα = - 3cos(α)/cos(β) . Equating both we get tan(α)=3tan(β).

Let the rods have length $2\ell$ . The horizontal force components at the ends of the rods are are equal to $F$ in opposite directions, resulting in torques in the same direction relative to each rod's center of mass; i.e. the torque due to the horizontal force components on each rod is equal to $\tau_x = 2F\ell\sin\theta.\ \ \ \ \ \ \ (\theta = \alpha,\beta)$ The vertical force components on each are equal to the weight of the rods suspended from each point, i.e. $2mg$ for the top joints, $mg$ for the central joints, and zero for the bottom joint. For each joint these vertical components have opposite directions, resulting in torques in the same sense. Thus $\tau_{y,\alpha} = -(2mg + mg)\ell\cos\alpha;\ \ \tau_{y,\beta} = -(mg + 0)\ell\cos\beta.$ Equilibrium requires that for every rod the torques $\tau_x,\tau_y$ are balanced, so that $3mg\ell\cos\alpha = 2F\ell\sin\alpha; \ \ \ \ \ mg\ell\cos\beta = 2F\ell\sin\beta.$ Therefore $\frac{\tan\alpha}{\tan\beta} = \frac{\sin\alpha\cos\beta}{\cos\alpha\sin\beta} = \frac{3mg\cdot 2F}{mg\cdot 2F} = \boxed{3}.$