In this challenge, the objective is to find the distance between 2 rods pulling elastics given that those elastics have equal heights. Now, because the elastics have different nominal lenghts and spring constants, they will only have equal lengths at a specific distance. This is because they must both be pulling with equal force. The force exerted by the two elastics is given by the first 2 equations below. They are equal, so we can solve them like a system of equations. We get that both are stretched to a length of 9 cm. The elastics are folded over twice, so their total length is thus only 4.5 centimetres. The combined force of the 2 elastics is 24 cmkg/s2. The weight resting upon the elastics will exert a downward force of 1 g * ≈9.807 m/s2 ≈ 9.807 mg/s2. In order for the elasitics to support it’s weight, the elastics must have an elasics must have this upwards force component. Let the angle below horizontal of the elastics be theta. θ ≈ asin(≈9.807/240), and the vertical distance between the rods is given by 4.5 * cos(asin(≈0.04086)). Using trigonometric identities, we simplify to the equation below. We get a distance of ≈8.992 centimetres.
\vec F_A=-{2kg\over s^2}(D-3cm)\\[8pt] \vec F_B=-{3kg\over s^2}(D-5cm)\\[16pt] \vec F_A=\vec F_B\\[8pt] {Dkg\over s^2}={9cmkg\over s^2}\\[8pt] D=9cm\\[32pt]F_A+F_B=24{cmkg\over s^2}\\[8pt] F_G\approx9.807{mg\over s^2}\\[16pt] x\approx4.5\cdot2\cdot\cos{\Bigg(\arcsin{\bigg({\approx9.807\over240}\bigg)}\Bigg)}\\[8pt] \approx\sqrt{1-\approx{0.04086}^2}\\[8pt] \approx8.992cm