Consider a system consisting of 21 identical point masses and 20 identical springs. They are arranged in a sequential chain as shown in the diagram below. The springs have a natural length of $1$ meter and a spring constant of $20 \, \text{N/m}$ . Gravity is $10 \, \text{m/s}^2$ downward. The first and last masses in the chain are fixed in place and cannot move.

At time $t = 0$ , the masses have the following properties (all quantities in SI units): $\begin{aligned} \text{subscript}:\ &k = 0\sim 20 \\ \text{Mass}:\ &m_k = 1 \\ \text{xpos}:\ &x_k = k \\ \text{ypos}:\ &y_k = 0 \\ \text{xvel}:\ &\dot{x}_k = 0 \\ \text{yvel}:\ &\dot{y}_k = 0. \end{aligned}$ The system moves over time until it eventually reaches stasis due to the effects of air resistance. When the system reaches stasis, how far below $y = 0$ is the lowest-hanging mass?

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Note:
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Drawing is not necessarily to scale.

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Hint:
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If the prospect of formally solving seems unpleasant, you can instead run a time-domain simulation and see what happens.

The answer is 33.826.

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Start at the center weight. Draw a freebody diagram, and look at the springs moving left (or right). With $\theta_1$ measured from the horizontal,

$F_1 \sin \theta_1 = \frac{m g}{2}$ where

mis the mass. Moving left to the second mass,$F_1 \cos \theta_1 = F_2 \cos \theta_2$ and

$F_2 \sin \theta_2 = \frac{3 m g}{2}$ . Then

$\cot \theta_2 = \frac{F_1 \cos \theta_1}{\frac{3 m g}{2}}$ so that

$\sin \theta_2 = \frac{\frac{3 m g}{2}}{\sqrt{F_1^2 \cos^2 \theta_1 + \left( \frac{3 m g}{2} \right) ^2 }} = \frac{\frac{3 m g}{2}}{\sqrt{F_1^2 (1-\sin^2 \theta_1) +\left( \frac{3 m g}{2} \right)^2 }} = \frac{\frac{3 m g}{2}}{\sqrt{F_1^2 + 2 (m g) ^2 }}$ and

$F_2 = \sqrt{F_1^2 + 2 (m g) ^2 }$

Continuing in this fashion we see that

$F_i \sin \theta_i = \frac{(2 i - 1) m g}{2}$

$F_i= \sqrt{F_{i-1}^2 + 2 (i-1) (m g) ^2 } = \sqrt{F_1^2 + i (i-1) (m g) ^2 }$

$F_i \cos \theta_i = F_{i-1} \cos \theta_{i-1}$

$l_i = \frac{F_i}{k} + l_0$ where

kis the spring constant and $l_0$ is the unstretched length.Now choose $\theta_1$ . Compute $F_1$ . Compute remaining $F_i$ 's, then $\theta_i$ 's, then $l_i$ 's.

Check the constraint $\sum _{ i =1}^{10 }{l_i \cos \theta_i }=10$ and iterate the choice of $\theta_1$ accordingly.

When $\sum _{ i =1}^{10 }{l_i \cos \theta_i }=10$ , $\sum _{ i =1}^{10 }{l_i \sin \theta_i }=33.826$