21 Masses and 20 Springs

Classical Mechanics Level 5

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 1 meter and a spring constant of 20 N/m 20 \, \text{N/m} . Gravity is 10 m/s 2 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 t = 0 , the masses have the following properties (all quantities in SI units): subscript : k = 0 ∼ 20 Mass : m k = 1 xpos : x k = k ypos : y k = 0 xvel : x ˙ k = 0 yvel : y ˙ k = 0. \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 y = 0 is the lowest-hanging mass?

Note: Drawing is not necessarily to scale.

Hint: 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.

Steven Chase by Steven Chase , 37, USA

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1 solution

G Silb
Nov 14, 2018

Start at the center weight. Draw a freebody diagram, and look at the springs moving left (or right). With θ 1 \theta_1 measured from the horizontal,

F 1 sin ⁡ θ 1 = m g 2 F_1 \sin \theta_1 = \frac{m g}{2} where m is the mass. Moving left to the second mass,

F 1 cos ⁡ θ 1 = F 2 cos ⁡ θ 2 F_1 \cos \theta_1 = F_2 \cos \theta_2 and

F 2 sin ⁡ θ 2 = 3 m g 2 F_2 \sin \theta_2 = \frac{3 m g}{2} . Then

cot ⁡ θ 2 = F 1 cos ⁡ θ 1 3 m g 2 \cot \theta_2 = \frac{F_1 \cos \theta_1}{\frac{3 m g}{2}} so that

sin ⁡ θ 2 = 3 m g 2 F 1 2 cos ⁡ 2 θ 1 + ( 3 m g 2 ) 2 = 3 m g 2 F 1 2 ( 1 − sin ⁡ 2 θ 1 ) + ( 3 m g 2 ) 2 = 3 m g 2 F 1 2 + 2 ( m g ) 2 \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 = F 1 2 + 2 ( m g ) 2 F_2 = \sqrt{F_1^2 + 2 (m g) ^2 }

Continuing in this fashion we see that

F i sin ⁡ θ i = ( 2 i − 1 ) m g 2 F_i \sin \theta_i = \frac{(2 i - 1) m g}{2}

F i = F i − 1 2 + 2 ( i − 1 ) ( m g ) 2 = F 1 2 + i ( 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 ⁡ θ i = F i − 1 cos ⁡ θ i − 1 F_i \cos \theta_i = F_{i-1} \cos \theta_{i-1}

l i = F i k + l 0 l_i = \frac{F_i}{k} + l_0 where k is the spring constant and l 0 l_0 is the unstretched length.

Now choose θ 1 \theta_1 . Compute F 1 F_1 . Compute remaining F i F_i 's, then θ i \theta_i 's, then l i l_i 's.

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

When ∑ i = 1 10 l i cos ⁡ θ i = 10 \sum _{ i =1}^{10 }{l_i \cos \theta_i }=10 , ∑ i = 1 10 l i sin ⁡ θ i = 33.826 \sum _{ i =1}^{10 }{l_i \sin \theta_i }=33.826

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