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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/7/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 7 / b / Solution by

Codex  0 2026-10-18

The Euler-Lagrange equations give

(m_{1} + m_{2} + m_{3}) \ddot{L}_{1} + m_{2} \ddot{L}_{2} + m_{3} \ddot{L}_{3} + k (L_{1} - L_{0}) m_{2} (\ddot{L}_{1} + \ddot{L}_{2}) + k (L_{2} - L_{0}) m_{3} (\ddot{L}_{1} + \ddot{L}_{3}) + k (L_{3} - L_{0}) = (m_{1} + m_{2} + m_{3}) g, = m_{2} g, = m_{3} g .

(1)

For example, differentiating

T

with respect to

\dot{L}_{1}

collects contributions from all three climbers, because changing

L_{1}

moves all of them. At equilibrium, all accelerations vanish, so

L_{1} = L_{0} + \frac{( m _{1} + m _{2} + m _{3} ) g}{k}, L_{2} = L_{0} + \frac{m _{2} g}{k}, L_{3} = L_{0} + \frac{m _{3} g}{k} .

(2)

The top rope supports the total weight, while each lower rope supports only its own climber. These extensions are positive, consistent with the taut-rope tension law.

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