A chain of length L has a total mass M and is suspended between two points a distance d apart. Determine the maximum tension and the sag in the chain. Given: L = 40 m M = 100 kg d = 10 m g = 9.81 m /s^2

Question

A chain of length L has a total mass [latex]M[/latex] and is suspended between two points a distance [latex]d[/latex] apart.Determine the maximum tension and the sag in the chain.

Given:

[latex]L[/latex] = [latex]40[/latex] [latex]m\quad \quad M[/latex] = [latex]100[/latex] [latex]kg\quad \quad d[/latex] = [latex]10[/latex] [latex]m\quad \quad g[/latex] = [latex]9.81[/latex] [latex]\frac { m }{ { s }^{ 2 } }[/latex]

Accepted Answer

[latex]w=M\frac { g }{ L }[/latex]

[latex]s=\left( \frac { { F }_{ H } }{ w } ight) \sinh { \left( \frac { w }{ { F }_{ H } } x ight) } \quad \quad y=\left( \frac { { F }_{ H } }{ w } ight) \left( \cosh { \left( \frac { w }{ { F }_{ H } } x ight) } -1 ight) \quad \quad \frac { d }{ dx } y=\sinh { \left( \frac { w }{ { F }_{ H } } x ight) }[/latex]

Guesses [latex]\quad { F }_{ H }=10N\quad \quad\quad\quad\quad\quad\quad \quad h=10M[/latex]

Given [latex]\quad \frac { L }{ 2 } =\left( \frac { { F }_{ H } }{ w } ight) \sinh { \left( \frac { w }{ { F }_{ H } } \frac { d }{ 2 } ight) } \quad\quad \quad h=\frac { { F }_{ H } }{ w } \left[ \cosh { \left( \frac { w }{ { F }_{ H } } \frac { d }{ 2 } ight) } -1 ight] \quad \quad\quad \left( \begin{matrix} { F }_{ H } \ h \end{matrix} ight) =Find\left( { F }_{ H },h ight) [/latex]

[latex]\quad\quad\quad{ heta }_{ max }=a an { \left[ \sinh { \left( \frac { w }{ { F }_{ H } } \frac { d }{ 2 } ight) } ight] } \quad \quad \quad \quad { T }_{ max }=\frac { { F }_{ H } }{ \cos { \left( { heta }_{ max } ight) } }[/latex]

[latex]\quad \quad { F }_{ H }=37.57N\quad \quad \quad \quad h=18.53m\quad \quad \quad { T }_{ max }=492N[/latex]

Question : A chain of length L has a total mass M and is suspended betw...

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