# Effort supported by a rope-bridge ?

Aug 30, 2016

Given
$L \to \text{Length of the rope} = 25 m$
$M \to \text{Mass of the man climbing} = 72 k g$
$T \to \text{Tension the rope can withstand} = 2.9 k N$

Let the sagging be x m when the man is at the mid point of the rope and the rope then subtends angle $\theta$ with the horizontal.

The two vertical components of tension will balance the weight of the man.

So

$2 T \sin \theta = M \times g$

$\sin \theta = \frac{M g}{2 T} \ldots . . \left(1\right)$

For $\theta$ being small $\sin \theta \approx \tan \theta = \frac{x}{\frac{L}{2}} = \frac{2 x}{L}$

Now imposing approximation the relation (1) becomes

$\frac{2 x}{L} = \frac{M \times g}{2 T}$

$\implies x = \frac{M \times g \times L}{4 T}$

$\implies x = \frac{72 \times 9.8 \times 25}{4 \times 2900} \approx 1.52 m$

Without approximation the realation (1) can be written as

$\frac{x}{\sqrt{{x}^{2} + {\left(\frac{L}{2}\right)}^{2}}} = \frac{M \times g}{2 T}$

$\implies \frac{{x}^{2} + {\left(12.5\right)}^{2}}{x} ^ 2 = {\left(\frac{2 \times 2900}{72 \times 9.8}\right)}^{2} = 67.57$

$\implies 1 + \frac{156.25}{x} ^ 2 = 67.57$

$\implies \frac{156.25}{x} ^ 2 = 67.57 - 1 = 66.57$

$\implies {x}^{2} = \frac{156.25}{66.57}$

$\implies x = \sqrt{\frac{156.25}{66.57}} \approx 1.53 m$

Another way of calculation

From (1)

$\theta = {\sin}^{-} 1 \left(\frac{M \times g}{2 T}\right) = {\sin}^{-} 1 \left(\frac{72 \cdot 9.8}{2 \cdot 2900}\right)$
$\theta = {6.99}^{\circ}$

So
$\frac{2 x}{L} = \tan \theta = \tan 6.99$

$x = \frac{25 \times \tan 6.99}{2} = 1.53 m$