Slacklining

Initially the rope is streched to the lenght $d$ due to the tensioning force $F_0$ , so that Hooke's law reads $\begin{aligned} F_0 &= K \frac{d - l_0}{l_0} \\ \Rightarrow \quad l_0 &= \frac{K}{F_0 + K} d = \frac{150}{155} \cdot 10 \,\text{m} = \frac{300}{31} \,\text{m} \approx 9.68 \,\text{m} \end{aligned}$ For the loaded slackline we can draw the following diagram:

With the man as additional load, the forces $\vec F_1$ and $\vec F_2$ along the rope must make up for weight force $\vec F_g$ , so that $\begin{aligned} \vec F_1 + \vec F_2 &= - \vec F_g \\ \Rightarrow \quad |\vec F_1|= |\vec F_2| &= F = \frac{F_g}{2 \sin \alpha} \end{aligned}$ with the deflection angle $\alpha$ . According to the diagram, the lenght of the rope results $\begin{aligned} l &= \frac{d}{\cos \alpha} \\ \Rightarrow F &= K \frac{l - l_0}{l_0} = 150,000 \,\text{N} \cdot \left(\frac{31}{30 \cos \alpha} - 1 \right) = \frac{750 \, \text{N}}{2 \sin \alpha} \end{aligned}$ This equation can be solved graphically and results an angle $\alpha = 0.07 \,\text{rad} = 4^\circ$ and a deflection $h = \frac{d}{2} \tan \alpha = 35\,\text{cm}$ An approximate solution can be obtained if we use the linear approximations $\cos \alpha \approx 1$ and $\sin \alpha \approx \alpha$ for small angles. $\begin{aligned} \frac{F}{K} &= \left(\frac{31}{30 \cos \alpha} - 1 \right) \approx \frac{1}{30} \\ &= \frac{1}{400 \sin \alpha} \approx \frac{1}{400 \alpha}\\ \Rightarrow \quad \alpha &\approx \frac{3}{40} \,\text{rad} = 0.075 \,\text{rad} = 4.3^\circ \\ \Rightarrow \quad h &= \frac{d}{2} \tan \alpha \approx 38 \,\text{cm} \end{aligned}$

I have used similar method as Markus one, but I use desmos to approximate it.

Let's take a look at right half, after the man standing on the rope, it form a right triangle which 90 degree is at upper-left corner. we have horizontal length $5$ , and let vertical length $x$ , hypotenuse $y$ .

First, we find the original length of the rope, I mean the right half ones, states as $l_0$ , You can easily find that as $l_0 = \frac{150}{31}$ , I didn't use approximation to this fraction, to retain accuracy of the result.

At the moment the man stand, let the force act on rope slightly changes from $F$ to $F'$ , so

$F' = K \frac{y-l_0}{l_0}$

Compute all of the constant will get

$F' = 31000y - 150000$

Then, it is necessary to know that the vertical component of $F'$ will equal to half of the man's weight, so it obtain:

$F' \frac{x}{y} = \frac{1}{2} F_g = 375$

Rearrange all of the above and finally add on simple Pythagoras Equation,

$y^2 = x^2 + 25 ------(1)$ $(31000y-150000)x=375y -------(2)$

Look at $x$ coordinate, we get $\boxed{0.349m}$ .

First, let's determine the total unstretched length of the rope

$F = K (l/l_0 - 1)$

$5000 = 150 \times 10^3 (10/l_0 - 1)$

$l_0 = 10 / (1 + 1/30) = 300 / 31$

Now, after applying the weight of the man on the rope, balance of forces dictates that

$F_g = 2 T \sin \theta$

i.e. $750 = 2 T \sin \theta$

where $\theta$ is the angle that the rope makes with the horizontal.

In addition, T and the new length of the rope obey Hooke's law,

The stretched length of half rope is $5 / \cos \theta$ , and the unstretched half rope is 150/31, hence

$T = 150 \times 10^3 ( \dfrac{(5 / \cos \theta )}{ (150 / 31 )} - 1 )$

Substituting this above in the first equation

$750 = 2 \times 150 \times 10^3 ( \dfrac{(5 / \cos \theta )}{ (150 / 31 ) } - 1 ) \sin \theta$

Simplifying

$15 \times 150 \cos \theta = 6 \times 10^3 (155 - 150 \cos \theta) \sin \theta$

simplifying further

$3 \cos \theta =40 ( 31 - 30 \cos \theta ) \sin \theta$

solving the above equation numerically yields $\theta = 0.069783 \text{ rad}$ .

Now $h = 5 \tan \theta = 0.349482 \text{ m } = 34.9 \text{ cm } \approx 35 \text{ cm }$

If we break the rope into 2 symmetrical triangles.... and assume that they are each supporting 1/2 of the human.

F=K * (l-lo) * (1/lo)

F = K* strain (% deformation)

F total = Ftension (Fo) + Fhuman (Fg)

Fo is constant @ 5000N - assume this is completely in x plane.

Fg applied to one side of rope = 1/2 Fg in y plane.

Ftotal = sqrt( (1/2*Fg)^2 + (Fo)^2 )

Ftotal = sqrt( 140625 + 25000000)

Ftotal = 5014.04278

Ftotal = K *strain

Ftotal/K = strain

5014.04278/150 = strain

0.033426952 = strain

strain = % deformatioon.

0.033426952*10m = 0.334269519m

Not exactly 35, and it makes a few assumptions, but gets pretty close.