I want to horizontally stretch a 50 meters wire rope and slide 100 kg object from side to side. What should be the minimum tensile strength (or carrying capacity) of the rope to be able to hold the object while it is placed in the middle of the 50 meters? We can assume the rope's weight is 100 grams/meter.
I am sure there's some straightforward formula for calculate this just need some help figuring it out. :)
Actually to get an exact expression for a hanging wire with a lumped mass somewhere in the middle is extremely hard (maybe impossible). Another thing to consider is the allowable sag of the rope. If the rope can sag a lot then the tension can be low, whereas if low sag is required the tension has to be really high. In addition as the tension increases the rope stretches increasing the sag but relieving the tension. It is a rather complex problem overall,
I can provide an approximate expression when the lumped weight $W$ is located in the middle of the span $S$. Also important is the unit weight $w = \frac{m g}{\ell}$. I also consider the rope to be inflexible.
The tension is split into the horizontal tension component $H$ which is constant thought the rope, and the total tangential tension $T$ which increases the further away you go from the sag point. The total sag amount is $D$ and the relationships between $H$, $D$ are $T$ are:
$$ \begin{align} T & = H \cosh\left( \frac{w S}{2 H} \right) + \frac{W}{2} \sinh\left( \frac{w S}{2 H} \right) \\ D & = \frac{H}{w} \left( \cosh \left( \frac{w S}{2 H} \right) -1 \right) + \frac{W}{2 w} \sinh \left( \frac{w S}{2 H} \right) \end{align} $$
A further approximation of the above can be down when $ H \gg \frac{w S}{2}$
$$ \begin{align} T & = \frac{8 H^2+w^2 S^2+2 S W w}{8 H} \\ D & = \frac{ \frac{w S^2}{8} + \frac{S W}{4}}{H} \end{align} $$
Or by solving the last one for the horizontal tension $H$
$$ T = w D + \frac{S}{4 D} \left(W - \frac{w S}{2} \right) $$
In your case $S=50$, $w = 0.1\times9.81$, $W=100\times 9.81$, so for $D=3$ meter sag, the tension is $T=3988$ newtons for example.
Your approximate design expression is $$\require{cancel} T = \frac{11,955}{D} + \left(\cancel{ 0.981 D }\right)$$
Edit 1
Upon further examination it can be said that worst tension occurs when the weight is at one end. There you add the vertical tension and the weight and vectorially combine the horizontal tension for the worst tension
$$ T = \sqrt{ (V+W)^2 + H^2 } $$
The vertical tension at the end of a cable is
$$ V = H \sinh \left(\frac{w S}{2 H} \right) $$
The horizontal tension is found (numerically) from the measured sag (without the weight).
$$ D = \frac{H}{w} \left( \cosh \left( \frac{w S}{2 H} \right) -1 \right) $$
Example
You string the rope with $D=1$ meter sag. From the above equation you find the horizontal tension to be $H=306.7$ newtons. You can use Wolfram Alpha, Excel or any numeric solver you can find for this step.
$$ 1 = \frac{H}{0.1 \times 8.81} \left( \cosh \left(\frac{0.1\times 9.81 \times 50}{2 H}\right) -1 \right) $$
The vertical tension on one end is then
$$ V = 306.7 \sinh \left(\frac{0.1\times 9.81 \times 50}{2\times 306.7}\right) = 24.55 $$
With a weight of $W = 100 \times 9.81 = 981$ the total tension on the end is
$$ T = \sqrt{(24.55+981)^2 + 306.7^2} = 1051.3 $$
