This is part of a homework question which I've been stuck on for several hours. I've tried googling "lowest point of rope", "lowest point of hanging cable", "lowest point of pulley", and a bunch of other combinations without luck.

> 
> Cable ABC has a length of 5m. The cable is attached to a wall on the
>  left at A, and attached to a wall on the right at C, 0.75m above the
>  vertical position of A (so C is attached at a higher location on the
>  wall). The distance between the walls is 3.5m. A 100kg sack is hanging
>  by a pulley on this cable at equilibrium, at B. Find the horizontal
>  distance x of the pulley from the left wall (neglect the size of the
>  pulley).
> 
> 
> 

Intuitively I think the the pulley would hang at the location where it's closest to the ground (hence "lowest point of loose cable..."). However, I have no idea how to calculate it. We're only allowed scientific calculators (no graphing) and whenever I try to set up an equation it blows up.

Once I figure out x it should be relatively easy to calculate the component forces for equilibrium.

I tried looking for examples in the textbook and internet for something like this without luck.

```
| ---- 3.5m -----|
---D-------------* <- C
|  |            /|
| <- 0.75m   /   |
* <- A    /      |
|\ |   /         F
| \|/
|  * <- B
E  |
  _______
 | 100kg |
|--| <- x
Length of cable: 5m

```

Here's a text diagram, as best as I could make it

**Update**: Found a hint from the textbook - `(3.5 - x)/cos(o) + x/cos(o) = 5`. Not quite sure what to make of it, but it does kinda remind me of an ellipse at a slant... <https://math.stackexchange.com/questions/108270/what-is-the-equation-of-an-ellipse-that-is-not-aligned-with-the-axis>

**Update 2**: Upon closer inspection of aufkag's angle-suggestion and the hint from the textbook, I believe he is correct about the angles being equal - the formula calculates the two segments of the rope from the adjacent sides `x` and `3.5 - x`. By the way, how can it be explained or "proved" or what's the law that says the angles between AB and the wall and AC and the wall in a setup like this are equal?

**Update 3**: (after solved, see comment for aufkag): Added D, E, F. ABD = BAE and CBD = BCF, but can anyone prove or point out the law that says ABD = CBD or BAE = BCF?

Anyways, the steps are:

```
o = angle AB and the horizontal or BC and the horizontal
x / cos(o) + (3.5 - x) / cos(o) = 5 (sum of segments of rope is 5)
tension in AB = tension in BC, therefore they share the same "load" of the mass, so we can calculate the tension in just one side
100 * 9.81 / 2 / sin(o) = 687N (approximately - first half of answer)
0.75 + xtan(o) = (3.5 - x)tan(o) (equal lengths for line segment BD)
solve for x to get 1.38m

```