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
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