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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
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closed as off-topic by Emilio Pisanty, Abhimanyu Pallavi Sudhir, Waffle's Crazy Peanut, Manishearth Sep 29 '13 at 7:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

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  • $\begingroup$ Perhaps try something like this. Define $x_A, y_A, x_B, y_B, x_C, y_C$ as the coordinates of the three points (measured from the top-left, $x$ going right and $y$ going down). You are looking for $x_B$. Use Pythagoras to compute the distance $D_{AB}$ between $A$ and $B$ and the distance $D_{BC}$ between $B$ and $C$. Set $D_{AB}+D_{BC}=5$. Write this equation out and fill in all the numbers you already know. Then maximise $y_B$ subject to this equation. Compute the corresponding $x_B$. $\endgroup$ – Keep these mind Sep 24 '13 at 20:11
  • $\begingroup$ @aufkag cool, I'll try that. This class (right now anyways) should not require calculus though, so it'd be great if someone could still figure out how to do it without it $\endgroup$ – Raekye Sep 24 '13 at 20:17
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    $\begingroup$ Without calculus? OK, you'll probably need physics then. My guess is that the angles of $AB$ and $BC$ are necessarily equal. Use that. Something like $(x_B-x_A)/(y_B-y_A)=-(x_C-x_B)/(y_C-y_B)$. $\endgroup$ – Keep these mind Sep 24 '13 at 20:29
  • $\begingroup$ @aufkag you mean the angles between AB and the wall, and BC and the wall? Hmmm, I'll try it out $\endgroup$ – Raekye Sep 24 '13 at 20:55
  • $\begingroup$ @aufkag alright I solved it using your tip that the angles must be equal. If you want to post an answer (preferably explaining why the angles must be equal - I see how the angles between a vertical line down B and AB/BC might be equal intuitively, but can't explain it) I'll mark it as accepted $\endgroup$ – Raekye Sep 24 '13 at 21:22
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You need to include the dynamical information about the pulley and the weight in your description, or you can never hope to solve this.

You should construct the free-body diagram for the pulley, with both sides of the rope plus the weight pulling on it. Since you know that

  • the tension on both sides is the same as they're the same rope, and
  • the horizontal components of the tension must cancel out so the pulley doesn't move to the side,

you should be able to derive that the angles of each side of the rope from the vertical are the same. This is your second independent geometrical constraint, which coupled with your ellipse gives a unique solution. (Note, though, that your textbook shows a way to pose this as a linear equation: use the angles to translate horizontal distances into diagonal rope lengths, and then add those up.)

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  • $\begingroup$ What is "dynamical information" about the pulley and weight? I did say that 1. the mass is 100kg (weight follows from it) and 2. neglect the size of the pulley (i.e. negligible size). I double checked (since I've solved it now) - and all the information I used is in the description. Other than that, the rest of your answer seems sound (I don't understand all of it, ellipses/math/physics aren't my forte :P ), but I'll give aufkag a chance to answer since he was the first to give the right tip (that the angles are the same) $\endgroup$ – Raekye Sep 24 '13 at 21:34
  • $\begingroup$ Talking about ellipses means dealing with the constant length of the rope, but you also need to express in equations that you have a pulley and that it's being forced by an external weight. The ellipse is the 'kinematics' (how things can move) but you need information about the forces present and the response of your system to them (i.e. the dynamics). $\endgroup$ – Emilio Pisanty Sep 24 '13 at 21:40
  • $\begingroup$ "A 100kg sack is hanging by a pulley on this cable at equilibrium, at B" <- doesn't that explain the dynamics? Just wondering if I'm missing anything $\endgroup$ – Raekye Sep 24 '13 at 23:43
  • $\begingroup$ You are missing it. It states the dynamics, but it does not attempt to implement them as a condition on the geometry. You had one equation but you were missing one constraint, which is right: if the force had a horizontal component (i.e. you were pulling it sideways) the pulley would move along the ellipse. You need a geometrical equation that implements the fact that the weight is a vertical force. $\endgroup$ – Emilio Pisanty Sep 24 '13 at 23:55
  • $\begingroup$ I checked my textbook and all the information it gave was already here. Nowhere does it imply that someone is pulling it or anything like that, and the others know what I/my textbook is saying. Anyways, aufkag was the most to the point with the final solution, but didn't post an answer. I can't select yours because while it mentions the information used to solve the question, I'd find it misleading if I were new to this problem (e.g. I still haven't found an elegant way to set up this ellipse because it's on a slant) $\endgroup$ – Raekye Sep 27 '13 at 11:52
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The following approach might be useful. You know that AB+BC=5m, that means that B lies on the ellipse with foci A and C and the sum of the distances to the foci for each point of the ellipse equal to 5m. You can write the equation of the ellipse and calculate the position of the horizontal straight line tangent to the ellipse. I suspect that you'll just need to solve a quadratic equation.

EDIT (9/24/2013): You can use the property of ellipse: in geometrical optics, the light propagating from one focus of the ellipse passes through the other focus after reflection at the boundary of the ellipse (see, e.g., http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Optical-property-of-an-ellipse.lesson ). That means that a straight line tangential to the ellipse at a certain point forms equal angles with the straight lines connecting the point with the foci of the ellipse - this was pointed out by aufkag.

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  • $\begingroup$ Okay I've been sorta getting somewhere (believe the a and b of the ellipse are 2.5 and 1.7455 if it were horizontal)... but stuck with the slant. I found a hint in the textbook, maybe you can help make sense of it? $\endgroup$ – Raekye Sep 24 '13 at 20:58
  • $\begingroup$ While it seems like a really cool solution, (in hindsight) there's a lot simpler linear solution and I still don't know how to set up this ellipse on a slant. Thanks for your input though $\endgroup$ – Raekye Sep 27 '13 at 11:53
  • $\begingroup$ You don't need to "set up this ellipse on the slant". You can choose a reference frame where AC is on the abscissa. Then the horizontal straight line tangential to the ellipse will be at an angle with respect to abscissa, but you'll have a canonical equation for the ellipse. $\endgroup$ – akhmeteli Sep 27 '13 at 13:18
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aufkag pointed out the necessary parts for the solution, but didn't make an answer. This walks through the problem using his tips

  1. The key is to realize that angles BAE and BCF are equal. By geometric laws DBA and DBC are equal to those other two too (notice parallel lines AE, DB, and CF). Proof by aufkag (quote from his last comment)

    Because the sack is just hanging there, the horizontal forces must cancel. Therefore the horizontal components of the tensions must cancel.[1] Because the pulley is free, the tension in both sides of the cable must be equal.[2] Combining these two facts, the vertical components of the tensions must be equal.[3] Therefore (2+3), both tensions must make equal opposite angles with DB.[4]

  2. So let $\theta$ be the angle between AB and the horizontal (same as BC and the horizontal)

  3. $\frac{x}{\cos(\theta)} +\frac{3.5 - x}{\cos(\theta)} = 5$ (sum of the segments of the rope is 5)

  4. Since the tension in the rope is the same on both segments, their horizontal and vertical components must be the same. If their vertical components are the same, they each share one half of the "load" of the pulley/weight. $\frac{100 \cdot 9.81}{ 2 \sin(\theta)} = 687 \, \mathrm{N}$ (approximately). This is the first half of the answer (tension in the cable)

  5. $0.75 + x \cdot \tan(\theta) = (3.5 - x) \cdot \tan(\theta)$ (equal lengths for the segment BD, calculated using geometry from the left hand side and the right hand side). Solve for $x$ to get 1.38m.
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