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David Morin's Rope around a Pole example

  • #1
1. A rope wraps an angle θ around a pole. You grab one end and pull with a tension T0. The other end is attached to a large object, say, a boat. If the coefficient of static friction between the rope and the pole is µ, what is the largest force the rope can exert on the boat, if the rope is to not slip around the pole?


The question I have is really just about the diagram. Here it shows the two tension forces at both ends, my question is how did he get the angle there to be sin(d(theta)/2)? Well, more importantly the d(theta)/2, I understand why its sin. I was unable to derive that myself, so I looked at the solution to see what angle I needed to use, and from there I was able to do the problem, but I still need to know how he got that angle.

My thoughts are that the angles on both sides of those tension forces contribute an angle of theta/2 and so that add up to being a total of theta. but I guess I still don't know why geometrically.
 

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  • #2
collinsmark
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Hello @BrandonBerisford,

Welcome to PF!

:welcome:

Try to think in terms of "similar triangles."

You haven't, as of yet, found what the other similar triangle actually is. :wink:

Suppose the pole has a radius r. Now imagine a right triangle with hypotenuse r (feel free to look at your diagram while doing so). How do the other legs and the angle θ fit in?
 
  • #3
Hello @BrandonBerisford,

Welcome to PF!

:welcome:

Try to think in terms of "similar triangles."

You haven't, as of yet, found what the other similar triangle actually is. :wink:

Suppose the pole has a radius r. Now imagine a right triangle with hypotenuse r (feel free to look at your diagram while doing so). How do the other legs and the angle θ fit in?
Ahh! so i cut the big arc in half with a radius of r, making two right triangles with each having a small angle of d(theta)/2 and drawing out the triangles from the sides with the tensions, it's a similar triangle with those and comparing angles we can say its the same?
 
  • #4
collinsmark
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Ahh! so i cut the big arc in half with a radius of r, making two right triangles with each having a small angle of d(theta)/2 and drawing out the triangles from the sides with the tensions, it's a similar triangle with those and comparing angles we can say its the same?
I think that should put you on the right track! :smile:

But to be sure, it wouldn't hurt to show the rest of your work for the remainder of the problem. There's some integration left, and perhaps a small-angle approximation that might fit in there somewhere.
 
  • #5
Ahh! so i cut the big arc in half with a radius of r, making two right triangles with each having a small angle of d(theta)/2 and drawing out the triangles from the sides with the tensions, it's a similar triangle with those and comparing angles we can say its the same?
I can't quite picture what you are saying because I seem to be drawing it wrong and not getting the same result. Can anyone supplement this with a picture, please?
 
  • #6
PeroK
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I can't quite picture what you are saying because I seem to be drawing it wrong and not getting the same result. Can anyone supplement this with a picture, please?
Note that the post is from last year.
 
  • #7
Note that the post is from last year.
Was I supposed to make a new topic about the drawing portion?
 
  • #8
haruspex
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I can't quite picture what you are saying because I seem to be drawing it wrong and not getting the same result. Can anyone supplement this with a picture, please?
Draw the bisector of the angle dθ and project the two arrows backwards so that these three lines meet. Do you see some similar triangles?
 
  • #9
Oh I get it now, thank you!
 

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