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Curiosity Question 1: Compression force from tension

  1. Mar 12, 2015 #1
    Here is some background for this post and for future "Curiosity question" posts.

    So last year in 10th grade, I took AP Calc AB and AP Physics B, and scored 5s on both, but a particularly high five on the physics, as I truly enjoy doing physics just for the bliss of it. Well this year, I stepped away from physics to do AP Chem, AP Macro, and AP USGov, until January when I learned about the US Physics Olympiad. After taking two F=ma past exams, and scoring within the cutoff for semis, I decided to go for it and do my best. FYI this was without having studied physics at all for 6 months and without even looking at any physics equation sheets (I just used my memory). Since then, I have been self-studying physics and calculus fairly intensely, as I just finished working through the AP Physics C Mechanics syllabus slightly less than a month after starting.

    All that to say, over the course of the past month, I have thought up a bunch of physics and math problems, which obviously don't have any published solutions. So I thought I would do a series of sorts with said problems for:
    a) the enjoyment of everyone who reads them
    b) work checking to see if they are riddled with mistakes or not
    c) approach analysis, asking people to give me feedback on how I worked it.

    Here is my first problem (hopefully of many). This one, like most of the ones I plan to include in the series I made up:
    1. The problem statement, all variables and given/known data
    If there is a rope binding a cylindrical object, what is the maximum amount of inward force, as a function of F(tension)? By the way, this is measuring the scalar force, not vector force, because obviously the net force is 0.

    2. Relevant equations
    I go about the solution by considering simpler regular polygon cross sections, generalizing, and then finding the case of the cylinder by treating its circular cross section as a regular infinity-gon. So there are not many equations needed except for:
    the internal angles of a regular polygon = ##180\frac{n-2}{n}##
    where n is the number of corners of the polygon.
    AND
    ##\theta = \frac{external angle - 180}{2}##
    3. The attempt at a solution
    FYI, there is a fuller solution in the scans, with some musings afterward, but here is the basic approach:

    A square, a regular 4-gon, has a rope bound around it. the compressive force can be restricted to the corners only, because its sides are straight. So at each corner, FT acts along the line of its side. Thus, eight FT forces are acting on the square. To find the force inward, I draw a line tangent to the corner and perpendicular to the origin. The inward force from each FT is ##F_T\sin(\theta)##, where theta is ##(external angle - 180) / 2##. It can also easily be shown that ##\theta =\frac{360}{2n}##, where n is the number of corners and 2 is the number of ##\theta##s per corner.

    Thus, Total inward force for a square = ##(4 corners)(2 F_Tper corner)(\sin\frac{360}{2*4 corners})##

    So for a regular n-gon, Total force = ##F_T n \sin\frac{180}{n}##
    So for a circle, and infinity-gon, Total force = ##\displaystyle\lim_{n\rightarrow +\infty} {n\sin(180/n)}##

    Numerically, the answer is 2pi, an interesting effect. But what I don't know how to do is to take that limit by non-numerical means. Any help would be appreciated.

    Again, if this solution was a bit too concise, or you want to see the diagrams, they are attached.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2015 #2

    BvU

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    $$\lim _{x\rightarrow 0} {\sin(x)\over x}$$ is a well-known limit. Its value is 1.
     
  4. Mar 13, 2015 #3
    Thanks for answering. So what you were saying was that n going to infinity is basically the same as 1/x where x is going to 0, and the same inside sine's argument.

    But it is not the same, because of the 180 degrees term inside. Now I made a mistake by missing the 2 inside the limit. It should have been: ##\displaystyle\lim_{n\rightarrow +\infty} {2n\sin(180/n)} ##
    If you do it on the calculator, you will get pi as the answer for ##\displaystyle\lim_{n\rightarrow +\infty} {n\sin(180/n)}## and 2pi (obviously) for ##\displaystyle\lim_{n\rightarrow +\infty} {2n\sin(180/n)} ##

    So my question is, why does that 180 degrees make the difference? Why is it that if you replace 180 with 360, the value doubles? Is this because, when the value of ##\theta## gets small, ##\theta=\sin\theta##?

    So when ##\theta## is small, multiplying ##\theta## by some value inside the argument of sine is basically the same as multiplying outside of it. ##\frac{\sin\theta}{\theta}## goes to ##\frac{\theta}{\theta}##, using radians of course.

    Also, does anyone have any suggestions on my approach to the problem or the analysis afterward (the "Compressive force problem page 3" file)?

    Thanks
     
  5. Mar 13, 2015 #4

    BvU

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    $$
    \lim _{x\rightarrow 0} {\sin(\pi x)\over x} = \pi \lim _{x\rightarrow 0} {\sin(\pi x)\over \pi x} = \pi \lim _{x'\rightarrow 0} {\sin( x')\over x'} = \pi
    $$
     
  6. Mar 15, 2015 #5

    BvU

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    The ##2\pi## comes from going full circle. That's ##2\pi## radians.

    If you write down the vector sum of the two forces from the two ends of a small piece of the rope

    upload_2015-3-15_23-45-46.png

    you get ##F_T\sin\theta\ ## inward, which for small theta goes to ## F_T\theta\ ##.
    The vertical components add up to ##F_T(1-\cos\theta) ## which goes to zero (like ##\tfrac {1}{2} \theta^2##, so faster than the other one)

    Add up ##2\pi## worth of small chunks of ##\theta## to go around the full circle and all these small ## F_T\theta\ ## together give you this factor ##2\pi##.

    So your Original question: What is the maximum inward force should be that there is no theoretical upper limit.

    What happens in reality depends on material properties: the strength of the rope and the material of the cylinder.
     
    Last edited: Mar 16, 2015
  7. Mar 16, 2015 #6
    Of course, infinite tension would produce infinite compressive force. But the point of the question, which has now been answered, was to find the maximum compressive force given a FT. Another way of stating it was to say, "Find the maximum compressive force in terms of FT."

    Thanks for your responses, and soon I should be able to post Curiosity Question 2
     
  8. Mar 16, 2015 #7

    Merlin3189

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    Watching this I have been trying to figure out what the physical significance of this inward force is. I know that in 3 dimensions the force per unit area is the pressure, so it comes into the pressure in a bubble or a pipe or tank, but I haven't worked out what the force itself summed over a complete surface, or here what the total force perpendicular to a closed line represents. Anyone any comments?
     
  9. Mar 16, 2015 #8

    BvU

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    No big deal. There was a nice thread asking at which ##\omega## the rope (in that case a chain) would fall down. Forgot which one it was.
     
  10. Mar 17, 2015 #9
    I thought of a practical example of the use of the compression force--tying up a rolled sleeping bag with a rope. It seems like the bag could be compressed the same amount by A) a rope around it with a tension FT, with total compressive force of ##2\pi F_T## or B) a direct force on one side of ##\pi F_T##. The normal force from the other side of the ##\pi F_T## would be equal and opposite, bringing the total compression to ##2\pi F_T##

    Hope this helps!
     
  11. Mar 17, 2015 #10

    haruspex

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    "Total inward force" does not seem to be very meaningful. Treating force as a vector it would be zero, but it's not clear how integrating it as a scalar produces anything useful.
    As for the net compression on the internals, it may be more instructive to consider slicing the cylinder along a diameter. The force from the rope on each half is 2T (because there are two rope ends each tension T acting on the same half). This method is routinely applied to spheres to figure out the pressure in a bubble.
    Adding the forces on the two opposite sides in a compression to get the total compression is quite wrong. If you push a block against a wall with force F, the wall reacts with force F, but the compression in the block is F, not 2F.
     
  12. Mar 19, 2015 #11
    @haruspex While I can see where you are coming from, I am still not totally convinced. Here is my line of reasoning:

    Though not the inspiration for this problem, I like to think of the example of the rolled up sleeping bag, a rough cylinder, which has a rope tied around it to compress it. Said rope has tension T, and compresses the sleeping bag. It seems that you agree that the total compression is not simply T, but some larger number, as you said above:
    I would like to suggest that the system cannot be simplified just to splitting the diagram in half and finding the net force on one half. What if, instead of just splitting it on one diameter, I decided to split it on two perpendicular diameters? I am sure this operation would be just as valid as finding the net force on one half. Well, as in the attached picture("Compression"), the net force on each corner is ##\sqrt{2}T##. This result for sure leads to a different answer for the compression force, whether you count it as a total of ##\sqrt{2}T, or 2\sqrt{2}T, or 4\sqrt{2}T##, the answer is still different from your previous calculation along a single diameter.

    Back to the sleeping bag: If I were to think that the total compression was 2T as suggested, that would lead to an interesting conclusion regarding the sleeping bag. The compression would be exactly the same as if the bag were set between two boards and the boards were connected by a rope with tension T on each side, as seen in the primitive paint drawing attached. But in reality we see that the cross sectional area of the real sleeping bag is round, because the rope introduces forces on the sides as well as the top and bottom.

    So if that reasoning is logical and sound, it would lead us to the conclusion that the only valid way to determine the compression (note: I am not totally sure if compression would be a scalar force or not, but I think it is a valid property to solve for as it does have real world significance, like the sleeping bag) would be to split the diagram along an infinite number of diagonals, which leads to the same conclusion as in the original post.

    Now regarding your final sentence,
    I tend to think that the reaction force is not to be dismissed, for this reason:
    Consider applying a constant force to a ball on the x axis, accelerating it linearly. As long as the ball has mass, it will be compressed (according to F=kΔL, where k is Young's modulus), but only by half the amount if that same force pinned the ball against a wall. Another experiment would be to place a spring of significant mass on a vertical rocket. Before the rocket fires up, an astronaut presses down on the spring with force F, and the floor responds with -F. Then, the rocket fires up and presses on the spring with a force F, accelerating it with the rocket. The compression would be half that of the time the astronaut pressed on the spring. Now naming the force F and 1/2F, or 2F and F, is mostly a matter of semantics, depending on how young's modulus is defined. But the point is that the reaction force does end up providing an extra bit of compression.

    tl;dr: Your argument is invalid.... just kidding. But I am looking forward to reading your response.
    Thanks!
     

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  13. Mar 19, 2015 #12

    haruspex

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    Then you have two forces T, at right angles, from the rope. Let the internal pressure be P and the cylinder length L. The 'cut' faces have forces PLR on each, where R is the radius of the cylinder. Again, these forces are at right angles, directly opposing the two T forces. Same result as for the half cylinder: T = PLR. More generally, if there are N turns of the rope, TN = PLR.
     
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