Either I have missed something obvious...

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oddy
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Either I have missed something obvious...

Postby oddy » Sun Jan 16, 2011 6:35 pm UTC

... or this is the most difficult A-level Mechanics 1 question of all time.

My friend is doing M1 next week (I've already sat it) and in some insane exam paper her teacher has set her, is this beast:

A rock climber has weight 550N. Rope attached to waist belt passes through ring which is fixed at point higher up cliff. She loses foothold and starts to move in direction 20 degrees to the horizontal. Tension in rope is 560N. Calculate angle rope makes with vertical.


I should mention that in M1 we use modelling assumptions of no resistive forces, weight is vertically downwards and that g=9.8, if that makes a difference.

Have a go at it, my solution was INSANELY RIDICULOUS, and though by using an iterative trigonometric formula I managed to get it, there must me an easier way. If you want to try, the mark scheme's answer was
Spoiler:
42.6 degrees.


My solution is below, if you think I may be exaggerating how convoluted my solution was, take a look. :mrgreen:

Spoiler:
So I started with a little force diagram, did a bit of resolving in x and y, and then created a force triangle with the weight and x and y components of tension included. With a bit of trig, with the angle set to 20 degrees, this gave us a little equation, which took some solving.

If you want to know how I did all the resolving etc, then you’ll find a nice picture of my working here: http://dl.dropbox.com/u/498402/mechanics-y%20part%20of%20solution.png.

For details of how I got from that equation to the one that I plopped into the equation solver, working is here: http://dl.dropbox.com/u/498402/maths-y%20part%20of%20solution.jpg.

By changing tan(20) and the fractions into decimals, I arrived at this equation which I felt inadequate to solve: [math]0.132-0.132cos^2(x)= -(1100/560)cosx+cos^2(x)+302500/313600[/math].

If we pop that equation into the equation solver http://www.numberempire.com/equationsolver.php then we get x = acos(5^(3/2)*sqrt(553465*cosx-234598)/sqrt(39869606)) .

If we use WolframAlpha to solve that, we get this: http://www.wolframalpha.com/input/?i=x+%3D+acos%285^%283%2F2%29*sqrt%28553465*cosx-234598%29%2Fsqrt%2839869606%29%29+.

If we use the bigger solution, x ~ 0.7432060359962328, and ask WolframAlpha to convert that to radians for us, then we get this: http://www.wolframalpha.com/input/?i=+0 ... to+degrees .

And strangely enough, it gives us 42.58 degrees... i.e. 42.6! So my ridiculous solution was right, but we must have missed something obvious that allowed us to do it.


I know I've missed something obvious, because there is no way that would be in a Mechanics 1 exam. If anyone has any blinding insights that my friend and I missed, please do tell. :D

oddy.

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Dopefish
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Re: Either I have missed something obvious...

Postby Dopefish » Sun Jan 16, 2011 7:08 pm UTC

Although I'm otherwise busy to actually try the problem myself to see if I end up in the same boat as you in terms what what equation(s) need to be solved, I am inclined to point out that the equation you felt unable to solve is simply a quadratic (let u=cos(x)), so you could solve that thing via the quadratic formula to get u, and then x=arccos(u). Assuming you have the use of a calculator, then it wouldn't be 'that' bad, even though the numbers involved might be kinda ugly.

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Re: Either I have missed something obvious...

Postby sikyon » Sun Jan 16, 2011 8:35 pm UTC

I don't understand this question. It seems to me the answer should be very straightforward - if the rope is of a fixed length, then an object must swing in a circle centered about the fixed point of the rope. The direction of movement will then always be tangent to this circle. Since we know the person moves at an angle 20 degrees to the horizontal, and a tangent makes an angle of 90 degrees to the radius (the direction of the rope), the angle of the rope to the verticle must be 20 degrees.

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Re: Either I have missed something obvious...

Postby oddy » Sun Jan 16, 2011 9:08 pm UTC

Dopefish - you're right, I didn't spot that. That would certainly make it plausible for an exam.

Sikyon - It say's they start to move at 20 degrees to the horizontal. That's (maybe not outside of M1) taken to mean accelerates at 20 degrees to the horizontal. If it were circular motion, the acceleration must be towards the centre of the circle, and it's at 20 degrees. I guess it's just poorly worded in that respect, my description is taken from what my friend typed into facebook chat so is probably not a faithful replica.

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Re: Either I have missed something obvious...

Postby thicknavyrain » Sun Jan 16, 2011 9:48 pm UTC

oddy wrote:Dopefish - you're right, I didn't spot that. That would certainly make it plausible for an exam.

Sikyon - It say's they start to move at 20 degrees to the horizontal. That's (maybe not outside of M1) taken to mean accelerates at 20 degrees to the horizontal. If it were circular motion, the acceleration must be towards the centre of the circle, and it's at 20 degrees. I guess it's just poorly worded in that respect, my description is taken from what my friend typed into facebook chat so is probably not a faithful replica.


You aren't expected to know about circular motion and acceleration being towards the centre of the circle until M3.
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Charlie!
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Re: Either I have missed something obvious...

Postby Charlie! » Mon Jan 17, 2011 12:10 am UTC

This question is horribly worded. Was this really the original wording? I can think of at least four different interpretations of "she starts to move in direction 20 degrees to the horizontal." I would tentatively guess that the cliff is meant to be a flat vertical surface, and she isn't directly under the ring, so she swings back and forth in a plane, and she is meant to start from rest.

So once interpreted this way (and interpreting "Tension in rope is 560N. Calculate angle rope makes with vertical." to mean "Calculate the angle the rope makes with the vertical when the tension on the rope is 560 N."), it isn't so hard. The tension in the rope is the component of gravity along the rope, plus the centrifugal force. Conservation of energy will give you the centrifugal force as a function of angle, and then that sum is possibly solvable for the angle.

For most interpretations, I think your answer (42 degrees) is physically impossible, since the starting point (and max if she starts at rest) is 20 degrees.
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Re: Either I have missed something obvious...

Postby pizzazz » Mon Jan 17, 2011 4:41 am UTC

Charlie! wrote:This question is horribly worded. Was this really the original wording? I can think of at least four different interpretations of "she starts to move in direction 20 degrees to the horizontal." I would tentatively guess that the cliff is meant to be a flat vertical surface, and she isn't directly under the ring, so she swings back and forth in a plane, and she is meant to start from rest.

So once interpreted this way (and interpreting "Tension in rope is 560N. Calculate angle rope makes with vertical." to mean "Calculate the angle the rope makes with the vertical when the tension on the rope is 560 N."), it isn't so hard. The tension in the rope is the component of gravity along the rope, plus the centrifugal force. Conservation of energy will give you the centrifugal force as a function of angle, and then that sum is possibly solvable for the angle.

For most interpretations, I think your answer (42 degrees) is physically impossible, since the starting point (and max if she starts at rest) is 20 degrees.


Huh? Presumably, since she is hanging from a point above, "20 degrees to the horizontal" seems to me like it should mean "20 degrees below the line perpendicular to the cliff/parallel to the ground" and so 42 degrees would be farther down, not up.

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Re: Either I have missed something obvious...

Postby jmorgan3 » Mon Jan 17, 2011 5:45 am UTC

sikyon wrote:I don't understand this question. It seems to me the answer should be very straightforward - if the rope is of a fixed length, then an object must swing in a circle centered about the fixed point of the rope. The direction of movement will then always be tangent to this circle. Since we know the person moves at an angle 20 degrees to the horizontal, and a tangent makes an angle of 90 degrees to the radius (the direction of the rope), the angle of the rope to the verticle must be 20 degrees.

That's what I though first as well. Then, however the problem is overconstrained. I think the problem lies in the fact that the rope isn't secured at the ring, merely passing through it. The rope is slipping through the ring as she falls, so it's not circular motion. It's a very odd way to compose a problem, but it works.
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sikyon
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Re: Either I have missed something obvious...

Postby sikyon » Mon Jan 17, 2011 4:48 pm UTC

jmorgan3 wrote:
sikyon wrote:I don't understand this question. It seems to me the answer should be very straightforward - if the rope is of a fixed length, then an object must swing in a circle centered about the fixed point of the rope. The direction of movement will then always be tangent to this circle. Since we know the person moves at an angle 20 degrees to the horizontal, and a tangent makes an angle of 90 degrees to the radius (the direction of the rope), the angle of the rope to the verticle must be 20 degrees.

That's what I though first as well. Then, however the problem is overconstrained. I think the problem lies in the fact that the rope isn't secured at the ring, merely passing through it. The rope is slipping through the ring as she falls, so it's not circular motion. It's a very odd way to compose a problem, but it works.


This reminds me of why I hated poorly worded physics questions. I remember on my grade 12 provincial physics exam, there was a question with a diagram. In this diagram, there was a piece specifically labeled as a "beam" but the wording referred to a "shelf". Of course, as it was a provincial exam the teachers could not clarify at all and I had to guess willy nilly (I did get it right though, but it was frustrating).

oddy
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Re: Either I have missed something obvious...

Postby oddy » Mon Jan 17, 2011 9:05 pm UTC

Charlie:
oddy wrote:...my description is taken from what my friend typed into facebook chat so is probably not a faithful replica.

It is not the original wording, as I said a few posts ago. :P

Sorry for all the confusion, mostly it's because of my [friends] bad transcription.

pizzazz wrote:Huh? Presumably, since she is hanging from a point above, "20 degrees to the horizontal" seems to me like it should mean "20 degrees below the line perpendicular to the cliff/parallel to the ground" and so 42 degrees would be farther down, not up.


Charlie! wrote:For most interpretations, I think your answer (42 degrees) is physically impossible, since the starting point (and max if she starts at rest) is 20 degrees.


The overall acceleration is in a direction 20 degrees below the horizontal, but the rope itself is at a greater angle. Your are right in your understanding of "20 degrees to the horizontal", pizzazz, but the idea is that the person is standing with their feet against the wall, like an abseiler. They slip and equilibrium is broken, and they accelerate in that direction. The angle we are trying to find is the angle that the rope makes with the (presumed) vertical cliff at the instant that they slip.

All the confusion seems to stem from the "begins to move" statement. I understand that it would describe circles (I've done M3 in which vertical circles do feature), but as it is M1 they do not know anything about circular motion, it could not be that. In vertical circles there is a radial acceleration [math]v^2/r=rw^2[/math] and a transverse acceleration [math]gsin (theta)[/math] In this particular instant, the resultant of the two accelerations is at 20 degrees below the horizontal. So the overall acceleration being 20 degrees below the horizontal doesn't imply that the rope is at 20 degrees.

However, if the the transverse acceleration were at 20 degrees to the horizontal, the rope would be at 20 degrees to the vertical. The transverse acceleration is at 42.6 degrees below the horizontal, the radial acceleration 'pulls it up', i.e. they all form a nice vector triangle, with the resultant at 20 degrees below the horizontal.

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Re: Either I have missed something obvious...

Postby Dopefish » Tue Jan 18, 2011 12:55 am UTC

(Just fyi, you can make a pretty theta in math tags by going \theta giving [imath]\theta[/imath]. Also, you can make a slightly prettier sin via \sin, giving [imath]\sin[/imath], otherwise it's typeset as the product of s*i*n. That last one doesn't matter too much but the symbols at least are straightforward enough you should learn them and love them. LaTeX is purdy. :D )


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