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 Posted: Tue Aug 22, 2006 7:45 pm
Recently I was helping someone else on gaia with a mathematics problem. I believe I recognize the form of the problem; we covered vectors as an extra topic after my AP Calculus class had finished its normal routine and the exam was over. I need to make sure I'm remembering things right. problem I would seriously glomp you and more if you could answer this. ;o; I need to know the answer to this: proposition recently passed, regarding the Blue Building. Since then, the structure is on its way to being rebuilt. A passing woman noticed a large, 2000 pound block, being held by two ropes that were attached to a horizontal beam. The rope on the left made a 30 degree angle, the rope on the right made a 40 degree angle (see image). What is the tension in the left cable? HINT: Round to the nearest whole number. Do not enter units. http://i4.photobucket.com/albums/y129/MoonFighter666/math.gif I decided that writing out my work and scanning it would be more helpful. I apologize if my handwriting is horrendous as I've been told.  I'd appreciate it if anyone could tell me if this is correct or not. This is at least partially correct, but I have a feeling I may have overlooked something.
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Posted: Tue Aug 22, 2006 8:43 pm
In the coordinates of the rope junction, the force vector components are F1 = [-F1 cos α; F1 sin α], F2 = [F2 cos β; F2 sin β], with the system F1 + F2 = [0;F], which is exactly what you have (F = 2000 lbf). The solution if (F1,F2) = (F cos β / sin(α+β), F sin α / sin(α+β)) = (1630 lbf, 1843 lbf). Your solution is correct in form in that F1 = F/(sin α + cos α tan β), but you have erred in the values of the trigonometric functions of the given angles. In particular, cos α = sqrt(3)/2, not sqrt(3)/3.
Something should have tipped you off in that you have a tension higher than the total load. Since the load is shared, it will be less for both ropes.
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Posted: Tue Aug 22, 2006 9:39 pm
VorpalNeko In the coordinates of the rope junction, the force vector components are F1 = [-F1 cos α; F1 sin α], F2 = [F2 cos β; F2 sin β], with the system F1 + F2 = [0;F], which is exactly what you have (F = 2000 lbf). The solution if (F1,F2) = (F cos β / sin(α+β), F sin α / sin(α+β)) = (1630 lbf, 1843 lbf). Your solution is correct in form in that F1 = F/(sin α + cos α tan β), but you have erred in the values of the trigonometric functions of the given angles. In particular, cos α = sqrt(3)/2, not sqrt(3)/3. Something should have tipped you off in that you have a tension higher than the total load. Since the load is shared, it will be less for both ropes. Yeah the value did strike me as odd, but I couldn't reconcile it. Ah! Right, the cosine of 30 degrees is The square root of 3 divided by 2. ><;; Dammit, how could I have mixed that up. I'll go and change that right now. Thanks for the help; is there anything else you can see I was mistaken on?
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Posted: Tue Aug 22, 2006 9:51 pm
Thanatos_M Ah! Right, the cosine of 30 degrees is The square root of 3 divided by 2. ><;; Dammit, how could I have mixed that up. I'll go and change that right now. Thanks for the help; is there anything else you can see I was mistaken on? Your solution was good right up until you started replacing the trigonometric functions with numerical values. The set-up and algebra are completely correct.
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Posted: Tue Aug 22, 2006 10:01 pm
VorpalNeko Thanatos_M Ah! Right, the cosine of 30 degrees is The square root of 3 divided by 2. ><;; Dammit, how could I have mixed that up. I'll go and change that right now. Thanks for the help; is there anything else you can see I was mistaken on? Your solution was good right up until you started replacing the trigonometric functions with numerical values. The set-up and algebra are completely correct. Thanks muchly; I'm starting to see why you're the Vice-Captain of this guild already.
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Posted: Sat Oct 14, 2006 7:24 pm
...whoa...
Flashback to Intro Physics....
I can't believe I already forgot how to do that (at first glance. After looking at it harder I understood it).
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