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Suspension cable statics calculus problem 
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#1
Mar3112, 12:37 AM

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#2
Mar3112, 08:45 AM

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You're on the right track. :)
You have 1 equation for point A. Can you make another equation for point B? 


#3
Mar3112, 03:51 PM

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#4
Mar3112, 05:04 PM

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Suspension cable statics calculus problem
And btw, you're using 15000 [lb] for ##w_0##, but ##w_0## is given to be 600 [lb/ft]. 


#5
Apr212, 02:34 AM

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#7
Apr512, 01:09 PM

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#8
Apr512, 03:36 PM

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If you're interested, I have a shorter version: ##20 x^2 = 30 (25  x)^2## ##2 x^2 = 3 (25  x)^2## Since x and (25  x) are both positive distances, we can take the square root and keep the positive versions: ##x \sqrt 2 = (25  x) \sqrt 3## ##x \sqrt 2 = 25 \sqrt 3  x \sqrt 3## ##x (\sqrt 2 + \sqrt 3) = 25 \sqrt 3## ##x = \frac {25 \sqrt 3} {\sqrt 2 + \sqrt 3} \approx 13.76## But consider that the tensional force is not pointing down, but along the rope. Since they ask for the tension in the rope in A and in B, you need that ##F_V = {dy \over dx} \cdot F_H## 


#9
Apr812, 11:36 AM

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#10
Apr812, 02:02 PM

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Good. Oh, but the unit is lb, and not l/ft. On support B you have the horizontal force Fh and this vertical force Fv. So what's the total force? 


#12
Apr912, 01:14 PM

P: 343

Don't you need to add those vectorially? (or did I miss something as I skimmed down through the solution?)
BTW, how did your gripper project turn out? 


#13
Apr912, 02:18 PM

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Ya know wee go n bada bing, bada boom n we ah atta thereah....day won't know that OldEngr63 hit um!
(You're not supposed to simply add up forces that are perpendicular to each other. ;) 


#14
Apr1012, 04:35 AM

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Ohhhhhhh Ok gotcha, now it makes perfect sense :)
I need to find the resultant vector for each! *smacks forehead* So (CALCULATION ATTACHED) Rb = 9086 lb Ra = 7735 lb BADABING BADABOOM I said! :) 


#15
Apr1012, 08:23 AM

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#16
Apr1012, 10:06 AM

P: 343

Thank you.



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