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## Suspension cable statics calculus problem

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I find myself with 2 unknowns, 1 equation.

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

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 You're on the right track. :)
I like you saying this lately :)

 You have 1 equation for point A. Can you make another equation for point B?
Oohhh....

Am I still on the right track?

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## Suspension cable statics calculus problem

 Quote by Femme_physics I like you saying this lately :)

 Oohhh.... Am I still on the right track?
I'm afraid that in the 2nd line you lost ##F_H##.

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

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 Quote by I like Serena I'm afraid that in the 2nd line you lost ##F_H##.
You mean purely because of math?

 And btw, you're using 15000 [lb] for ##w_0##, but ##w_0## is given to be 600 [lb/ft].
I see what you mean, since I don't know the length in each sectioning I can't use it like that. I'll jus use 600, with the units lb/ft. Yes?

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 Quote by Femme_physics You mean purely because of math?
Yes.

 I see what you mean, since I don't know the length in each sectioning I can't use it like that. I'll jus use 600, with the units lb/ft. Yes?
Yes.
 Recognitions: Gold Member I got it. My friend helped me with the math (the one who's registered as "niece of MD") :) We use X2 since the length can't be more than 25. Therefor we use FH2 as well.. But my big problem is relating the distance, w0 and FH, to A and B. I end up with this diagram...

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 Quote by Femme_physics I got it. My friend helped me with the math (the one who's registered as "niece of MD") :) We use X2 since the length can't be more than 25. Therefor we use FH2 as well.
Aha! So she does still do something every now and then! :)

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 my big problem is relating the distance, w0 and FH, to A and B. I end up with this diagram...
Looks good.
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##
 Recognitions: Gold Member I think I got it, but I don't know what to make out of Fv as far as each of the reaction forces at A and B.

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 Quote by Femme_physics I think I got it, but I don't know what to make out of Fv as far as each of the reaction forces at A and B.
You have the result Fv for point B here.
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?
 Recognitions: Gold Member Ahh.... Each has their own force. So, Badabing badaboom?
 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?
 Recognitions: Homework Help 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. ;)

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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