Suspension cable statics calculus problem

1. Mar 31, 2012

Femme_physics

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.

2. Mar 31, 2012

I like Serena

You're on the right track. :)

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

3. Mar 31, 2012

Femme_physics

I like you saying this lately :)

Oohhh....

Am I still on the right track?

4. Mar 31, 2012

I like Serena

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].

5. Apr 2, 2012

Femme_physics

You mean purely because of math?

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?

6. Apr 2, 2012

Yes.

Yes.

7. Apr 5, 2012

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..

But my big problem is relating the distance, w0 and FH, to A and B.

I end up with this diagram...

8. Apr 5, 2012

I like Serena

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:
$x \sqrt 2 = (25 - x) \sqrt 3$
$x \sqrt 2 = 25 \sqrt 3 - x \sqrt 3$
$x (\sqrt 2 + \sqrt 3) = 25 \sqrt 3$
$x = \frac {25 \sqrt 3} {\sqrt 2 + \sqrt 3} \approx 13.76$

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
$F_V = {dy \over dx} \cdot F_H$

Last edited: Apr 5, 2012
9. Apr 8, 2012

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.

10. Apr 8, 2012

I like Serena

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?

Last edited: Apr 8, 2012
11. Apr 9, 2012

Femme_physics

Ahh....

Each has their own force.

So,

12. Apr 9, 2012

OldEngr63

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. Apr 9, 2012

I like Serena

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. Apr 10, 2012

Femme_physics

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

Very, very, SLOWwwww... because of the teacher, not us. We're on our passover holiday right now. And I did eventually use a gripper's PDF guide to get ideas, and basically our main idea is a double-threaded spindle. But, right now, we're still awaiting orders and formulas.

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15. Apr 10, 2012

I like Serena

16. Apr 10, 2012

OldEngr63

Thank you.

17. Apr 11, 2012

Femme_physics

No, thank YOU! a lot. :)