## [SOLVED] Tension in Strings HELP thanks!--very simple.

 Mentor Blog Entries: 1 You need to learn how to find the components of a vector. If you have some vector with magnitude F that makes an angle $\theta$ with the x-axis (the horizontal), the components are:vertical (y) component = $F \sin\theta$ horizontal (x) component = $F \cos\theta$ I recommend looking at the examples on this page, especially the one with the dog on a leash: Trigonometric Method of Vector Resolution. In fact I strongly recommend that you spend a good bit of time exploring the rest of that site to better understand forces, components, and equilibrium.
 oh okay i think i get what you are saying. thank you! so what i basically have to do is this: 41.81*2*T= 156.96? is that what it means. i think?
 because the Tension in the x equals mass times gravity= 156.96 and the Tension in the y equals sin (theda) = 1.118 and the Tension in the x of the angle equals cos (theda) = 1/1.5
 Mentor Blog Entries: 1 No. What angle does each string make with the horizontal?
 48.187

Mentor
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 Quote by physicsbhelp 48.187
Good! Now use what I told you in post #36.

 so sin(48.187) + cos (48.187) = 0
 wait no sorry it should be: sin(48.187) + cos (48.187) = mg
 right?

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No. In post #36, I told you that:
 Quote by Doc Al vertical (y) component = $F \sin\theta$
In your problem, the F would represent the tension in the string. So write down the vertical component. (Just plug in the angle and compute the sin.)

Since you have two strings, you'll have to double it.

Remember: F, the tension, is an unknown. Once you get the equilibrium equation written, you will solve for F.

 oooo now i get what you mean so its 2F(sin(theda)) so basically i got 2*(sin(48.187))F which = 1.491F
 but what i don't get is how you have 2 STRINGS because i am jus ttlaking about diagram a) so i hope u are not trying to figure out the tension in both the pictures. or are you saying that since the big traingle can be divided into two triangles, there are two strings?

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 Quote by physicsbhelp oooo now i get what you mean so its 2F(sin(theda)) so basically i got 2*(sin(48.187))F which = 1.491F
Exactly! Now set that equal to what?

 Quote by physicsbhelp but what i don't get is how you have 2 STRINGS because i am jus ttlaking about diagram a) so i hope u are not trying to figure out the tension in both the pictures. or are you saying that since the big traingle can be divided into two triangles, there are two strings?
As far as the forces go, you can view the mass as being pulled by two strings--one left and one right. (Of course, it's just two parts of the same string, but that doesn't matter.)

 okay so now i totally get you. thanks so now the next step i set 1.491F=mg? or 1.491F=.6667--cos(48.187)

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 Quote by physicsbhelp so now the next step i set 1.491F=mg?
Yes! Then solve for the tension, F.
 or 1.491F=.6667--cos(48.187)
No!

 Blog Entries: 1 PhysicsB, though this may seem unrelated to this thread, what book are you working from? I know what course you are taking (The same one as myself) but what book are you working from?