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

by physicsbhelp
Tags: simple, solved, strings, tension, thanksvery
 P: 300 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?
 P: 300 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 P: 40,227 No. What angle does each string make with the horizontal?
 P: 300 48.187
Mentor
P: 40,227
 Quote by physicsbhelp 48.187
Good! Now use what I told you in post #36.
 P: 300 so sin(48.187) + cos (48.187) = 0
 P: 300 wait no sorry it should be: sin(48.187) + cos (48.187) = mg
 P: 300 right?
Mentor
P: 40,227
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.
 P: 300 oooo now i get what you mean so its 2F(sin(theda)) so basically i got 2*(sin(48.187))F which = 1.491F
 P: 300 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?
Mentor
P: 40,227
 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.)
 P: 300 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)
Mentor
P: 40,227
 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!
 P: 816 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?
 P: 300 THANK YOU ooo okay so lets see. (btw nice smiley face haha) okay so mg=156.96 so 1.491F=156.96 so F= 105.2963827Newtons? is that the correct answer?! THANK YOU
P: 300
 Quote by _Mayday_ 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?
i am not using a book. my physics teacher believes that we should not use books becuase it hinders your learning of physics. so he gives us random questions.
 P: 816 Yes but do you use a book in class to learn from or to use as a reference?

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