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Tension in Strings HELP thanks!-very simple.

by physicsbhelp
Tags: simple, solved, strings, tension, thanksvery
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physicsbhelp
#37
Feb11-08, 06:37 PM
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?
physicsbhelp
#38
Feb11-08, 06:40 PM
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
Doc Al
#39
Feb11-08, 06:56 PM
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No. What angle does each string make with the horizontal?
physicsbhelp
#40
Feb11-08, 06:58 PM
P: 300
48.187
Doc Al
#41
Feb11-08, 07:00 PM
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Quote Quote by physicsbhelp View Post
48.187
Good! Now use what I told you in post #36.
physicsbhelp
#42
Feb11-08, 07:03 PM
P: 300
so sin(48.187) + cos (48.187) = 0
physicsbhelp
#43
Feb11-08, 07:03 PM
P: 300
wait no sorry it should be: sin(48.187) + cos (48.187) = mg
physicsbhelp
#44
Feb11-08, 07:08 PM
P: 300
right?
Doc Al
#45
Feb11-08, 07:11 PM
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No. In post #36, I told you that:
Quote Quote by Doc Al View Post
vertical (y) component = [itex]F \sin\theta[/itex]
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.
physicsbhelp
#46
Feb11-08, 07:13 PM
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
physicsbhelp
#47
Feb11-08, 07:14 PM
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?
Doc Al
#48
Feb11-08, 07:26 PM
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Quote Quote by physicsbhelp View Post
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 Quote by physicsbhelp View Post
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.)
physicsbhelp
#49
Feb11-08, 07:30 PM
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)
Doc Al
#50
Feb11-08, 07:32 PM
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Quote Quote by physicsbhelp View Post
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!
_Mayday_
#51
Feb11-08, 07:32 PM
P: 815
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?
physicsbhelp
#52
Feb11-08, 07:34 PM
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
physicsbhelp
#53
Feb11-08, 07:35 PM
P: 300
Quote Quote by _Mayday_ View Post
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.
_Mayday_
#54
Feb11-08, 07:36 PM
P: 815
Yes but do you use a book in class to learn from or to use as a reference?


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