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

physicsbhelp

SOMEONE please help me with this problem!!! thanks i appreaciate all the effort/help!

Doc Al

You need to learn how to find the components of a vector. If you have some vector with magnitude F that makes an angle [itex]\theta[/itex] with the x-axis (the horizontal), the components are:

• vertical (y) component = [itex]F \sin\theta[/itex]
  
• horizontal (x) component = [itex]F \cos\theta[/itex]

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.

physicsbhelp

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

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

No. What angle does each string make with the horizontal?

physicsbhelp

48.187

Doc Al

Good! Now use what I told you in post #36.

physicsbhelp

so sin(48.187) + cos (48.187) = 0

physicsbhelp

wait no sorry it should be: sin(48.187) + cos (48.187) = mg

physicsbhelp

right?

Doc Al

No. In post #36, I told you that:
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

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

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

Exactly! Now set that equal to what?

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

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

Yes! Then solve for the tension, F.
No!

_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?