http://i137.photobucket.com/albums/q220/e3tiger/1a.jpg

Could someone give me a hint to start me on part a.

Thanks

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- Nov 30th 2010, 12:46 PMnugiboyTension in rope
http://i137.photobucket.com/albums/q220/e3tiger/1a.jpg

Could someone give me a hint to start me on part a.

Thanks - Nov 30th 2010, 12:54 PMAckbeet
So you're asked to show (using somewhat more intuitive notation) that

$\displaystyle \mathbf{T}_{HA}=\dfrac{T_{HA}}{13}\,\langle 0,-12,-5\rangle.$

Any vector whatsoever can be written as

$\displaystyle \mathbf{A}=A\hat{a},$

where $\displaystyle |\mathbf{A}|=A$ and $\displaystyle \hat{a}$ is a unit vector pointing in the same direction as $\displaystyle \mathbf{A}.$

Compare those two equations. What can you say? (There's actually a lot less to this problem than meets the eye.) - Nov 30th 2010, 01:15 PMnugiboy
Hmm ok so i understand that a vector can be written as the product of its magnitude and unit vector in the same direction.

So in this case, the vector of tension**T**ha can be written as its magnitude Tha multiplied by its unit vector. How do i find it's unit vector though? We have only been given the position vectors of the two ends

of the rope, which gives us the vector between H and A which is (0, -12, -5).

How does this vector relate to the force**T**ha?

Hope that makes sense - Nov 30th 2010, 01:16 PMAckbeet
Well, here are two thoughts.

1. In which direction is the tension in the HA rope acting?

2. What is the magnitude of the vector (0,-12,-5)? - Nov 30th 2010, 01:24 PMnugiboy
Ahh i get it now! Blindingly obvious now i see it. Thanks alot.

- Nov 30th 2010, 01:26 PMAckbeet
Yeah, I said there was less to this problem than meets the eye. You're welcome. Have a good one!