# Tension in rope

• Nov 30th 2010, 01:46 PM
nugiboy
Tension 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, 01:54 PM
Ackbeet
So you're asked to show (using somewhat more intuitive notation) that

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

Any vector whatsoever can be written as

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

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

Compare those two equations. What can you say? (There's actually a lot less to this problem than meets the eye.)
• Nov 30th 2010, 02:15 PM
nugiboy
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 Tha 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 Tha?
Hope that makes sense
• Nov 30th 2010, 02:16 PM
Ackbeet
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, 02:24 PM
nugiboy
Ahh i get it now! Blindingly obvious now i see it. Thanks alot.
• Nov 30th 2010, 02:26 PM
Ackbeet
Yeah, I said there was less to this problem than meets the eye. You're welcome. Have a good one!