Solving Cable Tension and Moment Magnitude Problem

[sapphire4770]

Homework Statement

http://books.google.ca/books?id=nYR... you want the magnitude of the moment&f=false
The above link has a copy of the question. It is question 4.65 on page 144

If you can't see the picture here is the needed information:
pointA(0,8,0)-on tree
pointB(0,0,10)
pointC(14,0,14)

There is a cable from A to B with a tension of 100lb and onother cable from A to C with an unknown tention. If the magnitude of the moment about the origin(0,0,0) is 1500 ft-lb, what is the tension on rope AC?

Homework Equations

M=r * F

The Attempt at a Solution

I know that the moment is equal to the position vector *(cross) the force in cartesian form, but how do you express the 100lb foce in rope AB in cartesian form? Do i even need to?

Also to find the final answer I am thining you have the equation:

1500 = the sum of both moment forces

 

[rl.bhat]

Hi sapphire4770, welcome to PF.
Position vector OA = 0i+8j+0k
Position vector OB = 0i +0j + 10k
Position vector OC = 14i + 0j + 14K
Vector AB = OB - OA. Similarly
Vector AC = OC - OA.
Now find the moments,

 

[sapphire4770]

Thanks for the help, but I'm still kinda confused.

Ok, so the vectors would be:
AB: [0,-8,10] or -8j +10k
AC: [14,-8,14] or 14i - 8i + 14k

To find the moment caused by AC do i simply multiply? 100[0,-8,10]? Or do i have to split up the 100 lb tension into its x y and z components and find the cross product?

Thanks again, any help is greatly appreciated!

 

[rl.bhat]

Find the vector OA. = 8(0i + j + 0k)
Find unit vector along AB and AC.
So the vector AB = 100/sqrt(164)[0i -8j + 10k]
The moment of AB about o is OAXAB.
Similarly find the vector AC and moment of AC about O. Since both of them are in the counterclockwise direction, add them to get net momentum.