Unit Vector Magnitudes and Forces

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nobodyuknow
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Homework Statement



http://prntscr.com/p7dxt

Here's a screenshot of the revision question.

Homework Equations


The Attempt at a Solution



Co-ordinates of each point
A, (0, 60, 0)
B, (40, 0, 0)
C, (-40, 0, 40)
D, (-60, 0, -60)

Position Vectors
rAB, 40i + -60j + 0k
rAC, -40i + -60j + 40k
rAD, -60i + -60j + -60k

Unit Vectors corresponding these position vectors
eAB, 0.5547i + -0.8321j + 0k
eAC, -0.4851i + -0.7276j + 0.4851k
eAD, -0.5774i + -0.5774j + -0.5774k

This is all I know, I'm not sure how to complete the other questions.

For question would I have to put the two position vectors and do a cross product?

Something like...

i j k
-0.4851 -0.7276 0.4851
-0.5774 -0.5774 -0.5774

[(-0.7276)(-0.5774) - (0.4851)(-0.5774)]i + [(-0.4851)(-0.5774) - (0.4851)(-0.5774)]j + [(-0.4851)(-0.5774) - (-0.7276)(-0.5774)]k

If not, I'm totally lost, and need help!
 
Last edited:
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I see no reason to consider cross products here. It's a simple statics question.
What components of the forces in the cables are of interest? What equations can you write down which say they balance out?
 
So essentially, balance out the forces...

Basically something like...

ForceTotal = ForceAB + ForceAC + ForceAD

2kN(eAB) = ForceAB = 1.1094i - 1.6642j + 0k
 
Last edited:
nobodyuknow said:
So essentially, balance out the forces...

Basically something like...

ForceTotal = ForceAB + ForceAC + ForceAD

2kN(eAB) = ForceAB = 1.1094i - 1.6642j + 0k
Forces in the y direction are not interesting. Whatever they add up to in tensions will be balanced by compression in the tower.
Your resolution of the 2kN into i and j looks right, but I'd rather you stuck with the algebraic symbols, like 'cos(θ)', not plugging in actual numbers until the end. It makes it much easier to follow what you're doing and spot any errors.
Create unknowns for the other tensions, write out their resolutions into i, j, k and hence the balance of forces equation.
 
So do you mean like...
ForceTotal = (eABi |FAB| + eACi |FAC| + eADi |FAD|)i + (eABj |FAB| + eACj |FAC| + eADj |FAD|)j + (eABi |FAB| + eACk |FAC| + eADk |FAD|)k

Which then becomes something like...

eACi |FAC| + eADi |FAD|)i = eABi |FAB|
eACj |FAC| + eADj |FAD|)j = eABj |FAB|
eACk |FAC| + eADk |FAD|)k = eABk |FAB|
 
Yes, except that as I mentioned you cannot write a useful equation for the vertical forces. That would involve the compression in the tower, which you don't care about.