Unit Vector Magnitudes and Forces

In summary: So the equations you need to solve are just the two that you wrote in your last reply. (But again, I discourage the use of actual numbers until the end. Keep the unknowns in symbolic form as much as possible.)In summary, the conversation involves a question about finding the position vectors and unit vectors for specific points and the corresponding forces in a statics problem. The solution involves writing equations for the balance of forces and using unknowns for the tensions in the cables. Actual numbers should not be used until the end.
  • #1
nobodyuknow
64
0

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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  • #2
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?
 
  • #3
So essentially, balance out the forces...

Basically something like...

ForceTotal = ForceAB + ForceAC + ForceAD

2kN(eAB) = ForceAB = 1.1094i - 1.6642j + 0k
 
Last edited:
  • #4
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.
 
  • #5
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|
 
  • #6
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.
 

Related to Unit Vector Magnitudes and Forces

1. What is a unit vector?

A unit vector is a vector with a magnitude of 1. It is used to represent direction and does not change when multiplied by a scalar.

2. How is a unit vector calculated?

To calculate a unit vector, divide each component of the vector by its magnitude. This will result in a vector with a magnitude of 1 and the same direction as the original vector.

3. What is the significance of unit vector magnitudes?

Unit vector magnitudes are important because they allow us to represent direction without being affected by the scale of the vector. This makes it easier to compare and analyze vectors.

4. How are unit vectors used in force calculations?

Unit vectors are used in force calculations to represent the direction of a force. By breaking down a force vector into its components, we can use unit vectors to determine the direction and magnitude of each component and calculate the overall force.

5. Can unit vectors be negative?

No, unit vectors cannot be negative. Since their magnitude is always 1, they only represent direction and do not have a negative or positive value.

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