1. Space Vectors Problem 3.19
Determine the moment about the origin of the coordinate system, given a force vector and its distance from the origin. This requires vector multiplication!
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2. Problem 3.19 Determine the moment about the origin.
Given a force vector, F = 6i + 4j - 1k
This force vector acts on a point, A.
Given the distance from the origin to point A, this distance is represented as vector, r = -2i + 6j + 3k

3. y x o
Begin with the coordinate system and the positive axes x, y, and z. z

4. y
Draw the position vector for r that defines the distance from the origin, O, to point A.
r = -2i + 6j + 3k -2 o x
Beginning at the origin, on the x -axis, move 2 units to the left for the negative direction. z

5. y +6
r = -2i + 6j + 3k -2 x
Now move 6 units up in the positive y -direction. z

6. y +3 +6
r = -2i + 6j + 3k -2 x
Then move 3 units parallel to the z -axis in the positive direction. z

7. y A r
r = -2i + 6j + 3k x O
This locates point A. Represent this distance from the origin to point A as vector r. z

8. y y
Follow the same procedure to locate a force, F. A
F = 6i + 4j - 1k r x x O
+6i
Along the x -axis, move 6 units to the right in the positive direction. z z

9. y y A
+4j r x x O
+6i
F = 6i + 4j - 1k
Parallel to the y -axis, move 4 units up in the positive direction. z z

10. y y A
-1k
+4j r x x O
+6i
F = 6i + 4j - 1k
Parallel to the z -axis, move 1 unit toward the back, in the negative direction. z z

11. y y A
-1k F
+4j r x x O
+6i
F = 6i + 4j - 1k
This represents the position for the force vector, F. z z

12. y y A F r x x O
Note the orientation for the force vector, F, and its relative position with the distance vector, r. z z

13. y y A F r x x O
Keeping the same orientation, move the force vector, F, toward point A. z z

14. y y A F r x x O
Slide the force vector, F, until it passes through point A. z z

15. y y F A r x x O
Move the force, F, so that it acts through point A. Note the orientation of F and r. z z

16. y y
This force has components that are a perpendicular distance from the origin. This force then produces a moment about the fixed origin, which acts as the pivot or hinge point. F r x x O
The moment about the origin is determined by vector multiplication, or the CROSS PRODUCT of these two vectors. z z

17. y y
To determine the moment, think about the starting point and moving toward the force.
M = r x F F A r x x O
Distance from origin, O, to point A.
r = OA = -2i + 6j + 3k
Force acting at point A,
F = 6i + 4j - 1k z z

18. M = r x F where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k i j k -2 6 3
-2 6 3
6 4 -1

19. M = r x F ( i ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k i j k
-2 6 3
6 4 -1 =
6 3
4 -1
( i )

20. M = r x F ( i ) - ( j ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k
-2 6 3
6 4 -1 =
6 3
4 -1
-2 3
6 -1
( i )
- ( j )

21. M = r x F ( i ) + ( k ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k
-2 6 3
6 4 -1 =
6 3
4 -1
-2 3
6 -1
-2 6
6 4
( i )
- ( j )
+ ( k )

22. M = r x F ( i ) + ( k ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k
-2 6 3
6 4 -1 =
6 3
4 -1
-2 3
6 -1
-2 6
6 4
( i )
- ( j )
+ ( k )
Now cross multiply the minor matrices.

23. M = r x F ( i ) + ( k ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k
-2 6 3
6 4 -1
For each minor matrix, take the product of these terms on this diagonal ... =
6 3
4 -1
-2 3
6 -1
-2 6
6 4
( i )
- ( j )
+ ( k )
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }

24. M = r x F ( i ) + ( k ) where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k
-2 6 3
6 4 -1
… then subtract the product of these terms on this diagonal. =
6 3
4 -1
-2 3
6 -1
-2 6
6 4
( i )
- ( j )
+ ( k )
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }

25. M = r x F ( i ) The final solution appears below. + ( k ) where
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
M = r x F
i j k
-2 6 3
6 4 -1
The final solution appears below. =
6 3
4 -1
-2 3
6 -1
-2 6
6 4
( i )
- ( j )
+ ( k )
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }
M = - 18i j k