Space Vectors Problem 3.19

1. Space VectorsProblem 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! (Slides will advance automatically - or hit the space bar to advance slides.)

2. Problem 3.19Determine 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 Aas 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 pointand 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. where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k M = r x F i j k -2 6 3 6 4 -1

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

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

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

22. where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k M = r x F i j k -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. where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k M = r x F i j k -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. where r = OA = -2i + 6j + 3k F = 6i + 4j - 1k M = r x F i j k -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. 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 + 16j - 44k