Download F = 6i + 4j

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!
(Slides will advance automatically - or
hit the space bar to advance slides.)
1
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
2
y
o
x
Begin with the coordinate system and
the positive axes x, y, and z.
z
3
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
4
y
+6
r = -2i + 6j + 3k
-2
x
Now move 6 units up in the positive
y -direction.
z
5
y
+3
+6
r = -2i + 6j + 3k
-2
x
Then move 3 units parallel to the
z -axis in the positive direction.
z
6
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
7
y
Follow the same procedure to
locate a force, F.
A
F = 6i + 4j - 1k
r
O
z
+6i
x
Along the x -axis, move 6 units to the
right in the positive direction.
8
y
A
+4j
r
O
z
x
+6i
F = 6i + 4j - 1k
Parallel to the y -axis, move 4 units up
in the positive direction.
9
y
A
-1k
+4j
r
O
z
x
+6i
F = 6i + 4j - 1k
Parallel to the z -axis, move 1 unit
toward the back, in the negative
direction.
10
y
A
-1k
F
r
O
z
+4j
x
+6i
F = 6i + 4j - 1k
This represents the position for the
force vector, F.
11
y
A
F
r
O
z
x
Note the orientation for the force
vector, F, and its relative position
with the distance vector, r.
12
y
A
F
r
O
z
x
Keeping the same orientation,
move the force vector, F, toward
point A.
13
y
A
F
r
O
z
x
Slide the force vector, F, until it
passes through point A.
14
y
F
A
r
O
z
x
Move the force, F, so that it acts through
point A. Note the orientation of F and r.
15
y
F
r
O
z
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.
x
The moment about the origin is
determined by vector
multiplication, or the CROSS
PRODUCT of these two vectors.
16
y
F
A
To determine the moment,
think about the starting point
and moving toward the force.
M =r x F
r
O
x
Distance from origin, O,
to point A.
r = OA = -2i + 6j + 3k
z
Force acting at point A,
F = 6i + 4j - 1k
17
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
18
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
(i)
i
j
k
-2
6
3
6
4
-1
6
3
4
-1
=
19
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
(i)
i
j
k
-2
6
3
6
4
-1
6
3
4
-1
=
- (j)
-2
3
6
-1
20
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
6
(i)
4
=
3
-1
-2
- (j)
6
3
-1
+ (k)
-2
6
6
4
21
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
6
(i)
4
=
3
-1
-2
- (j)
6
3
-1
+ (k)
-2
6
6
4
Now cross multiply the minor matrices.
22
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
6
(i)
4
3
-1
For each minor matrix, take the
product of these terms on this
diagonal ...
=
-2
- (j)
6
3
-1
+ (k)
-2
6
6
4
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }
23
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
6
(i)
4
3
-1
… then subtract the product of
these terms on this diagonal.
=
-2
- (j)
6
3
-1
+ (k)
-2
6
6
4
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }
24
where
M =r x F
r = OA = -2i + 6j + 3k
F = 6i + 4j - 1k
i
j
k
-2
6
3
6
4
-1
6
(i)
4
The final solution appears
below.
=
3
-2
-1
- (j)
6
3
-1
+ (k)
-2
6
6
4
(i) { (6x-1) - (3x4) } - (j) { (-2x-1) - (3x6) } + (k) { (-2x4) - (6x6) }
M = - 18i + 16j - 44k
25