engineering mechanics statics - rami zakaria · 3. force vectors; position vector & dot product...
TRANSCRIPT
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3. Force Vectors;
Position Vector & Dot product
Engineering Mechanics
Statics
Mechanical Systems Engineering -2016
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Position Vector
A position vector is a fixed vector
that locates a point in space relative
to another point.
Consider two points, A and B, in a 3-D
space. Let their coordinates be (xA, yA, zA)
and (xB, yB, zB ), respectively.
The position vector directed from A to B, r AB , is defined as:
Note that B is the ending point and A is the starting point. ALWAYS subtract the
“tail” coordinates from the “tip” coordinates!
kzzjyyixxr ABABABAB
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Dot Product
The dot product of vectors A and B is defined as:
The angle is the smallest angle between the two
vectors and is always in a range of 0º to 180º.
cosBABA
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Dot Product Characteristics:
1. The result of the dot product is a scalar (a positive or negative
number).
2. The units of the dot product will be the product of the units of the
A and B vectors.
By definition:
zzyyxx
zyxzyx
BABABA
kBjBiBkAjAiABA
kkjjii
kjkiji
1,1,1
0,0,0
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The dot product can be used to determine the angle between two vectors
For the given two vectors in the Cartesian form, one
can find the angle by:
a) Finding the dot product,
b) Finding the magnitudes (||A|| & ||B||) of the vectors A & B, and
c) Using the definition of dot product and solving for , i.e.,
= cos-1 [(A • B) / (||A|| ||B||)], where 0º 180º .
Question . Find the dot product of the two vectors P and Q.
P = {5 i + 2 j + 3 k} m
Q = {-2 i + 5 j + 4 k} m
A) -12 m B) 12 m C) 12 m 2
D) -12 m 2 E) 10 m 2
What is the
angle between
the two vectors?
zzyyxx BABABABA
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Force Vector Directed Along a Line
a) Find the position vector, rAB , along two points on that line.
b) Find the unit vector describing the line’s direction,
uAB = (rAB / ||rAB||).
c) Multiply the unit vector by the magnitude of the force,
F = ||F|| uAB .
If a force is directed along a line,
then we can represent the force
vector in Cartesian coordinates by
using a unit vector and the force’s
magnitude. So we need to:
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Example:
Plan:
1. Find the position vector rAC and then the unit vector uAC.
2. Obtain the force vector as: FAC = (420 N) (uAC ).
Given: The 420 N force
along the cable AC.
Find: The force FAC in the
Cartesian vector form.
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Solution:
(We can also find rAC by subtracting the
coordinates of A from the coordinates of C.)
||rAC||= (22 + 32 + 62)1/2 = 7 m
Now: uAC = rAC / ||rAC|| and FAC = {420 uAC } N = {420 (rAC / ||rAC||)} N
So FAC = 420{ (2 i + 3 j 6 k) / 7 } N
= {120 i + 180 j - 360 k } N
From the figure, when relating A to C, we
will have to go 2 m in the x-direction, 3
m in the y-direction, and -6 m in the z-
direction. Hence,
rAC = {2 i + 3 j 6 k} m.
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Plan:
1. Find position vector rAO
2. Find the angle = cos-1{(F • rAO)/(||F|| ||rAO||)}
3. Find the projection via ||FAO||= ||F|| cos
Given: The force acting on the hook
at point A.
Find: (1)The angle between the
force vector and the line AO,
(2) The magnitude of the
projection of the force
along the line AO.
Example:
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Solution:
rAO = {1 i + 2 j 2 k} m
||rAO || = (12 + 22 + 22)1/2 = 3 m
F = { 6 i + 9 j + 3 k} kN
||F|| = (62 + 92 + 32)1/2 = 11.22 kN
= cos-1{(F • rAO) / (||F|| ||rAO||)}
= cos-1 {18 / (11.22 * 3)} = 57.67°
F • rAO = ( 6)(1) + (9)(2) + (3)(2) = 18 kN.m
|| FAO ||= ||F|| cos = 11.22 cos (57.67°) = 6.00 kN
Or:
uAO = rAO / ||rAO|| = {( 1/3) i + (2/3) j + ( 2/3) k}
|| FAO ||= F • uAO = ( 6)( 1/3) + (9)(2/3) + (3)( 2/3) = 6.00 kN
Vector
magnitude
Vector
magnitude
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1. The dot product of two vectors P and Q is
defined as
A) ||P|| ||Q|| cos B) ||P|| ||Q|| sin
C) ||P|| ||Q|| tan D) ||P|| ||Q|| sec
P
Q
Questions:
2. The dot product of two vectors results in a _________
quantity.
A) Scalar B) Vector
C) Complex D) Zero
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
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Engineering Mechanics: Statics, Twelfth Edition
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Engineering Mechanics: Statics, Twelfth Edition
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Engineering Mechanics: Statics, Twelfth Edition
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler
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Engineering Mechanics: Statics, Twelfth Edition
Russell C. Hibbeler