ENGR-36_Lec-05_Dot_Product_H13e
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Transcript ENGR-36_Lec-05_Dot_Product_H13e
Engineering 36
Chp 4: Force
Resultants (2)
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
Scalar (Dot) Product of 2 Vectors
The SCALAR Product or
DOT Product Between Two
Vectors P and Q Is Defined As
P Q PQ cos
scalar result
Scalar Product Math Properties
• ARE Commutative
• ARE Distributive
• Are NOT Associative
P Q Q P
P Q 1 Q 2 P Q 1 P Q 2
P Q S undefined
– Undefined as (P•Q) is NO LONGER a Vector
Engineering-36: Engineering Mechanics - Statics
2
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
Scalar Product – Cartesian Comps
Scalar Products With Cartesian Unit
Components
P Q Px i Py j Pz k Q x i Q y j Q z k
i i 1
j j 1
k k 1
i j 0
jk 0
Thus
P Q Px Q x P y Q y Pz Q z
2
2
2
2
P P Px P y Pz P
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
k i 0
Scalar Product - Applications
Angle Between Two Vectors
P Q PQ cos Px Q x Py Q y Pz Q z
cos
Px Q x Py Q y Pz Q z
PQ
Projection Of A Vector On
A Given Line
POL P cos projection
P Q PQ cos
P Q
P cos POL
Q
of P along OL
For Any Axis Defined By A Unit Vector
POL
P ˆ
is
the unit vecto
r along OL
Px cos x Py cos y Pz cos z
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
Vector Magnitude by DOT
A vector DOTed with itself reveals the
Square of the
Phythagorean Length
P P Px Py Pz P
2
2
2
2
Thus the Vector Magnitude
P
PP
This is IDEAL
forMATLAB
Engineering-36: Engineering Mechanics - Statics
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P P P
2
x
2
y
2
z
P
2
>> Pv = [-7 3 11] % [Px*i Py*j Pz*k]
Pv =
-7
3
11
>> Pm = sqrt(dot(Pv,Pv))
Pm =
13.3791
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
DOT-Prod Application Summary
Given Two intersecting
Vectors or Lines
AB
AB
arccos
0 180
Parallel & Perpendicular
Components
• Given Vector VAB, and line
AC find the || & ┴
Components of VAB, VAD & VDB,
relative to line AC
Engineering-36: Engineering Mechanics - Statics
6
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
DOT-Prod Application Summary
First Calc θ by method of
the previous slide
arccos
AB AC AB AC
Then Simply Use Trig
on Right-Triangle ADB
V AD V AB cos
V DB V AB sin
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
Example: P2-120 by MATLAB
Determine the
magnitudes of the
components of
F = 600N acting along
and perpendicular to
segment DE of the pipe
assembly
Notes
• The Angle θ between
DE & EB (the
direction of F)
appears to be
OBTUSE
• Fpar
F|| F cos
• Fperp
F F sin
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
Example: P2-120 by MATLAB
% Bruce Mayer, PE
% ENGR36 * 18Jul2
% ENGR36_parNperp_Projection_H13e_P2_120_1207.m
%
% Magnitude of a vector by ANON fcn
MagV = @(z) sqrt(dot(z,z))
%
% Find unit vector along EB, the Force Direction
EBv = [-4 -3 2] % in m => [delX*i delY*j delZ*k]
EVm = MagV(EBv)
uEB = EBv/EVm
%
% Find unit Vector along Pipe Segment DE
DEv = [0 3 0]
DEm = MagV(DEv)
uDE = DEv/DEm
%
% Angle between the unit vectors
Q = acosd(dot(uEB,uDE))% in °
%
Fm = 600 % in Newtons
%
% the PARALLEL projection of F on DE
Fpar = Fm*cosd(Q)
% the PERPENDICULAR projection of F on DE
Fperp = Fm*sind(Q)
%
disp(' ')
disp('======================================')
disp('Chk by finding F against ED (the opposite of DE)')
% Find unit Vector along Pipe Segment DE
EDv = [0 -3 0]
EDm = MagV(EDv)
uED = EDv/EDm
%
Qchk = acosd(dot(uEB,uED))% in °
FparChk = Fm*cosd(Qchk)
FperpChk = Fm*sind(Qchk)
Engineering-36: Engineering Mechanics - Statics
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Q =
123.8545
Fpar =
-334.2516
Fperp =
498.2729
====================================
Chk by finding F against ED (the
opposite of DE)
Qchk =
56.1455
FparChk =
334.2516
FperpChk =
498.2729
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
WhiteBoard Work
1
4
Let’s Work
Some “Angle”
Problems
2
3
Engineering-36: Engineering Mechanics - Statics
10
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
1800
2400
1050
TBC =
5.3 kN
1200
Engineering-36: Engineering Mechanics - Statics
11
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt
18 00
2400
10 50
12 00
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt