ENGR-36_Lec-05_Dot_Product_H13e

Report
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
 
jk  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
3
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
4
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 
PP 
 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
 AB 

 AB 
  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
7
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
8
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
12
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt

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