ENGR-36 Lec-05 Fa12 Dot Product H13e
description
Transcript of ENGR-36 Lec-05 Fa12 Dot Product H13e
[email protected] • ENGR-36_Lec-05_Force_Resultants-2.ppt 1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PE Licensed Electrical & Mechanical Engineer
Engineering 36
Chp 4: Force
Resultants (2)
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Scalar (Dot) Product of 2 Vectors
The SCALAR Product or DOT Product Between Two Vectors P and Q Is Defined As
resultscalarcosPQQP
PQQP
2121 QPQPQQP
undefined SQP
Scalar Product Math Properties
• ARE Commutative
• ARE Distributive
• Are NOT Associative
– Undefined as (P•Q) is NO LONGER a Vector
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Scalar Product – Cartesian Comps
Scalar Products With Cartesian Unit Components
Thus
000111 ikkjjikkjjii
kQjQiQkPjPiPQP zyxzyx
2222PPPPPP
QPQPQPQP
zyx
zzyyxx
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Scalar Product - Applications Angle Between Two Vectors
PQ
QPQPQP
QPQPQPPQQP
zzyyxx
zzyyxx
cos
cos
OL
OL
PPQ
QP
PQQP
OLPPP
cos
cos
along of projection cos
zzyyxx
OL
PPP
PP
coscoscos
ˆ
OL alongr unit vecto theis
Projection Of A Vector On A Given Line
For Any Axis Defined By A Unit Vector
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Vector Magnitude by DOT
A vector DOTed with itself reveals the Square of the Phythagorean Length
Thus the Vector Magnitude
This is IDEAL forMATLAB
2222PPPPPP zyx
2222PPPP zyx PPP
>> Pv = [-7 3 11] % [Px*i Py*j Pz*k] Pv = -7 3 11 >> Pm = sqrt(dot(Pv,Pv)) Pm = 13.3791
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
DOT-Prod Application Summary
Given Two intersecting Vectors or Lines
Parallel & Perpendicular Components
• Given Vector VAB, and line AC find the || & ┴ Components of VAB, VAD & VDB, relative to line AC
1800arccos BA
BA
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
DOT-Prod Application Summary
First Calc θ by method of the previous slide
Then Simply Use Trig on Right-Triangle ADB
ACABACAB arccos
sin
cos
ABDB
ABAD
VV
VV
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
• Fperp
cos|| FF
sinFF
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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)
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
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s Work Some “Angle”
Problems
1
2
3
4
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
1050
1800
2400
1200
TBC = 5.3 kN
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Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
1050
1800
2400
1200
1050
1800
2400
1200