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

[email protected]

Engineering 36

Chp 4: Force

Resultants (2)



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



Scalar Product – Cartesian Comps

Scalar Products With Cartesian Unit Components

Thus

000111 ikkjjikkjjii

kQjQiQkPjPiPQP zyxzyx

2222PPPPPP

QPQPQPQP

zyx

zzyyxx



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



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



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



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



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



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



WhiteBoard Work

Let’s Work Some “Angle”

Problems

1

2

3

4



1050

1800

2400

1200

TBC = 5.3 kN



1050

1800

2400

1200

1050

1800

2400

1200

engr-36 lec-05 fa12 dot product h13e

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