Vector Blueprints Slides Module 01
Vector Blueprints
Bridging the gap between geometric intuition and algebraic precision in physical systems.
ENG-PHYS // VECTOR ANALYSIS // 1.1
The Navigation Dilemma
An aircraft aims for a heading of 090° (Due East) at an airspeed of 250 knots.
"A crosswind of 40 knots is blowing from the North (180°)."
Why can't we just add 250 + 40?
Scalars only account for magnitude.
Physical reality is directional.
Thought Experiment
Sketch the resultant path. Where does the plane actually end up?
What is a Vector?
Geometric vs. Algebraic Identity
The Geometric View
A directed line segment with a specific magnitude (length) and direction.
NOTATION
\[ \vec{A} \] or \[ \mathbf{A} \]
The Algebraic View
An ordered set of numbers (components) representing shifts along coordinate axes.
NOTATION
\[ \langle A_x, A_y, A_z \rangle \]
The Blueprint Conversion
Polar → Cartesian
Horizontal Component
\[ A_x = |\vec{A}| \cos(\theta) \]
Vertical Component
\[ A_y = |\vec{A}| \sin(\theta) \]
The Standard Angle Rule
In mathematics, \( \theta \) is measured counter-clockwise from the positive x-axis.
In engineering, we often define angles relative to surfaces or gravity. Context is king.
"Geometry tells you what to draw; Algebra tells you how to compute."
Reconstructing the Whole
Cartesian → Polar
Magnitude (Norm)
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2} \]
The Pythagorean Theorem in action.
Direction (Angle)
\[ \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right) \]
Careful with the quadrant! (use atan2)
Coordinate Independence
The vector itself exists in physical space. Our components are just one way to describe it based on our choice of axes.
Physics doesn't care about your x-axis.
Vector Foundations Worksheet Vector Blueprints
PHYS-201: Vector Analysis & Mechanics
Name:
Date:
Technical Specifications
Vectors are the language of physical systems. In this worksheet, you will practice the translation between geometric (visual) and algebraic (numerical) descriptions of 2D vectors. Ensure all angles are measured from the standard positive x-axis unless otherwise specified.
1
Scalar vs. Vector Inventory
Classify each physical quantity below and provide a brief justification (e.g., "Direction matters" or "Magnitude only").
A plane flying at 450 mph West.
Scalar
Vector
The 15 kg mass of a steel girder.
Scalar
Vector
A force of 200 N pushing downwards.
Scalar
Vector
2
Component Resolution (Polar → Cartesian)
Calculate the horizontal (\(A_x\)) and vertical (\(A_y\)) components for the following vectors. Round to two decimal places.
Vector A
\( |\vec{A}| = 12.0 \text{ units}, \theta = 30^\circ \)
\( A_x \)
\( A_y \)
Vector B
\( |\vec{B}| = 8.5 \text{ units}, \theta = 135^\circ \)
\( B_x \)
\( B_y \)
3
Vector Reconstruction (Cartesian → Polar)
Determine the magnitude (\(M\)) and standard angle (\(\theta\)) for the vectors given by their components.
Vector C
\( \vec{C} = \langle 4, -3 \rangle \)
Magnitude \( |\vec{C}| \)
Direction \( \theta \) (degrees)
Vector D
\( \vec{D} = \langle -10, -10 \rangle \)
Magnitude \( |\vec{D}| \)
Direction \( \theta \) (degrees)
4
Geometric Synthesis
On the grid provided, sketch the vector \( \vec{R} \) where:
Assume each grid square is 1 unit. Clearly label the origin and the terminal point.
Verification Calculations
Show your mathematical breakdown for the vector sketched to the left.
End of Technical Blueprint Phase 1 // System Verification Required
Vector Foundations Answer Key Answer Key
Vector Foundations Worksheet // Instructor Reference
Solutions Guide
1
Scalar vs. Vector Inventory
450 mph West
Scalar Vector
Includes both magnitude and a specific spatial direction.
15 kg mass
Scalar Vector
Mass has no direction; it is a scalar quantity of matter.
200 N downwards
Scalar Vector
Force is inherently directional (weight in this case).
2
Component Resolution
\( |\vec{A}| = 12.0, \theta = 30^\circ \)
\( A_x = 12 \cos(30^\circ) = 10.39 \)
\( A_y = 12 \sin(30^\circ) = 6.00 \)
\( |\vec{B}| = 8.5, \theta = 135^\circ \)
\( B_x = 8.5 \cos(135^\circ) = -6.01 \)
\( B_y = 8.5 \sin(135^\circ) = 6.01 \)
3
Vector Reconstruction
\( \vec{C} = \langle 4, -3 \rangle \)
\( |\vec{C}| = \sqrt{4^2 + (-3)^2} = 5.00 \)
\( \theta = \tan^{-1}(-3/4) = -36.87^\circ \text{ or } 323.13^\circ \)
\( \vec{D} = \langle -10, -10 \rangle \)
\( |\vec{D}| = \sqrt{(-10)^2 + (-10)^2} = 14.14 \)
\( \theta = \tan^{-1}(-10/-10) + 180^\circ = 225^\circ \) (Quadrant III check)
4
Geometric Synthesis
Vector R (pointing SW)
\( R_x = 15 \cos(210^\circ) = -12.99 \)
\( R_y = 15 \sin(210^\circ) = -7.50 \)
The vector should be drawn starting at origin (0,0) and ending at approximately (-13, -7.5). It is located in the 3rd Quadrant.
Vector Assembly Slides Module 02
Vector Assembly Line
Standardizing operations: Arithmetic, Scalar Scaling, and the Unit Vector "Quality Control".
The "Unit" Standard
What is a Unit Vector?
A vector with a magnitude of exactly 1. Its only job is to point.
\[ \hat{u} = \frac{\vec{v}}{|\vec{v}|} \]
"Dividing a vector by its own length 'normalizes' it, stripping away the scale but keeping the orientation."
î
Points along the positive x-axis.
ĵ
Points along the positive y-axis.
k̂
Points along the positive z-axis.
Assembling the Resultant
Geometric: Tip-to-Tail
Place the tail of the second vector at the tip of the first. The resultant starts at the beginning and ends at the finish line.
Algebraic: Component-Wise
\[ \vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} \]
This is the "Assembly Line" way. Add each piece independently.
Why Algebraic?
Scales to 100+ vectors easily.
Precision over drawing errors.
Computers love component addition.
Scaling Operations
Scaling & Reflection
Magnify
If \( s > 1 \), the vector stretches. Direction remains same.
\[ s\vec{A} \text{ (stretch)} \]
Contract
If \( 0 < s < 1 \), the vector shrinks. Direction remains same.
\[ s\vec{A} \text{ (shrink)} \]
Reflect
If \( s < 0 \), the vector flips 180°.
\[ -1\vec{A} \text{ (flip)} \]
Resultant Challenge
Five tugboats are pulling a stranded tanker. Each pulls with a different force and at a different angle.
Your Mission
"Design a single 'Mega-Tug' that could replace all fiveboats, maintaining the exact same force and direction."
Vector Assembly Worksheet The Assembly Line
OPERATIONS // UNIT VECTORS // ARITHMETIC
Operator ID:
QC-01
Quality Control: Vector Normalization
"Convert each raw vector into a standard unit vector \( \hat{u} \). Check that its magnitude is exactly 1."
RAW VECTOR A
\( \vec{A} = 3\hat{i} - 4\hat{j} \)
Step 1: Magnitude
Step 2: Unit Vector \( \hat{a} \)
RAW VECTOR B
\( \vec{B} = \langle 1, 1 \rangle \)
Step 1: Magnitude
Step 2: Unit Vector \( \hat{b} \)
ASSEMBLY-02
Linear Assembly (Arithmetic)
Given the base components: \( \vec{U} = 5\hat{i} + 2\hat{j} \) and \( \vec{V} = -3\hat{i} + 8\hat{j} \). Perform the following operations.
\( \vec{R}_1 = \vec{U} + \vec{V} \)
Calculation Area
\( \vec{R}_2 = 2\vec{U} - \vec{V} \)
Calculation Area
\( |\vec{R}_2| \)
Final Magnitude Result
STRESS-03
Industrial Stress Test
The Mooring Problem
A vessel is held by three mooring lines. Each exerts a force vector:
\( F_1 = \langle 400, 200 \rangle \text{ N} \)
Task: Find the Total Resultant Force (\( \vec{F}_{net} \)) and its direction.
CONFIDENTIAL // ASSEMBLY LINE PRODUCTION // UNIT 2
Vector Assembly Answer Key Assembly Solution Guide
OPERATIONS // UNIT VECTORS // ARITHMETIC
QC-01
Normalization Solutions
\( \vec{A} = 3\hat{i} - 4\hat{j} \)
1. \( |\vec{A}| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5 \)
2. \( \hat{a} = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} = 0.6\hat{i} - 0.8\hat{j} \)
\( \vec{B} = \langle 1, 1 \rangle \)
1. \( |\vec{B}| = \sqrt{1^2 + 1^2} = \sqrt{2} \approx 1.414 \)
2. \( \hat{b} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \approx 0.707\hat{i} + 0.707\hat{j} \)
ASSEMBLY-02
Arithmetic Solutions
Base vectors: \( \vec{U} = 5\hat{i} + 2\hat{j} \) and \( \vec{V} = -3\hat{i} + 8\hat{j} \)
\( \vec{R}_1 = \vec{U} + \vec{V} \)
\( (5 - 3)\hat{i} + (2 + 8)\hat{j} = \mathbf{2\hat{i} + 10\hat{j}} \)
\( \vec{R}_2 = 2\vec{U} - \vec{V} \)
\( 2(5\hat{i} + 2\hat{j}) - (-3\hat{i} + 8\hat{j}) = (10 + 3)\hat{i} + (4 - 8)\hat{j} = \mathbf{13\hat{i} - 4\hat{j}} \)
\( |\vec{R}_2| \)
\( \sqrt{13^2 + (-4)^2} = \sqrt{169 + 16} = \sqrt{185} \approx \mathbf{13.60 \text{ units}} \)
STRESS-03
Mooring Problem Solution
Component Summation
\( \Sigma F_x = 400 - 100 - 200 = 100 \text{ N} \)
\( \Sigma F_y = 200 + 600 - 300 = 500 \text{ N} \)
\( \vec{F}_{net} = 100\hat{i} + 500\hat{j} \text{ N} \)
Magnitude & Angle
\( |\vec{F}_{net}| = \sqrt{100^2 + 500^2} = \sqrt{260,000} \approx \mathbf{509.9 \text{ N}} \)
\( \theta = \tan^{-1}(500/100) = \mathbf{78.69^\circ} \)
Power Projection Slides Module 03
The Power Projection
Dot Products: Decoding the interaction between vectors and the geometry of work.
Interaction & Alignment
Why do we multiply vectors this way?
The Dot Product measures how much one vector aligns with another.
The Geometric Definition
\[ \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos(\theta) \]
"If vectors are perpendicular (\(90^\circ\)), the product is zero. They don't 'see' each other."
The Algebraic Definition
\[ A_x B_x + A_y B_y + A_z B_z \]
SCALAR
The Output
VECTOR
The Input
The Shadow Principle
Projection is finding the "shadow" of one vector onto another.
The Scalar Projection (Length)
\[ \text{comp}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]
The Vector Projection
\[ \text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b} \]
Decomposition
This allows us to split a force into parallel and perpendicular components relative to any surface.
Application: The Inclined Plane
1
Gravity pulls straight down (\( -mg\hat{j} \)).
2
The surface only feels the normal projection (perpendicular).
3
The object only accelerates from the tangential projection (parallel).
Efficiency Check
"Why push down on a lawnmower handle?"
Work = \(\vec{F} \cdot \vec{d}\)
Only the component of force in the direction of motion does work.
The Angle Detective
The dot product is the fastest way to find the angle between two arbitrary vectors in space.
\[ \theta = \cos^{-1}\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \right) \]
Dot = 0 → 90°
Dot > 0 → Acute
Dot < 0 → Obtuse
Vector Projections Worksheet The Power Projection
MODULE 03 // DOT PRODUCTS // WORK // ENERGY
Team:
Efficiency Rating: [ ]%
Mission Briefing
Not all forces contribute equally to a system's motion. In this lab, you will use the dot product to determine the efficiency of force application and calculate the "work" performed along specific trajectories.
03.1
Angle Detective
Calculate the dot product of the following vector pairs and use it to find the angle between them.
PAIR A
\( \vec{U} = 4\hat{i} + 3\hat{j} \)
\( \vec{V} = 2\hat{i} - 5\hat{j} \)
1. Dot Product (\( \vec{U} \cdot \vec{V} \))
2. Angle (\( \theta \))
PAIR B
\( \vec{A} = \langle 10, 0 \rangle \)
\( \vec{B} = \langle -3, 6 \rangle \)
1. Dot Product (\( \vec{A} \cdot \vec{B} \))
2. Angle (\( \theta \))
03.2
Work & Flow
Scenario: Cargo Transport
A crate is pulled 15 meters along a flat floor (Displacement \( \vec{d} = 15\hat{i} \)). The cable pulling the crate exerts a force of \( \vec{F} = 40\hat{i} + 30\hat{j} \text{ Newtons} \).
Analytical Task
Calculate the total work done by the force. Recall that \( W = \vec{F} \cdot \vec{d} \).
Efficiency Discussion
What is the "wasted" component of the force? (The component that does zero work).
03.3
The Inclined Plane (Projection)
The Setup:
A block with weight \( \vec{G} = \langle 0, -100 \rangle \text{ N} \) sits on a ramp. The ramp surface points in the direction of the unit vector \( \hat{u}_{ramp} = 0.866\hat{i} + 0.5\hat{j} \).
Find the Scalar Projection (\( \text{comp}_{\hat{u}} \vec{G} \)) of gravity onto the ramp.
Sketch the force decomposition here showing the parallel (\( F_{||} \)) and perpendicular (\( F_{\perp} \)) components of gravity.
Power Analysis Protocol // System Energy Verified // Module 03 Complete
Vector Projections Answer Key Power Solutions
MODULE 03 // DOT PRODUCTS // WORK // ANSWER KEY
Instructor Key
03.1
Angle Detective Solutions
PAIR A (\( \vec{U} = 4\hat{i} + 3\hat{j}, \vec{V} = 2\hat{i} - 5\hat{j} \))
Dot Product: \( 4(2) + 3(-5) = 8 - 15 = \mathbf{-7} \)
Magnitudes: \( |\vec{U}|=5, |\vec{V}|=\sqrt{29} \approx 5.385 \)
Angle: \( \cos\theta = \frac{-7}{5 \cdot 5.385} \approx -0.26 \)
\( \theta = \mathbf{105.07^\circ} \) (Obtuse)
PAIR B (\( \vec{A} = \langle 10, 0 \rangle, \vec{B} = \langle -3, 6 \rangle \))
Dot Product: \( 10(-3) + 0(6) = \mathbf{-30} \)
Magnitudes: \( |\vec{A}|=10, |\vec{B}|=\sqrt{45} \approx 6.708 \)
Angle: \( \cos\theta = \frac{-30}{10 \cdot 6.708} \approx -0.447 \)
\( \theta = \mathbf{116.57^\circ} \)
03.2
Work & Flow Solutions
Analytical Calculation
\( \vec{F} = \langle 40, 30 \rangle, \vec{d} = \langle 15, 0 \rangle \)
\( W = 40(15) + 30(0) = \mathbf{600 \text{ Joules}} \)
Only the horizontal component contributes to work.
Efficiency Discussion
The \( 30\hat{j} \) component is wasted.
This component is perpendicular to the displacement (\( 90^\circ \)). Work done is zero. It only serves to reduce the normal force on the crate.
03.3
Inclined Plane Solutions
Calculation:
\( \vec{G} = \langle 0, -100 \rangle \text{ N}, \hat{u} = \langle 0.866, 0.5 \rangle \)
\( \text{comp}_{\hat{u}} \vec{G} = \vec{G} \cdot \hat{u} = 0(0.866) + (-100)(0.5) = \mathbf{-50 \text{ N}} \)
The negative sign indicates the force acts opposite to the direction of \(\hat{u}\) (down the ramp).
Graphic Confirmation:
Gravity (vertical) is decomposed into a component along the ramp (parallel) and into the ramp (normal). The parallel component is \( G \sin(30^\circ) \approx 50 \text{ N} \).
Dynamic Navigators Slides Module 04
Dynamic Navigators
Mastering the Vector Equation of Motion: Relative Velocity and Shifting Reference Frames.
Tracking Active
The Frame Chain
Motion is never absolute. It depends on who is watching.
The Relative Velocity Equation
\[ \vec{v}_{ac} = \vec{v}_{ab} + \vec{v}_{bc} \]
"Velocity of A relative to C equals velocity of A relative to B plus B relative to C."
Aviation
Groundspeed = Airspeed + Wind
Maritime
Track = Heading + Current
Interception
Closing Velocity = \( \vec{v}_{target} - \vec{v}_{chaser} \)
The Power of Subtraction
Finding the velocity of Object A as seen by Object B.
\[ \vec{v}_{rel} = \vec{v}_a - \vec{v}_b \]
"Final minus Initial in the frame of reference."
Geometric Method
1 Draw both vectors from a common origin.
2 Draw the resultant from the tip of the second to the tip of the first.
The Navigation Triangle
Wind Correction Angle
The angle between the course (where you want to go) and the heading (where you point the nose).
Drift
The displacement caused by the medium (wind/water) over time.
Components are Essential
\( v_{gx} = v_{ax} + v_{wx} \)
\( v_{gy} = v_{ay} + v_{wy} \)
Break everything into x and y. Add the pieces. Find the new magnitude and direction.
Pilot's Error
A pilot sets a course for an airport 200km East. They point the nose East and fly for 1 hour.
A 50km/h wind was blowing from the North the whole time.
Discussion
How many kilometers off-course is the pilot? In what direction must they look to see the airport?
Vector Navigation Worksheet Dynamic Navigators
MODULE 04 // RELATIVE VELOCITY // FRAMES
Clearance Level Flight-Ready
A
The Frame Shift
"Object A moves with velocity \( \vec{v}_A = 80\hat{i} + 20\hat{j} \text{ km/h} \) relative to the ground. Object B moves with velocity \( \vec{v}_B = -30\hat{i} + 50\hat{j} \text{ km/h} \) relative to the ground."
1. Relative Vector
Calculate the velocity of A relative to B (\( \vec{v}_{AB} \)).
2. Interception Speed
Calculate the speed (magnitude) at which the distance between A and B is changing.
B
Flight Planning: The Wind Triangle
Flight Data
> Desired Course: Due North (\( 0^\circ \))
> Desired Groundspeed: 400 km/h
> Current Wind: 60 km/h from the West (\( 270^\circ \))
A pilot must determine the heading and airspeed required to maintain a perfect ground track of 400 km/h Due North despite the crosswind.
Recall: \( \vec{v}_{ground} = \vec{v}_{air} + \vec{v}_{wind} \)
1. Component Decomposition
Identify the x and y components of the Ground Velocity and the Wind Velocity.
\( \vec{v}_{ground} = \)
\( \vec{v}_{wind} = \)
2. Vector Solution
Solve for the required Air Velocity vector (\( \vec{v}_{air} \)) and calculate its magnitude and heading angle.
C
Graphical Verification
Sketch the Wind Triangle for Part B. Use the grid to represent your vectors tip-to-tail.
Safety Check
If the pilot fails to correct for wind and simply points North, where will they be after 30 minutes? Calculate the error displacement.
Navigation Log Verified // Reference Frame: Earth // Tracking Complete
Vector Navigation Answer Key Navigator Keys
MODULE 04 // RELATIVE VELOCITY // SOLUTIONS
Reference Document Instructor Final
A
The Frame Shift Solutions
1. Relative Vector Calculation
\( \vec{v}_{AB} = \vec{v}_A - \vec{v}_B \)
\( \vec{v}_{AB} = (80 - (-30))\hat{i} + (20 - 50)\hat{j} \)
\( \vec{v}_{AB} = \mathbf{110\hat{i} - 30\hat{j} \text{ km/h}} \)
2. Interception Speed
\( |\vec{v}_{AB}| = \sqrt{110^2 + (-30)^2} \)
\( |\vec{v}_{AB}| = \sqrt{12100 + 900} = \sqrt{13000} \)
\( |\vec{v}_{AB}| \approx \mathbf{114.02 \text{ km/h}} \)
B
Wind Triangle Solutions
1. Vector Equation Setup
\( \vec{v}_g = \vec{v}_a + \vec{v}_w \implies \vec{v}_a = \vec{v}_g - \vec{v}_w \)
\( \vec{v}_g \)
\( \langle 0, 400 \rangle \)
\( \vec{v}_w \)
\( \langle 60, 0 \rangle \)
\( \vec{v}_a \)
\( \langle -60, 400 \rangle \)
Required Airspeed
\( |\vec{v}_a| = \sqrt{(-60)^2 + 400^2} \approx \mathbf{404.47 \text{ km/h}} \)
Required Heading
\( \theta = \tan^{-1}(400/-60) \approx 98.53^\circ \)
Magnetic Heading: 351.47° (Correction: 8.53° Left)
C
Safety Check Solution
The Error Displacement:
If they point North with no correction:
\( \vec{v}_g = \vec{v}_a + \vec{v}_w = \langle 0, 400 \rangle + \langle 60, 0 \rangle = \langle 60, 400 \rangle \text{ km/h} \)
After 30 minutes (\( t = 0.5 \text{ h} \)):
\( \Delta \vec{s} = \vec{v}_g \cdot t = \langle 30, 200 \rangle \text{ km} \)
The pilot is 30 km East of their intended course.
Static Equilibrium Slides Module 05
Static Structural Equilibrium
Transitioning to 3D: Direction Cosines, Systems of Equations, and the Science of Holding Still.
The Third Dimension
In 3D space, we describe vectors using direction cosines.
Components
\[ A_x = |\vec{A}| \cos(\alpha) \]
\[ A_y = |\vec{A}| \cos(\beta) \]
\[ A_z = |\vec{A}| \cos(\gamma) \]
Fundamental Identity
\[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \]
"The square of the direction cosines must sum to unity. This defines the unit vector in 3D space."
\( \hat{u} = \cos\alpha \hat{i} + \cos\beta \hat{j} + \cos\gamma \hat{k} \)
The Law of Stability
Statics 101
Equilibrium
An object is in static equilibrium if the vector sum of all forces acting upon it is zero.
Vector Equations to Linear Systems
\[ \sum \vec{F} = \vec{0} \]
\[ \sum F_x = 0 \]
\[ \sum F_y = 0 \]
\[ \sum F_z = 0 \]
"One vector equation in 3D equals three scalar equations."
The Suspended Load
A 500kg traffic light is suspended by three cables anchored at different points in space.
Identify anchor coordinates.
Define unit vectors for each cable.
Set up the \(3 \times 3\) system.
System Matrix
\[ \begin{bmatrix} u_{1x} & u_{2x} & u_{3x} \\ u_{1y} & u_{2y} & u_{3y} \\ u_{1z} & u_{2z} & u_{3z} \end{bmatrix} \begin{bmatrix} T_1 \\ T_2 \\ T_3 \end{bmatrix} = \begin{bmatrix} 0 \\ W \\ 0 \end{bmatrix} \]
Where \( T_n \) are the unknown tensions and \( W \) is weight.
Final Review
MOD 01
Representations
MOD 02
Arithmetic
MOD 03
Products
MOD 04
Dynamics
"The language of vectors allows us to build bridges that don't fall, navigate oceans we can't see across, and understand the invisible forces of the universe."
Sequence Complete
Static Equilibrium Worksheet Structural Equilibrium
MODULE 05 // 3D VECTORS // STATIC STABILITY
Project Type Analysis
The Traffic Signal Problem
A large traffic signal with a mass of 80 kg is suspended at the origin O (0, 0, 0) by three cables. The weight of the signal acts in the negative y-direction. The cables are anchored at the following points:
Cable A
Anchor: (-2, 4, 3)
Cable B
Anchor: (3, 5, 2)
Cable C
Anchor: (0, 6, -4)
01
Unit Vector Specification
Find the unit vector \( \hat{u} \) for each cable pointing from the origin toward the anchor.
Cable A
Magnitude \( L_A \)
Unit Vector \( \hat{u}_A \)
Cable B
Magnitude \( L_B \)
Unit Vector \( \hat{u}_B \)
Cable C
Magnitude \( L_C \)
Unit Vector \( \hat{u}_C \)
02
Equilibrium Equations
Write the three equilibrium equations based on the principle \( \sum \vec{F} = 0 \). Use \( T_A, T_B, T_C \) as the unknown tensions.
Sum of Forces in X-Axis
Sum of Forces in Y-Axis (Include Weight!)
Note: Weight \( W = 80 \text{ kg} \times 9.81 \text{ m/s}^2 \approx 785 \text{ N} \)
Sum of Forces in Z-Axis
03
System Synthesis
Arrange your equations into a matrix form \( [A][T] = [B] \).
[ M ] [ T ] = [ W ]
Discussion
"Which cable do you predict will have the highest tension? Justify your answer based on its orientation relative to the anchors."
Structural Integrity Verified // Final Module Complete // vector-analysis-systems
