CIV ENG 2P04: Lecture 4 - CH2: Force & Position Vectors Part iii

CIV ENG 2P04: Statics and Strength of Materials - Lecture 4: CH2: Force & Position Vectors Part iii

Course Overview

This course is divided into two main sections:

Statics (Weeks 1 to 6)

• CH: 1, 2, 3: Particle Equilibrium, Force Vectors, Position Vectors, Force System Resultants, and 3D Force Vectors.

• CH: 4: Equilibrium of Rigid Bodies. Explores static (in)determinacy, stability, and free body diagrams.

• CH: 5: Centre of Gravity, Centroids & Moment of Inertia.

• CH: 6: Structural Analysis. Covers internal forces and the analysis of trusses.

Mechanics of Materials (Weeks 7 to 13)

• CH: 7: Stress & Strain.

• CH: 8: Mechanical Properties of Materials.

• CH: 9: Axial Load.

• CH: 10: Torsion. Focuses on shear stress & strain, and torsion of solid and hollow circular sections.

• CH: 11: Bending.

• CH: 12: Transverse Shear.

• CH: 13: Combined Loading.

• CH: 14: Stress Transformation. Includes principal stresses and Mohr's circle representation.

Learning Outcomes (CH2)

By the end of this chapter, students should be able to:

• Add forces and resolve them into components.

• Express force and position as Cartesian vectors.

• Utilize the dot product to determine the angle between two vectors or the projection of one vector onto another.

Lecture Outline

This lecture covers:

• Cartesian Vectors

• Addition of Cartesian Vectors

• Position Vectors

• Force Vectors Directed Along a Line

• Dot Product

2.9. Dot Product

The dot product is a fundamental tool for analyzing vector relationships, particularly useful in statics for determining angles and projections.

Applications and Conceptual Questions

• Cable Angle Problem: If the physical locations of four cable ends are known, the dot product can be used to calculate the angle ($\alpha$) between the cables and their common anchor point. Factors controlling this angle include the spatial coordinates of the anchor and cable ends.

• Wrench Problem: To determine the component of a force ($F$) applied to a wrench at point A that actually helps turn the bolt, the dot product would be used to find the projection of $F$ along the direction that causes rotation (e.g., perpendicular to the wrench arm and aligned with the axis of the bolt's rotation).

• Real-world connection: The Gordie Howe International Bridge (connecting Windsor, Ont., and Detroit, Mich., across the Detroit River, expected to open Fall 2025) provides a relevant example of structural engineering where vector analysis and dot products would be crucial for design and analysis.

Definition and Characteristics

The dot product of two vectors, $\mathbf{A}$ and $\mathbf{B}$, is defined as:

$\mathbf{A} \cdot \mathbf{B} = AB \cos \theta$

Where:

• $A$ and $B$ are the magnitudes of vectors $\mathbf{A}$ and $\mathbf{B}$, respectively.

• $\theta$ is the smallest angle between the two vectors, always in the range of $0^[\circ]$ to $180^[\circ]$.

Key Characteristics:

1. Scalar Result: The outcome of a dot product is always a scalar (a positive or negative number), not another vector.

2. Units: The units of the dot product are the product of the units of the vectors $\mathbf{A}$ and $\mathbf{B}$.

Laws of Dot Product

1. Commutative Law: The order of the vectors does not affect the result.
   $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$

2. Multiplication by a Scalar: A scalar multiplier can be applied to either vector or the entire dot product.
   $a(\mathbf{A} \cdot \mathbf{B}) = (a\mathbf{A}) \cdot \mathbf{B} = \mathbf{A} \cdot (a\mathbf{B})$

3. Distributive Law: The dot product can be distributed over vector addition.
   $\mathbf{A} \cdot (\mathbf{B} + \mathbf{D}) = (\mathbf{A} \cdot \mathbf{B}) + (\mathbf{A} \cdot \mathbf{D})$

Dot Product of Cartesian Unit Vectors

For standard orthogonal unit vectors ( $\mathbf{i}, \mathbf{j}, \mathbf{k}$ ):

• Orthogonal Unit Vectors: The dot product of two different unit vectors is zero because the angle between them is $90^[\circ]$, and $\cos(90^[\circ]) = 0$.
  $\mathbf{i} \cdot \mathbf{j} = \mathbf{i} \cdot \mathbf{k} = \mathbf{j} \cdot \mathbf{k} = 0$

• Parallel Unit Vectors: The dot product of a unit vector with itself is one because the angle between them is $0^[\circ]$, and $\cos(0^[\circ]) = 1$.
  $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1$

Finding the Angle Between Two Vectors in Cartesian Form

Given two vectors $\mathbf{A}$ and $\mathbf{B}$ in Cartesian (component) form, the angle $\theta$ between them can be found using the following steps:

1. Calculate the dot product:
   $\mathbf{A} \cdot \mathbf{B} = A<em>x B</em>x + A<em>y B</em>y + A<em>z B</em>z$

2. Find the magnitudes of the vectors:
   $A = \sqrt{A<em>x^2 + A</em>y^2 + A<em>z^2}$ $B = \sqrt{B</em>x^2 + B<em>y^2 + B</em>z^2}$

3. Use the definition of the dot product to solve for $\theta$:
   $\theta = \cos^{-1} \left[ \frac{\mathbf{A} \cdot \mathbf{B}}{AB} \right]$, where $0^[\circ] \leq \theta \leq 180^[\circ]$.

Determining Components of a Vector Parallel and Perpendicular to a Line

The dot product is essential for finding the scalar and vector components of a vector that are parallel and perpendicular to a given line (or another vector).

Steps to find the scalar projection ($A_a$) of vector $\mathbf{A}$ along a line $aa$:

1. Find the unit vector ($\mathbf{u}_a$) along line $aa$. This vector defines the direction of the line.

2. Calculate the scalar projection ($A<em>a$) of $\mathbf{A}$ along line $aa$. This is done by taking the dot product of $\mathbf{A}$ with the unit vector $\mathbf{u}</em>a$. This projection represents the magnitude of the component of $\mathbf{A}$ that lies along the line $aa$.
   $A<em>a = \mathbf{A} \cdot \mathbf{u}</em>a = A<em>x u</em>x + A<em>y u</em>y + A<em>z u</em>z$
   Alternatively, if the angle $\theta$ between $\mathbf{A}$ and $\mathbf{u}<em>a$ is known: $A</em>a = A \cos \theta$

The vector component of $\mathbf{A}$ parallel to line $aa$ is then $\mathbf{A}<em>a = A</em>a \mathbf{u}<em>a$. The vector component perpendicular to line $aa$ ($\mathbf{A}</em>\perp$) can be found using vector subtraction:
$\mathbf{A}<em>\perp = \mathbf{A} - \mathbf{A}</em>a$

Example: Angle and Projection of a Force Along a Line

Problem Statement: Determine the angle between a given force vector and line AO, and determine the magnitude of the projection of the force along the line AO.

Given:

• Force vector: $\mathbf{F} = {-6\mathbf{i} + 9\mathbf{j} + 3\mathbf{k}} \text{ kN}$

• Position vector from origin to A (representing line AO): $\mathbf{r}_{AO} = {-1\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}} \text{ m}$

Plan:

1. Find the dot product of $\mathbf{F}$ and $\mathbf{r}_{AO}$.

2. Calculate the magnitudes of $\mathbf{F}$ and $\mathbf{r}_{AO}$.

3. Use the dot product definition to find the angle $\theta$.

4. Find the unit vector along line AO ($\mathbf{u}_{AO}$).

5. Calculate the scalar projection of $\mathbf{F}$ along $\mathbf{u}_{AO}$.

Solution Steps:

1. Calculate Magnitudes:

   • Magnitude of $\mathbf{r}<em>{AO}$: $r</em>{AO} = \sqrt{(-1)^2 + (2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \text{ m}$

   • Magnitude of $\mathbf{F}$:
     $F = \sqrt{(-6)^2 + (9)^2 + (3)^2} = \sqrt{36 + 81 + 9} = \sqrt{126} \approx 11.22 \text{ kN}$

2. Calculate the Dot Product of $\mathbf{F}$ and $\mathbf{r}<em>{AO}$:
   $\mathbf{F} \cdot \mathbf{r}</em>{AO} = (-6)(-1) + (9)(2) + (3)(-2) = 6 + 18 - 6 = 18 \text{ kN} \cdot \text{m}$

3. Determine the Angle ($\theta$) between $\mathbf{F}$ and $\mathbf{r}<em>{AO}$:
   Using the formula $\theta = \cos^{-1} \left[ \frac{\mathbf{F} \cdot \mathbf{r}</em>{AO}}{F r_{AO}} \right]$:
   $\theta = \cos^{-1} \left[ \frac{18}{(11.22)(3)} \right] = \cos^{-1} \left[ \frac{18}{33.66} \right] \approx 57.67^[\circ]$

4. Find the Unit Vector along AO ($\mathbf{u}<em>{AO}$):
   $\mathbf{u}</em>{AO} = \frac{\mathbf{r}<em>{AO}}{r</em>{AO}} = \frac{{-1\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}} \text{ m}}{3 \text{ m}} = \left(-\frac{1}{3}\right)\mathbf{i} + \left(\frac{2}{3}\right)\mathbf{j} + \left(-\frac{2}{3}\right)\mathbf{k}$

5. Calculate the Scalar Projection of $\mathbf{F}$ along Line AO ($F<em>{AO}$):
   Using the formula $F</em>{AO} = \mathbf{F} \cdot \mathbf{u}<em>{AO}$: $F</em>{AO} = (-6)\left(-\frac{1}{3}\right) + (9)\left(\frac{2}{3}\right) + (3)\left(-\frac{2}{3}\right) = 2 + 6 - 2 = 6.00 \text{ kN}$
   Alternatively, using the angle:
   $F_{AO} = F \cos \theta = 11.22 \cos (57.67^[\circ]) \approx 6.00 \text{ kN}$

Summary of Key Concepts (Revisited)

This lecture reinforced understanding of:

• Cartesian Vectors

• Addition of Cartesian Vectors

• Position Vectors

• Force Vectors Directed Along a Line

• Dot Product