Comprehensive Notes on Vectors and Equilibrium

Learning Objectives for Vectors and Equilibrium

  • Understand and use the rectangular coordinate system.
  • Understand the conceptual ideas of unit vectors, null vectors, and position vectors.
  • Represent a vector as two perpendicular rectangular components.
  • Understand the rules of vector addition and extend these rules to adding vectors using rectangular components.
  • Understand the multiplication of vectors and solve associated math problems.
  • Define the moment of force, also known as torque.
  • Appreciate the application and use of torque due to a force.
  • Develop an understanding that a system is in equilibrium when there is no resultant force (F=0\sum \mathbf{F} = 0) and no resultant torque (τ=0\sum \boldsymbol{\tau} = 0).
  • Appreciate the applications of the principle of moments.
  • Apply comprehensive knowledge to solve problems in statics.

Basic Concepts of Vectors

  • Definition: Physical quantities that possess both numerical (magnitude) and directional properties are called vectors.
  • Physical Quantities: Common examples of vectors encountered in physics include velocity, acceleration, and force. These require both magnitude and direction for a complete description.
  • Denotation in Books: In printed textbooks, vectors are typically denoted by boldface characters such as A\mathbf{A}, d\mathbf{d}, r\mathbf{r}, and v\mathbf{v}.
  • Denotation in Handwriting: In handwritten notes, vectors are represented by putting an arrowhead over the letter (e.g., A\vec{A}).
  • Magnitude Only: To refer specifically to the magnitude of a vector d\mathbf{d}, a light face type such as d\text{d} is used.
  • Graphical Representation: A vector is represented graphically by a directed line segment with an arrowhead. The length of the line segment corresponds to the magnitude of the vector according to a chosen scale.

Rectangular Coordinate System

  • Definition: A system consisting of two reference lines drawn at right angles to each other is known as a coordinate axes system. The point of intersection is called the origin.
  • Terminology: This setup is also referred to as the Cartesian or rectangular coordinate system.
  • Axes: One line is the vertical yy-axis (positive direction upward), and the other is the horizontal xx-axis (positive direction to the right).
  • Direction in a Plane: The direction of a vector in a plane is denoted by the angle θ\theta which the representative line of the vector makes with the positive xx-axis in the anti-clockwise direction.
  • Notation: A point PP in a plane is marked (a,b)(a, b). This means starting at the origin, moving a\text{a} units along the positive xx-axis and b\text{b} units along the positive yy-axis.
  • Direction in Space: This requires a third axis, the zz-axis, which is at a right angle to both the xx and yy axes. The location is specified by three angles (α\alpha, β\beta, and γ\gamma) made with the xx, yy, and zz axes, respectively. The point is denoted as P(a,b,c)P(a, b, c).

Addition and Subtraction of Vectors

  • Head to Tail Rule: To add two vectors A\mathbf{A} and B\mathbf{B}, draw their representative lines so that the tail of vector B\mathbf{B} coincides with the head of vector A\mathbf{A}. The vector joining the tail of A\mathbf{A} to the head of B\mathbf{B} represents the resultant vector R=A+B\mathbf{R} = \mathbf{A} + \mathbf{B}.
  • Commutative Property: Vector addition is commutative, meaning A+B=B+A\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}. The result is the same regardless of the order of addition.
  • Resultant Vector: A single vector that has the same effect as all the original vectors taken together (e.g., a single force vector representing multiple concurrent forces).
  • Vector Subtraction: Subtraction is equivalent to the addition of a vector with its direction reversed. To subtract B\mathbf{B} from A\mathbf{A}, calculate A+(B)\mathbf{A} + (-\mathbf{B}), where (B)(-\mathbf{B}) is the negative vector of B\mathbf{B}.

Multiplication of a Vector by a Scalar

  • Definition: The product of a vector A\mathbf{A} and a number n>0n > 0 is a new vector nAn\mathbf{A} with the same direction but a magnitude multiplied by nn.
  • Directional Change: If the vector is multiplied by a negative number, its direction is reversed.
  • Scalar Quantity Multiplication: If nn represents a scalar physical quantity, the product results in a new physical quantity with dimensions equal to the product of the dimensions of the two original quantities.
  • Example: Multiplying scalar mass m\text{m} by velocity v\mathbf{v} results in the vector quantity momentum p\mathbf{p}, with dimensions based on mass and velocity.

Unit, Null, and Equal Vectors

  • Unit Vector: A vector with a magnitude of exactly 11 in a given direction, used to represent direction. It is written with a 'hat' (e.g., A^\hat{A}). Formula: A^=AA\hat{A} = \frac{\mathbf{A}}{\text{A}}.
    • i^\hat{i} represents the orientation along the xx-axis.
    • j^\hat{j} represents the orientation along the yy-axis.
    • k^\hat{k} represents the orientation along the zz-axis.
    • r^\hat{r} represents the direction of the vector r\mathbf{r}.
    • n^\hat{n} represents the direction of a normal line drawn on a specific surface.
  • Null Vector: A vector with zero magnitude and an arbitrary direction. Example: Adding a vector and its negative results in a null vector (A+(A)=0\mathbf{A} + (-\mathbf{A}) = 0).
  • Equal Vectors: Two vectors are equal if they possess the same magnitude and direction, regardless of where their initial points are located. Parallel vectors with the same magnitude are equal.

Rectangular Components of a Vector

  • Definition: A component is the effective value of a vector in a specific direction. Rectangular components are those resolved along mutually perpendicular directions.
  • Resolution of Vector A: For a vector A\mathbf{A} at angle θ\theta to the xx-axis:
    • $x$-component: Ax=Acos(θ)A_x = A\cos(\theta)
    • $y$-component: Ay=Asin(θ)A_y = A\sin(\theta)
  • Vector Form: A=Axi^+Ayj^\mathbf{A} = A_{x}\hat{i} + A_{y}\hat{j}.
  • Determination from Components:
    • Magnitude: A=Ax2+Ay2A = \sqrt{A_{x}^2 + A_{y}^2}
    • Direction: θ=tan1(AyAx)\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)

Position Vector

  • Definition: The position vector r\mathbf{r} describes the location of a point relative to the origin. Its tail is at the origin (0,0)(0,0) and its head is at point P(a,b)P(a, b).
  • Two Dimensions: r=ai^+bj^\mathbf{r} = a\hat{i} + b\hat{j}, with magnitude r=a2+b2r = \sqrt{a^{2} + b^{2}}.
  • Three Dimensions: For point P(a,b,c)P(a, b, c), the vector is r=ai^+bj^+ck^\mathbf{r} = a\hat{i} + b\hat{j} + c\hat{k}, with magnitude r=a2+b2+c2r = \sqrt{a^2 + b^2 + c^2}.

Vector Addition by Rectangular Components

  • Process for Resultant R:
    1. Find xx and yy components of all given vectors.
    2. Calculate the total xx-component of the resultant: Rx=Ax+Bx+Cx+R_x = A_x + B_x + C_x + \dots
    3. Calculate the total yy-component of the resultant: Ry=Ay+By+Cy+R_y = A_y + B_y + C_y + \dots
    4. Find the magnitude: R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}
    5. Find the direction: ϕ=tan1RyRx\phi = \tan^{-1}\left|\frac{R_y}{R_x}\right|
  • Determining the Quadrant for angle θ\theta:
    • 1st Quadrant: Both RxR_x and RyR_y are positive (++). θ=ϕ\theta = \phi.
    • 2nd Quadrant: RxR_x is negative (-), RyR_y is positive (++). θ=180ϕ\theta = 180^{\circ} - \phi.
    • 3rd Quadrant: Both RxR_x and RyR_y are negative (-). θ=180+ϕ\theta = 180^{\circ} + \phi.
    • 4th Quadrant: RxR_x is positive (++), RyR_y is negative (-). θ=360ϕ\theta = 360^{\circ} - \phi.

Scalar or Dot Product

  • Definition: The scalar product of A\mathbf{A} and B\mathbf{B} is denoted AB\mathbf{A} \cdot \mathbf{B} and equals ABcos(θ)AB\cos(\theta), where θ\theta is the angle between them.
  • Physical Interpretation: AB=A×(projection of B on A)=A(Bcos(θ))\mathbf{A} \cdot \mathbf{B} = A \times (\text{projection of B on A}) = A(B\cos(\theta)).
  • Work Application: Work Done=Fd=Fdcos(θ)\text{Work Done} = \mathbf{F} \cdot \mathbf{d} = Fd\cos(\theta).
  • Key Characteristics:
    1. Commutative: AB=BA\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}.
    2. Perpendicular Vectors: If θ=90\theta = 90^{\circ}, AB=0\mathbf{A} \cdot \mathbf{B} = 0. Consequently, i^j^=j^k^=k^i^=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0.
    3. Parallel Vectors: If θ=0\theta = 0^{\circ}, AB=AB\mathbf{A} \cdot \mathbf{B} = AB. Consequently, i^i^=j^j^=k^k^=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1.
    4. Antiparallel Vectors: If θ=180\theta = 180^{\circ}, AB=AB\mathbf{A} \cdot \mathbf{B} = -AB.
    5. Self Product: AA=A2\mathbf{A} \cdot \mathbf{A} = A^{2}.
    6. Component Form: AB=AxBx+AyBy+AzBz\mathbf{A} \cdot \mathbf{B} = A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}.
    7. Angle Calculation: cos(θ)=AxBx+AyBy+AzBzAB\cos(\theta) = \frac{A_{x}B_{x} + A_{y}B_{y} + A_{z}B_{z}}{AB}.

Vector or Cross Product

  • Definition: The vector product A×B\mathbf{A} \times \mathbf{B} is a vector defined as A×B=ABsin(θ)n^\mathbf{A} \times \mathbf{B} = AB\sin(\theta)\hat{n}, where n^\hat{n} is a unit vector perpendicular to the plane containing A\mathbf{A} and B\mathbf{B}.
  • Right Hand Rule: Rotate A\mathbf{A} into B\mathbf{B} through the smaller angle; curl right-hand fingers in that direction. The erect thumb indicates the direction of the product vector.
  • Key Characteristics:
    1. Non-Commutative: A×B=(B×A)\mathbf{A} \times \mathbf{B} = -(\mathbf{B} \times \mathbf{A}).
    2. Perpendicular Vectors: Magnitude is maximum (ABAB). i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}.
    3. Parallel/Antiparallel Vectors: A×B=0\mathbf{A} \times \mathbf{B} = 0. Self product A×A=0\mathbf{A} \times \mathbf{A} = 0; i^×i^=j^×j^=k^×k^=0\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0.
    4. Area Interpretation: The magnitude of A×B\mathbf{A} \times \mathbf{B} equals the area of the parallelogram formed with A\mathbf{A} and B\mathbf{B} as adjacent sides.
    5. Determinant Form:     A×B=i^j^k^AxAyAzBxByBz\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}
  • Applied Examples:
    • Torque: τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}.
    • Magnetic Force: Force on a charge qq with velocity v\mathbf{v} in field B\mathbf{B} is F=q(v×B)\mathbf{F} = q(\mathbf{v} \times \mathbf{B}).

Torque (Moment of Force)

  • Conceptual Definition: The turning effect of a force. It depends on the applied force and the length of the handle (moment arm).
  • Magnitude: τ=lF\tau = lF, where ll is the moment arm (perpendicular distance from the pivot to the line of action of the force).
  • General Formula: τ=rFsin(θ)\tau = rF\sin(\theta), where θ\theta is the angle between position vector r\mathbf{r} and force F\mathbf{F}.
  • Vector Definition: τ=r×F\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}.
  • Units: SI unit is the Newton-meter (NmN\,m).
  • Pivot Point Influence: If the line of action passes through the pivot, moment arm l=0l = 0, so torque τ=0\tau = 0.
  • Rotational Analogue: Torque is to rotational motion what force is to linear motion; it determines angular acceleration.

Equilibrium of Forces and Torques

  • General Definition: A body is in equilibrium if it is at rest (static) or moving with uniform velocity (dynamic).
  • First Condition of Equilibrium: The vector sum of all forces acting on a body must be zero.
    • F=0\sum \mathbf{F} = 0
    • In components: Fx=0\sum F_x = 0 and Fy=0\sum F_y = 0.
    • Convention: Rightward/Upward (+), Leftward/Downward (-).
  • Second Condition of Equilibrium: For a body to have zero angular acceleration, the vector sum of all torques about any arbitrary axis must be zero.
    • τ=0\sum \boldsymbol{\tau} = 0
    • Convention: Counter-clockwise torques are positive (+); clockwise torques are negative (-).
  • Complete Equilibrium: Both conditions must be satisfied.
    • Translational equilibrium: Linear acceleration is zero.
    • Rotational equilibrium: Angular acceleration is zero.

Numerical Examples and Case Studies

  • Aeroplane Position (Ex 2.1): Aeroplane A at (2,3,4)(2, 3, 4) and B at (5,6,7)(5, 6, 7).
    • OA=2i^+3j^+4k^OA = 2\hat{i} + 3\hat{j} + 4\hat{k}
    • OB=5i^+6j^+7k^OB = 5\hat{i} + 6\hat{j} + 7\hat{k}
    • AB=OBOA=3i^+3j^+3k^AB = OB - OA = 3\hat{i} + 3\hat{j} + 3\hat{k}
    • Distance=32+32+32=275.2km\text{Distance} = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} \approx 5.2\,km.
  • Two Force Resultant (Ex 2.2): Forces of 10N10\,N and 20N20\,N at 3030^{\circ} and 6060^{\circ}.
    • Fx=10cos(30)+20cos(60)=18.66NF_x = 10\cos(30^{\circ}) + 20\cos(60^{\circ}) = 18.66\,N
    • Fy=10sin(30)+20sin(60)=22.32NF_y = 10\sin(30^{\circ}) + 20\sin(60^{\circ}) = 22.32\,N
    • F=18.662+22.322=29NF = \sqrt{18.66^2 + 22.32^2} = 29\,N; θ=tan1(22.3218.66)=50\theta = \tan^{-1}\left(\frac{22.32}{18.66}\right) = 50^{\circ}.
  • Resultant Magnitude Equality (Ex 2.3): If F1=F2=R|\mathbf{F}_1| = |\mathbf{F}_2| = |\mathbf{R}|, the angle between forces is 120120^{\circ}.
  • Work Calculation (Ex 2.4): For F=2i^+3j^\mathbf{F} = 2\hat{i} + 3\hat{j} and movement from (1,3)(1, 3) to (5,7)(5, 7).
    • d=(51)i^+(73)j^=4i^+4j^\mathbf{d} = (5-1)\hat{i} + (7-3)\hat{j} = 4\hat{i} + 4\hat{j}.
    • Work=(2×4)+(3×4)=8+12=20\text{Work} = (2\times 4) + (3\times 4) = 8 + 12 = 20 units.
  • Suspended Load (Ex 2.7): A load at point PP supported by two cords at 6060^{\circ} and 2020^{\circ}. If max tension is 50N50\,N, the maximum weight suspended is 52N52\,N.
  • Diving Board Reaction (Ex 2.9): A 300N300\,N boy stands on the edge of a 200N200\,N diving board (4.0m4.0\,m long). Calculation shows pedestal reaction forces R1=1.1kNR_1 = -1.1\,kN (downward) and R2=1.6kNR_2 = 1.6\,kN (upward).

Practical Curiosities and Tips

  • Stability on a Bus: Keeping legs far apart while standing in the aisle of a bumpy-riding bus increases stability by widening the base of support.
  • Door and Nose Experiment: Touching your nose to the end of a door and trying to rise on your toes is difficult because it shifts your center of gravity in a way that requires more torque to balance than your base can provide.
  • Chinese Acrobats: Their balancing acts are a live demonstration of complex equilibrium where F=0\sum \mathbf{F} = 0 and τ=0\sum \boldsymbol{\tau} = 0 across their entire bodies.