5.1 Math Review: Scalars, Vectors, and Trigonometry

Math Review: Scalar and Vector Quantities

This section reviews fundamental mathematical concepts, distinguishing between scalar and vector quantities and outlining methods for vector manipulation.

Scalar Quantities

Scalar quantities are physical measurements that possess magnitude only. They describe 'how much' or 'how many' but do not include a directional component. Key examples include:

• Mass

• Volume

• Pressure

• Density

• Speed

Vector Quantities

Vector quantities are physical measurements that possess both magnitude and direction. They describe both 'how much' and 'in which way' a quantity acts. Key examples include:

• Force

• Weight

• Pressure

• Torque

• Velocity

Note: Pressure is listed under both scalar and vector quantities. In some contexts (e.g., hydrostatic pressure in a fluid at a point), it might be considered scalar, while in others (e.g., pressure exerted by a force over an area), its directional aspect becomes relevant when considering specific axes or surfaces.

Vector Resolution

Vector resolution is the process of breaking down a single vector into two or more component vectors that are perpendicular to each other. These components, when combined (composed), will yield the original vector. This is often done along horizontal (x) and vertical (y) axes for analysis.

• Concept: Any vector can be represented as the sum of its perpendicular components.

• Illustration: A resultant vector can be decomposed into an x-component and a y-component, each acting along its respective axis.

Vector Composition

Vector composition is the process of combining two or more vectors to find a single resultant vector that has the same effect as all the original vectors combined.

Composition of Vectors in the Same Direction

When vectors act in the same direction, their magnitudes are simply added together to find the resultant vector's magnitude.

• Example: A $50 ext{ N}$ force acting in one direction combined with another $50 ext{ N}$ force acting in the same direction results in a $100 ext{ N}$ force in that same direction.

Composition of Vectors in Opposite Directions

When vectors act in opposite directions, their magnitudes are subtracted. The direction of the resultant vector will be the direction of the vector with the larger magnitude.

• Example: A $50 ext{ N}$ force in one direction combined with a $20 ext{ N}$ force in the opposite direction results in a $30 ext{ N}$ force in the direction of the original $50 ext{ N}$ force.

Vector Algebra: Tip-to-Tail Method

The tip-to-tail method is a graphical technique for adding vectors. To add two or more vectors using this method:

1. Place the tail of the second vector at the tip (head) of the first vector.

2. If adding more vectors, continue placing the tail of the next vector at the tip of the previous one.

3. The resultant vector is drawn from the tail of the very first vector to the tip of the very last vector.

Mathematical Approach: Trigonometry and Pythagorean Theorem

When dealing with right-angled triangles formed by vectors and their components, trigonometric functions and the Pythagorean Theorem are essential tools.

Right Triangle Terminology

• Hypotenuse: The side opposite the right angle, representing the magnitude of the resultant vector.

• Opposite: The side opposite the angle $heta$ (often used for the vertical component).

• Adjacent: The side next to the angle $heta$ (often used for the horizontal component).

Trigonometric Functions for Vector Resolution

• Sine (sin): Relates the opposite side to the hypotenuse.

  • $ext{sin } heta = rac{ ext{opposite}}{ ext{hypotenuse}}$

  • $ext{opposite} = ext{hypotenuse} imes ext{sin } heta$

• Cosine (cos): Relates the adjacent side to the hypotenuse.

  • $ext{cos } heta = rac{ ext{adjacent}}{ ext{hypotenuse}}$

  • $ext{adjacent} = ext{hypotenuse} imes ext{cos } heta$

• Tangent (tan): Relates the opposite side to the adjacent side.

  • $ext{tan } heta = rac{ ext{opposite}}{ ext{adjacent}}$

Pythagorean Theorem for Vector Composition

The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b).

• $ext{Opposite}^2 + ext{Adjacent}^2 = ext{Hypotenuse}^2$

Solved Examples

Example 1: Vector Resolution (Finding Components)

Problem: A long jumper takes off with a resultant velocity ($V_R$) of $9 ext{ m/s}$ at an angle of $25^{ ext{o}}$ to the horizontal. How fast is the jumper moving in the vertical and horizontal directions?

Given:

• Resultant Velocity ($V_R$ or Hypotenuse) = $9 ext{ m/s}$

• Angle ($heta$) = $25^{ ext{o}}$

Unknowns:

• Vertical velocity ($V_v$ or Opposite)

• Horizontal velocity ($V_H$ or Adjacent)

Calculations:

• Vertical Velocity ($V_v$):

  • Using the sine function: $V<em>v = V</em>R imes ext{sin } heta$

  • $V_v = 9 ext{ m/s} imes ext{sin } 25^{ ext{o}}$

  • $V_v hickapprox 9 ext{ m/s} imes 0.4226$

  • $V_v hickapprox 3.8 ext{ m/s}$

• Horizontal Velocity ($V_H$):

  • Using the cosine function: $V<em>H = V</em>R imes ext{cos } heta$

  • $V_H = 9 ext{ m/s} imes ext{cos } 25^{ ext{o}}$

  • $V_H hickapprox 9 ext{ m/s} imes 0.9063$

  • $V_H hickapprox 8.2 ext{ m/s}$

Example 2: Vector Composition (Finding Resultant)

Problem: A high jumper takes off with a vertical velocity of $4.3 ext{ m/s}$ and a horizontal velocity of $2.5 ext{ m/s}$. What is the resultant velocity of the jumper and the take-off angle?

Given:

• Vertical velocity ($V_v$ or Opposite) = $4.3 ext{ m/s}$

• Horizontal velocity ($V_H$ or Adjacent) = $2.5 ext{ m/s}$

Unknowns:

• Resultant velocity ($V_R$ or Hypotenuse)

• Take-off angle ($heta$)

Calculations:

• Resultant Velocity ($V_R$):

  • Using the Pythagorean Theorem: $ext{Opposite}^2 + ext{Adjacent}^2 = ext{Hypotenuse}^2$

  • $(4.3)^2 + (2.5)^2 = V_R^2$

  • $18.49 + 6.25 = V_R^2$

  • $24.74 = V_R^2$

  • $V_R = ext{sqrt{24.74}} hickapprox 4.97 ext{ m/s}$

• Take-off Angle ($heta$):

  • Using the tangent function: $ext{tan } heta = rac{ ext{opposite}}{ ext{adjacent}}$

  • $ext{tan } heta = rac{4.3}{2.5}$

  • $ext{tan } heta = 1.72$

  • $heta = ext{tan}^{-1}(1.72) hickapprox 59.8^{ ext{o}}$