1.1.1 Scalars & Vectors

In mechanics, it's important to differentiate between scalars and vectors, as they form the foundation of many calculations.

Scalars

• Definition: A scalar is a quantity that has magnitude only, without any direction.

Vectors

• Definition: A vector is a quantity that has both magnitude and direction.

Vectors are represented by arrows, where:

• The length of the arrow indicates the magnitude.
• The direction of the arrow indicates the direction of the vector.

Key Concepts

1. Addition of Vectors:

• Graphically: Add vectors by placing them head-to-tail and then drawing the resultant vector from the tail of the first vector to the head of the last.
• Mathematically: Vectors can be added by breaking them into components (usually along the $x$ and $y$ axes), adding the respective components, and then recombining them.

2. Resultant Vector:

• The vector representing the combined effect of two or more vectors.
• Example: If you push a box with a force of $10$ $N$ $east$ and another force of $10$ $N$ $north$, the resultant vector will have a magnitude of 14.1 $N$ (using Pythagoras' theorem) and a direction of 45° northeast.

3. Scalar Multiplication:

• Multiplying a vector by a scalar changes its magnitude but not its direction.
• Example: Doubling the speed (a scalar) of a car moving north (a vector) doubles its velocity.

Example Problem

Solution:

4. Identify the vectors:

• First displacement vector, $A$: $4$ $km$ $north$.
• Second displacement vector, $B$: $3$ $km$ $east$.

5. Graphical representation:

• Draw vector $A$ ($4$ $km$ $north$).
• From the head of vector $A$, draw vector $B$ ($3$ $km$ $east$).

6. Calculate the resultant vector:

• The resultant displacement vector, $R$, is the diagonal of the right triangle formed by $A$ and $B$.
• Use Pythagoras' theorem to find the magnitude of $R$:

$R = \sqrt{A^2 + B^2} = \sqrt{(4 \, \text{km})^2 + (3 \, \text{km})^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{km}$

7. Determine the direction:

• The direction is given by the angle $θ$ relative to the north (vector $A$).
• Use trigonometry ($tan θ = opposite/adjacent$):

$\tan \theta = \frac{3 \, \text{km}}{4 \, \text{km}} = 0.75$

• Find $θ$ using the inverse tangent function:

$\theta = \tan^{-1}(0.75) \approx 36.9^\circ$

So, the hiker's resultant displacement is 5 km at an angle of 36.9° $east$ $of$ $north$.

This example illustrates how vectors (displacement in this case) can be combined to find a resultant vector, taking into account both magnitude and direction.

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