# SCALAR AND VECTOR QUANTITIES

## Scalar Quantities

A **scalar quantity** is one which has magnitude only, i.e. it has only size. A scalar quantity is completely defined by stating its magnitude (and unit). Scalar quantities include: length, mass, time, temperature, amount of substance.

## Vector Quantities

A **vector quantity** is one which has not only magnitude (size), but also direction.

A vector quantity is completely defined only when its magnitude, unit, and direction are stated.

A vector quantity may be positive or negative, depending on direction of motion or measurement.

If taken as positive in a certain direction, a vector quantity must be taken as negative in the opposite direction.

Vector quantities include: displacement, velocity, acceleration, momentum, force, etc.

## Common Scalar vs. Vector Comparisons

**Distance vs. Displacement****Speed vs. Velocity****Mass vs. Weight**

## Combination of Scalars and Vectors

### Scalars

Scalar quantities can be combined by simply adding or subtracting their magnitudes.

N.B. Only scalar quantities of the same units may be added or subtracted.

Example:

Adding/subtracting two (i) masses, (ii) lengths, and (iii) times.

m1 = 10kg

m2 = 15kg

m1 + m2 = 25kg

m2 – m1 = 5kg

### Vectors

Vectors may be represented graphically by using arrows.

The length of the arrow indicates the vector’s magnitude and the arrowhead the vector’s direction.

Unlike scalars, which can be simply added or subtracted, special rules are applied to the combination (addition) of vectors. Vectors are added to produce a resultant – a single vector which produces the same effect of the combining vectors.

### Rules for adding vectors

#### Parallel vectors – vectors acting in the same direction along a straight line.

Rule: If two or more vectors act in the same direction along a straight line, their magnitudes are simply added.

Example:

[insert: PICTURE 2-1]

Resultant force = 15N (in same direction of two forces)

[insert: PICTURE 2-2]

Resultant Displacement = 18m

#### Anti-parallel vectors – vectors in a straight line but in opposite directions

Rule: If two vectors act in opposite directions along a straight line, one is given a negative value (usually the smaller) and the vectors are then added.

Example:

[insert: PICTURE 2-3]

Resultant force = 5N (in direction of 10N force)

[insert: PICTURE 2-4]

Resultant displacement = 2m (to the left of A)

#### Non-parallel vectors – vectors acting at an angle

Two non-parallel vectors may be added by using the parallelogram law which states:

If two vectors acting at a point are represented in both magnitude and direction by the sides of a parallelogram drawn to scale and from the point, their resultant is given by the length of the diagonal drawn from the same point.

N.B. The parallelogram law is usually applied to forces but can be applied to any two vectors which act an angle.

Example:

Two forces of magnitudes 4N and 5N respectively act at an angle of 45°. Find the magnitude and direction of their resultant.

Solution:

Using the scale 1cm = 1N and letting 5N be a horizontal force.

[insert: PICTURE 2-5]

Length of diagonal = 8.35cm

Resultant force = 8.35N

Angle θ = 20°

Resultant force = 8.35N at 20° to the 5N force

#### Perpendicular vectors – the parallelogram law still applies, but perpendicular vectors are more accurately solved using Pythagoras’ Theorem.

Example:

Two forces of magnitudes 5N and 6N respectively act at a point perpendicularly to each other. Find the magnitude and direction of their resultant by (i) scale drawing, and (ii) Pythagoras’ Theorem.

Solution:

Using the scale 1cm = 1N and letting 6N be a horizontal force.

[insert: PICTURE 2-6]

Length of diagonal = 7.85cm

Resultant force = 7.85N

Angle θ = 40°

R = 7.85N at 40° to 6N force

[insert: PICTURE 2-7]

By Pythagoras’ Theorem:

R2 = 52 + 62 = 25 + 36 = 61

R = √61 = 7.81N

Tan θ = 5/6 🡺 θ = Tan-1 (5/6)

Θ = 39.81°

Resultant force = 7.81N at 39.8° to the 6N force (the horizontal)

### Resolving a Vector

As shown, two perpendicular vectors can be combined to give a single vector, called the resultant, which acts at some angle to either of the two vectors.

The reverse can also be done, i.e. a single vector inclined at an angle to the horizontal (or vertical) can be used to obtain two perpendicular vectors. When this is done, we say that the single vector has been resolved into its components.

N.B. The sides of a triangle may be represented by vectors, i.e. by forces, displacements, velocities, etc.

Example:

Suppose a force F acts at appoint at an angle θ to the horizontal as shown.

[insert: PICTURE 2-8]

F will have components Fx and Fy shown below, where:

Fx = component of F in the x-direction (horizontal component)

Fy = component of F in the y-direction (vertical component)

[insert PICTURE 2-9]

Applying Trig ratios (sine and cosine):

Cos θ = Fx/F 🡺 Fx = cos θ

Sin θ = Fy/F 🡺 Fy = sin θ

Example:

If a car drives up a 30° slope with a velocity v, express the components of the car’s velocity in terms of v. If v = 50ms-1, calculate the components of the car’s velocity.

Solution:

[insert PICTURE 2-10]

Vertical component:

VV = v sin 30

Horizontal component:

VH = v cos 30

If v = 50ms-1, then

VV = 50 sin 30 = 25ms-1

VH = 50 cos 30 = 43.3ms-1

N.B. √ (252 + 43.32) = √2500 = 50ms-1 = v

Consider a body of mass (m) and has weight (mg) resting on a slope which is inclined at an angle (θ) to the horizontal as shown.

[insert PICTURE 2-11]

The body’s weight (mg) has components

perpendicular to the slope: mg cos θ

parallel to the slope: mg sin θ