2.3.1 - Scalars and vectors

Scalar or vector

A physical quantity tells you something measurable about a system. For this lesson, the key question is whether direction is part of that quantity.

Examples of scalar quantities include mass, time, temperature, energy, distance, speed, density and pressure. If a runner travels a total distance of $400\,\mathrm{m}$, the statement is complete without saying north, south, left or right.

Examples of vector quantities include displacement, velocity, acceleration, force, weight, momentum and electric field strength. If a force is $12\,\mathrm{N}$, that is only the magnitude. A complete force might be $12\,\mathrm{N} \text{vertically upwards}$ or $12\,\mathrm{N} \text{to the left}$.

The difference between distance and displacement is a useful first test. Distance is the total length of the route travelled, so it is a scalar. Displacement is the straight-line change in position from start to finish, with direction, so it is a vector.

Worked example: scalar or vector?

A cyclist rides $300\,\mathrm{m}$ east and then $300\,\mathrm{m}$ west, returning to the starting point.

• Total distance travelled = $600\,\mathrm{m}$. This is a scalar.
• Displacement = $0\,\mathrm{m}$. This is a vector quantity, and here the magnitude is zero because the final position is the starting position.

Do not use the words "has a positive value" as the definition of a scalar. Some scalar changes can be written with a sign using a chosen convention, but a scalar still has no spatial direction.

Adding and subtracting vectors

Vector quantities add by direction as well as magnitude. If two forces of $5\,\mathrm{N}$ act in the same direction, the resultant is $10\,\mathrm{N}$ in that direction. If they act in opposite directions, the resultant is $0\,\mathrm{N}$. The numbers alone are not enough.

For a diagram method, use head-to-tail addition:

1. Draw the first vector to scale in its correct direction.
2. From the head of the first vector, draw the second vector to scale in its correct direction.
3. Draw the resultant from the tail of the first vector to the head of the second vector.

[DIAGRAM: asset_slug: 2_03_1_scalars_vectors_and_components__diagram_01; caption: Head-to-tail vector addition also handles subtraction by reversing the vector being subtracted.; description: Head-to-tail addition and subtraction with the reversed second vector.]

Vector subtraction means adding the negative of a vector:

The vector $-\vec{B}$ has the same magnitude as $\vec{B}$ but the opposite direction. This is why subtracting $4\,\mathrm{N} \text{east}$ is the same as adding $4\,\mathrm{N} \text{west}$.

Worked example: collinear vectors

A force of $18\,\mathrm{N}$ acts to the right. A second force of $7\,\mathrm{N}$ acts to the left.

Choose right as positive. The resultant is:

The positive sign means the resultant is to the right, so the answer is $11\,\mathrm{N} \text{to the right}$.

Resultant by vector triangle

Two coplanar vectors lie in the same plane. Their resultant can be found from a vector triangle by calculation or by scale drawing.

For a scale drawing, choose a sensible scale such as $1\,\mathrm{cm}$ represents $2\,\mathrm{N}$, draw the two vectors head-to-tail with a ruler and protractor, then measure the resultant length and direction. The scale must be stated or clear. Small drawing errors can change the answer, so calculation is preferred when the numbers make calculation straightforward.

When the two vectors are perpendicular, the vector triangle is a right-angled triangle. Use Pythagoras for the magnitude and trigonometry for the direction.

Worked example: perpendicular displacement vectors

A student walks $40\,\mathrm{m} \text{east}$ and then $30\,\mathrm{m} \text{north}$. Determine the magnitude and direction of the resultant displacement.

The two displacements are perpendicular, so:

Let $\theta$ be the angle north of east:

So the resultant displacement is $50\,\mathrm{m} \text{at} 37 {}^{\circ} \text{north of east}$ to two significant figures.

For two vectors of $6.0\,\mathrm{N}$ east and $4.0\,\mathrm{N}$ at $60^\circ$ north of east, choose $1\,\mathrm{cm}$ to represent $1\,\mathrm{N}$. Draw a 6.0 cm east arrow, then a 4.0 cm arrow at $60^\circ$ from its head. The resultant measures about 8.7 cm at $23^\circ$ north of east: about $8.7\,\mathrm{N}$. The angle inside this head-to-tail triangle is $120^\circ$, not the original $60^\circ$; Pythagoras cannot be used on this triangle.

For non-perpendicular vectors, the same head-to-tail idea still applies. You can use a scale drawing, or resolve the vectors into perpendicular components and calculate from the components. The next part teaches that component method.

Resolving into components

Resolving a vector means replacing it with perpendicular component vectors. The components have the same combined effect as the original vector.

[DIAGRAM: asset_slug: 2_03_1_scalars_vectors_and_components__diagram_02; caption: The horizontal component is adjacent to the stated angle and the vertical component is opposite.; description: Vector resolution: the horizontal component is adjacent to the stated angle; the vertical component is opposite.]

Here $F$ is the magnitude of the original vector, $F_x$ is the component along the x-axis, $F_y$ is the component along the y-axis, and $\theta$ is the angle between the vector and the x-axis. The component formulae are given in the data booklet, but you must still choose the correct angle.

The safest rule is not "cos is horizontal" and "sin is vertical". The safe rule is:

• cosine gives the side adjacent to the angle in the right-angled triangle
• sine gives the side opposite the angle in the right-angled triangle

If the angle is measured from the vertical instead of from the horizontal, the adjacent and opposite sides swap. Use degree mode for angles in degrees and radian mode for radians: $\pi/3\,\mathrm{rad}=60^\circ$, so both give cosine $0.5$.

Choose positive axes before resolving. A force to the left has a negative horizontal component; a downward force has a negative vertical component. Add signed components, then use $R=\sqrt{R_x^2+R_y^2}$. An inverse-tangent value alone does not identify the quadrant: use the signs of both components to state the actual compass direction.

For instance, a $10\,\mathrm{N}$ force at $60^\circ$ north of west has components $-5.0\,\mathrm{N}$ eastwards and $+8.66\,\mathrm{N}$ northwards. Combining it with $3.0\,\mathrm{N}$ east gives $R_x=-2.0\,\mathrm{N}$, $R_y=8.66\,\mathrm{N}$. The resultant is $8.9\,\mathrm{N}$, directed $13^\circ$ west of north, not into the south-east quadrant.

Worked example: resolving a force

A force of $20\,\mathrm{N}$ acts at $35 {}^{\circ}$ above the horizontal. Calculate its horizontal and vertical components.

The angle is measured from the horizontal x-axis, so:

To a sensible number of significant figures, the components are $16\,\mathrm{N} \text{horizontally}$ and $11\,\mathrm{N} \text{vertically upwards}$.

Worked example: adding by components

A force of $10\,\mathrm{N} \text{east}$ acts with a second force of $6.0\,\mathrm{N} \text{at} 60 {}^{\circ} \text{north of east}$. Determine the resultant.

Resolve the angled force:

Add like components:

Now calculate the resultant magnitude:

The direction north of east is:

The resultant is therefore $14\,\mathrm{N} \text{at} 22 {}^{\circ} \text{north of east}$.