Scalars and vectors
Which way?
- Walk $3\ \text{km}$ out and $3\ \text{km}$ back: you travel $6\ \text{km}$, but end up where you started.
- The distance is $6\ \text{km}$; the displacement is $0$.
- The difference is direction — and that is the whole idea here.
Scalars and vectors
- A scalar has size only.
- A vector has size and direction.
You walk $4\ \text{km}$ east, then $4\ \text{km}$ back west. What is your displacement, in km?
Displacement is a vector — start to finish. You end where you began, so it is $0$. The distance travelled is $8\ \text{km}$.
Which is which?
- Scalars: mass, time, temperature, energy, work, power, distance, speed, density.
- Vectors: displacement, velocity, acceleration, force (incl. weight), momentum.
Select all the quantities that are vectors.
Vectors have a direction: displacement, velocity, force and momentum. Mass, energy and speed are scalars (speed is the size of velocity, with no direction).
A quick test
- Ask: "in which direction?"
- "In which direction is the temperature?" makes no sense → temperature is a scalar.
- "In which direction is the velocity?" makes sense → velocity is a vector.
Temperature is a vector.
You cannot ask "in which direction is the temperature?", so it is a scalar.
Vectors are arrows
- Draw a vector as an arrow.
- Its length (to scale) shows the magnitude; its direction shows the direction.
Adding vectors: tip to tail
- Draw the arrows tip to tail.
- The resultant goes from the first tail to the last tip.
Two forces act at a point: $3\ \text{N}$ east and $4\ \text{N}$ north. What is the size of the resultant?
They are at right angles, so $|R| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\ \text{N}$.
Subtracting vectors
- To find $\vec{X} - \vec{Y}$, add the reverse of $\vec{Y}$: $\vec{X} + (-\vec{Y})$.
- The reverse has the same size but points the opposite way.
To find $\vec{X} - \vec{Y}$ by drawing arrows, you:
$\vec{X} - \vec{Y} = \vec{X} + (-\vec{Y})$ — reverse $\vec{Y}$ (same size, opposite direction) and add tip to tail.
Resolving into components
- Any vector splits into two perpendicular parts.
- For a vector $v$ at angle $\theta$ to the horizontal: $v_{\text{H}} = v\cos\theta$, $v_{\text{V}} = v\sin\theta$.
A velocity of $10\ \dfrac{\text{m}}{\text{s}}$ points $60^{\circ}$ above the horizontal. What is its horizontal part?
Horizontal part $= v\cos\theta = 10 \times \cos 60^{\circ} = 10 \times 0.5 = 5\ \dfrac{\text{m}}{\text{s}}$.
On a slope
- On a slope of angle $\theta$, split the weight along and across the slope.
- Along: $W_{\parallel} = W\sin\theta$. Across: $W_{\perp} = W\cos\theta$.
On a slope of angle $\theta$, the part of the weight acting along the slope is $W$ ____ $\theta$.
Along the slope, $W_{\parallel} = W\sin\theta$; the part pressing into the slope is $W_{\perp} = W\cos\theta$.
When do you resolve?
- Whenever you need how much of a vector acts in one direction.
- Example: the part of a force pulling along a ramp, or the up and across parts of a thrown ball's velocity.
You've got it
- a vector has direction, a scalar does not (ask "in which direction?")
- add vectors tip to tail; at right angles $|R| = \sqrt{X^{2}+Y^{2}}$
- resolve a vector with $v\cos\theta$ (horizontal) and $v\sin\theta$ (vertical)