Adding and subtracting vectors

Two ropes, one box

Two children pull a box with ropes. One pulls with $30\ \text{N}$, the other with $40\ \text{N}$. How hard is the box pulled altogether?

If both ropes point the same way, the answer is $70\ \text{N}$. If they pull in opposite directions, it is only $10\ \text{N}$. At any angle in between, the answer lies somewhere between $10\ \text{N}$ and $70\ \text{N}$. To find it, we need a rule for adding arrows.

The triangle law: head to tail

Think back to the walk: $3\ \text{km}$ east, then $4\ \text{km}$ north. You did one displacement, and then from where it ended you did the next one. That is exactly how vectors add.

To find $\vec{R} = \vec{A} + \vec{B}$:

1. Draw $\vec{A}$.
2. Draw $\vec{B}$ with its tail on the head of $\vec{A}$.
3. Draw $\vec{R}$ from the tail of $\vec{A}$ to the head of $\vec{B}$.

$\vec{R}$ is the resultant. It does the same job as $\vec{A}$ and $\vec{B}$ together.

With more than two vectors, keep going head to tail. The resultant joins the first tail to the last head. This is the polygon law. If the last head lands on the first tail, the resultant is zero. Three equal pulls at $120^\circ$ to each other cancel out in exactly this way.

The parallelogram law: tail to tail

Sometimes both vectors start at the same point, like two ropes tied to one hook. Draw them from that point, and complete the parallelogram. The diagonal from the common point is $\vec{A} + \vec{B}$. It is the same answer as the triangle law, just drawn differently.

Two facts follow from the picture:

• $\vec{A} + \vec{B} = \vec{B} + \vec{A}$. The order does not matter.
• $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$. The grouping does not matter either.

Subtracting: add the opposite

How do we take $\vec{B}$ away from $\vec{A}$? We add the opposite vector:

In the parallelogram, $\vec{A} - \vec{B}$ is the other diagonal. It goes from the head of $\vec{B}$ to the head of $\vec{A}$.

You will subtract vectors most often to find a change. The change in velocity is $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$: final minus initial.

Think first. A ball hits a wall at $8\ \text{m/s}$ and bounces straight back at $6\ \text{m/s}$. Is the change in velocity $2\ \text{m/s}$?

No. Take the direction away from the wall as positive. Then $\vec{v}_1 = -8\ \text{m/s}$ and $\vec{v}_2 = +6\ \text{m/s}$, so $\Delta v = 6 - (-8) = 14\ \text{m/s}$, away from the wall. Turning round costs a lot of velocity.

Play with both diagonals

Here $\vec{A}$ (blue) and $\vec{B}$ (amber) start from the same point. The green arrow is the sum and the red arrow is the difference: the two diagonals of the parallelogram.

How long is the resultant?

Drawing is slow. We want a formula. Look at the triangle made by $\vec{A}$, $\vec{B}$ and $\vec{R}$. If the angle between $\vec{A}$ and $\vec{B}$ (drawn tail to tail) is $\theta$, the angle inside the triangle is $180^\circ - \theta$. The cosine rule then gives

The plus sign comes from $\cos(180^\circ - \theta) = -\cos\theta$. The direction of $\vec{R}$, as the angle $\alpha$ from $\vec{A}$, is

For the difference, $-\vec{B}$ makes the angle $180^\circ - \theta$ with $\vec{A}$, so the sign flips: $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta}$.

Special cases to know by heart:

• $\theta = 0^\circ$, same direction: $R = A + B$. This is the largest possible value.
• $\theta = 90^\circ$, at right angles: $R = \sqrt{A^2 + B^2}$.
• $\theta = 180^\circ$, opposite directions: $R = |A - B|$. This is the smallest possible value.
• Two equal vectors of size $A$: $R = 2A\cos\dfrac{\theta}{2}$. At $120^\circ$ this gives $R = A$.

So the size of a sum always lies between $|A - B|$ and $A + B$. That is a quick check on any answer.

Worked example. Forces of $8\ \text{N}$ and $5\ \text{N}$ act on a ring, with $60^\circ$ between them. Find the resultant.

1. $R^2 = (8\ \text{N})^2 + (5\ \text{N})^2 + 2 \times 8\ \text{N} \times 5\ \text{N} \times \cos 60^\circ = 129\ \text{N}^2$, so $R = \sqrt{129\ \text{N}^2} = 11.4\ \text{N}$.
2. $\tan\alpha = \dfrac{5\sin 60^\circ}{8 + 5\cos 60^\circ} = \dfrac{4.33}{10.5} = 0.412$, so $\alpha = 22.4^\circ$ from the $8\ \text{N}$ force.
3. Check: $11.4\ \text{N}$ lies between $3\ \text{N}$ and $13\ \text{N}$. Good.

Watch out. The $\theta$ in the formula is the angle between the vectors drawn tail to tail. In a head-to-tail drawing, the angle you see at the joint is $180^\circ - \theta$. Using it by mistake changes the plus sign into a minus sign.

Practise with new numbers

Try it in Python

The program finds the size of the sum of a $30\ \text{N}$ and a $40\ \text{N}$ force, for angles from $0^\circ$ to $180^\circ$.

Remember these

• Triangle law: head to tail. Parallelogram law: tail to tail, take the diagonal.
• $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$, with $\theta$ the angle between the vectors drawn tail to tail.
• $|A - B| \le R \le A + B$.
• $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, and a change is always final minus initial.

Check yourself

Try each one on your own first. If you are stuck, open a hint. When you have an answer, check it with Show the answer.

1. Two forces of $5\ \text{N}$ and $12\ \text{N}$ act at right angles. How big is the resultant?

2. What are the largest and the smallest possible resultants of two vectors of sizes $7$ and $3$?

3. Two equal forces of $10\ \text{N}$ act on a point. At what angle between them is the resultant also $10\ \text{N}$?

4. A car's velocity changes from $20\ \text{m/s}$ east to $20\ \text{m/s}$ north. How big is the change in velocity, and which way does it point?