Why 3 Newtons Plus 4 Newtons Can Equal 5 Newtons

Two bags of $3$ kg and $4$ kg always weigh $7$ kg together. But two ropes pulling a cart with $3$ N and $4$ N can give **anything from $1$ N to $7$ N — it depends on the directions of the pulls.

Quantities with direction need their own mathematics: vectors.** They are the language of all motion in a plane, and of forces, fields and torque later.

This part covers scalars and vectors, adding vectors, resolving vectors, and the dot and cross products.

**A scalar has only magnitude, while a vector has magnitude and direction and adds by the triangle law; in a plane, a vector is written as $x\hat{i}+y\hat{j}$ using unit vectors along the axes.

- Scalars — mass, time, distance, speed, work, energy, temperature
- Vectors — displacement, velocity, acceleration, force, momentum

Position and displacement.** The position vector of $(x,y)$ is $\vec{r}=x\hat{i}+y\hat{j}$, and displacement is $\Delta \vec{r}={\vec{r}}_2-{\vec{r}}_1$.

Worked example. A body moves from $P(2,3)$ m to $Q(5,7)$ m.

$$\Delta \vec{r}=(5-2)\hat{i}+(7-3)\hat{j}=3\hat{i}+4\hat{j}\text{ m}|\Delta \vec{r}|=\sqrt{9+16}=5\text{ m}$$

Unit vector in that direction:

$$\hat{n}=\frac{3\hat{i}+4\hat{j}}{5}=0.6\hat{i}+0.8\hat{j}$$

An everyday example. Telling a friend to walk 3 km will not get them to your house; 3 km north will.

The substance. Electric current has a direction but is a scalar, because currents meeting at a junction add as plain numbers, not by the triangle law.

**Place vectors head to tail and join the first tail to the last head (triangle and polygon laws), or draw them from one point as sides of a parallelogram whose diagonal is the resultant $R=\sqrt{A^2+B^2+2AB\cos \theta }$.**

The direction of $\vec{R}$ from $\vec{A}$ is $\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$. To subtract, add the reversed vector: $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$. Two vectors are equal only if they have the same magnitude and the same direction.

**Worked example 1 — at $90°$.** $A=3$ N and $B=4$ N:

$$R=\sqrt{9+16}=5\text{ N}\tan \alpha =\frac{4}{3}\Rightarrow \alpha \approx 53.1°$$

**Worked example 2 — at $60°$.**

$$R=\sqrt{9+16+2(3)(4)(0.5)}=\sqrt{37}\approx 6.08\text{ N}\tan \alpha =\frac{4(0.866)}{3+4(0.5)}\approx 0.693\Rightarrow \alpha \approx 34.7°$$

$$|\vec{A}-\vec{B}|=\sqrt{9+16-12}=\sqrt{13}\approx 3.61\text{ N}$$

Worked example 3 — polygon law. Walking $4$ m east, $3$ m north, then $4$ m west gives a resultant of **$3$ m north.

An everyday example. Two people pulling a handcart with ropes at an angle move it along the diagonal of the parallelogram formed by their pulls.

The substance. The resultant always lies between $|A-B|$ and $A+B$** — here between $1$ N and $7$ N.

**A vector $\vec{A}$ at angle $\theta$ to the x-axis has components $A_x=A\cos \theta$ and $A_y=A\sin \theta$; its magnitude is $\sqrt{A_x^2+A_y^2}$ and its direction is ${\tan }^{-1}\frac{A_y}{A_x}$.

Worked example 1 — resolving.** A $10$ N force at $30°$ to the x-axis:

$$A_x=10\cos 30°\approx 8.66\text{ N}{A}_y=10\sin 30°=5\text{ N}$$

Worked example 2 — adding by components. $\vec{A}=3\hat{i}+4\hat{j}$ and $\vec{B}=5\hat{i}-2\hat{j}$:

$$\vec{R}=8\hat{i}+2\hat{j}R=\sqrt{64+4}=\sqrt{68}\approx 8.25\theta ={\tan }^{-1}\frac{2}{8}\approx 14.0°$$

Worked example 3 — rebuilding. Components $A_x=-6$ and $A_y=8$ give $A=10$; since $A_x<0$ and $A_y>0$, the vector lies in the second quadrant, at $180°-53.1°=126.9°$ from the x-axis.

An everyday example. Pulling a trolley bag by its tilted handle: only the horizontal component of your pull moves it forward; the vertical component just lifts it slightly.

The substance. Adding components is far quicker than the parallelogram formula when three or more vectors are involved.

**The dot product $\vec{A}\cdot \vec{B}=AB\cos \theta$ is a scalar that measures how much one vector points along the other; the cross product has magnitude $AB\sin \theta$, points perpendicular to both by the right-hand rule, and equals the area of the parallelogram they form.

Dot product.** $\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$; it is zero for perpendicular vectors.

Worked example 1. $\vec{A}=2\hat{i}+3\hat{j}-\hat{k}$, $\vec{B}=\hat{i}-\hat{j}+2\hat{k}$:

$$\vec{A}\cdot \vec{B}=2-3-2=-3\cos \theta =\frac{-3}{\sqrt{14}\sqrt{6}}\approx -0.327\Rightarrow \theta \approx 109.1°$$

Worked example 2 — work. $\vec{F}=5\hat{i}+2\hat{j}$ N moves a body through $\vec{s}=3\hat{i}+4\hat{j}$ m: $W=\vec{F}\cdot \vec{s}=15+8=23$ J.

Cross product. $\hat{i}\times \hat{j}=\hat{k}$, and $\vec{A}\times \vec{B}=-\vec{B}\times \vec{A}$.

Worked example 3. $\vec{A}=2\hat{i}+3\hat{j}$, $\vec{B}=\hat{i}+4\hat{j}$:

$$\vec{A}\times \vec{B}=(2\times 4-3\times 1)\hat{k}=5\hat{k}\text{parallelogram area}=5\text{ square units}$$

An everyday example. Tightening a nut with a spanner works best with a perpendicular push — the cross product is largest at $90°$.

The substance. **$\vec{A}\cdot \vec{A}=A^2$ but $\vec{A}\times \vec{A}=0$.**