# Scalars and Vectors

A scalar is any quantity that can be represented mathematically by a single real number. A real number $$x$$ is defined by two properties, its modulus $$\lvert \mathbf{x} \rvert$$ and its sign. Thus the magnitude of a scalar is $$\lvert x \rvert$$.

A vector is any quantity that has a magnitude and a direction in space. Vectors are denoted as $$\mathbf{x}$$.

## Using Arrows to Represent Vectors

We can represent vectors diagramatically. An arrow has a length that corresponds to the length of the vector, and points in a direction according to the vector. We often us the Cartesian coordinate system.

The zero vector is the unique vector with length 0. It has no direction. It is denotade as $$\mathbf{0}$$.

Two vectors are equal if they have the same magnitude and direction.

## Vector Algebra

We can scale any vector by using scalar multiplication.

Exercise 3
1. $$-\mathbf{v}$$
2. The vector $$-1.5\mathbf{v}$$ is 150% the size of $$\mathbf{v}$$, but in the opposite direction.
3. $$\frac{1}{\lvert \mathbf{v} \rvert} \mathbf{v}$$ is in the same direction as $$\mathbf{v}$$, but is of length 1.

For any non-zero vector $$\mathbf{v}$$, the unit vector in the direction of $$\mathbf{v}$$ is the vector

$\mathbf{\hat{v}} = \frac{1}{\lvert \mathbf{v} \rvert} \mathbf{v}$

Two common unit vectors are $$\mathbf{i}$$ and $$\mathbf{j}$$, which correspond to the x- and y- axis in the Cartesian plane. They are known as Cartesian unit vectors.

Exercise 4
1. $$-35\mathbf{j}$$
2. $$-112\mathbf{i}$$

Vector addition is commutative, which can be seen by the Parallelogram Rule

Exercise 7

$4(\mathbf{a} - \mathbf{c}) + 3(\mathbf{c} - \mathbf{b}) + 2(2\mathbf{a} - \mathbf{b} - 3\mathbf{c}) \\ = 4\mathbf{a} - 4\mathbf{c} + 3\mathbf{c} - 3\mathbf{b} + 4\mathbf{a} - 2\mathbf{b} - 6\mathbf{c} \\ = 8\mathbf{a} - 5\mathbf{b} -7\mathbf{c}$

# Cartesian Components and Products

## Vectors in Three Dimensions

### A Three Dimensional Cartesian Coorodinate System

The 3-dimensional coordinate system consists of the $$x$$- and $$y$$- axis as before, with a third axis perpendicular to these - the $$z$$-axis. Alternatively, the coordinate system can be described by three planes - the $$(x, y)$$ plane, the $$(x, z)$$ and the $$(y, z)$$ plane. A point $$P$$ in 3 dimensional space can be described as a triple of its perpendicular distances to these three planes - the Cartesian coordinates of $$P$$.

The right and left hand rules help us determine which direction the $$z$$-axis points. In this system, the positive $$x$$- and $$y$$-axis follow the direction of the thumb and index finger, while the middle finger - perpendicular to these - points in the direction of the positive $$z$$ axis.

Exercise 8
1. Left-handed
2. Right-handed
3. Right-handed
4. Right-handed

### Component Form of a Three-Dimensional Vector

The position vector of a point $$A$$ relative to the origin $$O$$ is the displacement of $$A$$ from $$O$$.

A vector $$\mathbf{a} = \vec{PQ}$$ where $$P$$ is the point $$(p_1, p_2, p_3)$$ and $$Q$$ is the point $$(q_1, q_2, q_3)$$ has component form

$\mathbf{a} = a_i\mathbf{i} + a_j\mathbf{j} + a_k\mathbf{k}$

where $$a_i = q_1 - p_1$$, $$a_j = q_2 - p_2$$, $$a_k = q_3 - p_3$$. The scalars $$a_i, a_j, a_k$$ are the (Cartesian) components of $$\mathbf{a}$$.

Exercise 9
1. $\begin{array}{rl} \mathbf{d} &= 2\mathbf{a} - 3\mathbf{b} \\ &= 2(\mathbf{i} + \mathbf{j} + \mathbf{k}) - 3(2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) \\ &= 2\mathbf{i} + 2\mathbf{j} + 2\mathbf{k} - 6\mathbf{i} + 9\mathbf{j} + 3\mathbf{k} \\ &= -4\mathbf{i} + 11\mathbf{j} + 5\mathbf{k} \end{array}$

$\begin{array}{rl} \mathbf{e} &= \mathbf{a} - 2\mathbf{b} + 4 \mathbf{c}\\ &= (\mathbf{i} + \mathbf{j} + \mathbf{k}) - 2(2\mathbf{i} - 3\mathbf{j} - \mathbf{k}) + 4(3\mathbf{i} + \mathbf{k}) \\ &= \mathbf{i} + \mathbf{j} + \mathbf{k} - 4\mathbf{i} + 6\mathbf{j} + 2\mathbf{k} + 12\mathbf{i} + 4\mathbf{k} \\ &= 9\mathbf{i} + 7 \mathbf{j} + 7\mathbf{k} \end{array}$

2. $\lvert \mathbf{d} \rvert = \sqrt{(-4)^2 + 11^2 + 5^2} = 9\sqrt{2}$ $\lvert \mathbf{e} \rvert = \sqrt{9^2 + 7^2 + 7^2} = \sqrt{179}$

3. $\lvert \mathbf{a} \rvert = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ $\mathbf{\hat{a}} = \frac{1}{\sqrt{3}}\mathbf{a}$

4. Expanding, we have

$\mathbf{i} + \mathbf{j} + \mathbf{k} + a_i \mathbf{i} + a_j \mathbf{j} + a_k \mathbf{k} = 2\mathbf{i} - 3\mathbf{j} - \mathbf{k}$

Comparing co-efficients, we have

$\begin{cases} 1 + a_i = 2 \\ 1 + a_j = -3 \\ 1 + a_k = -1 \end{cases}$

Thus $$a_i = 1$$, $$a_j = -4$$, $$a_k = -2$$, and therefore $$\mathbf{x} = \mathbf{i} - 4 \mathbf{j} - 2 \mathbf{k}$$

Exercise 10

$$\mathbf{0} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}\quad \mathbf{i} = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \quad \mathbf{j} = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} \quad \mathbf{k} = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$$

### Vector Equation of a Straight Line

If $$A$$ and $$B$$ are any two distinct points on a straight line in space with position vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ respectively, then the vector equation of a straight line is

$\mathbf{r} = (1 - s)\mathbf{a} + s \mathbf{b}$

where $$\mathbf{r}$$ represents the position vector of any point on the line.

In Cartesian coordinate form, with

$\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$ $\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}$ $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$

the equation of the line is

$x = a_1 + s (b_1 - a_1)\quad,\quad y = a_2 + s (b_2 - a_2)\quad,\quad z = a_3 + s (b_3 - a_3)$

This is the parametric form of the straight line, where coordinates are expressed in terms of a parameter $$s$$.

Exercise 11
1. $\mathbf{r} = x\mathbf{i} + y \mathbf{j} + z \mathbf{k}$ where $x = 1 + s \quad , \quad y = 1 + 2s \quad , \quad z = 2 - s$

2. The point cuts the $$(x, z)$$-plane when $$y = 0$$. That is,

$0 = 1 + 2s$

Thus $$s = -\frac{1}{2}$$. At this parameter the coordinates are

$\mathbf{r} = \frac{1}{2}\mathbf{i} + 0\mathbf{j} + \frac{5}{2}\mathbf{k}$

Exercise 12
1. $\lvert \mathbf{a} \rvert = \sqrt{2^2 + (-1)^2} = \sqrt{5}$ $\lvert \mathbf{b} \rvert = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{35}$

2. $\mathbf{a} + \mathbf{b} = 2\mathbf{i} - \mathbf{j} + \mathbf{i} + 3\mathbf{j} + 5\mathbf{k} = 3\mathbf{i} + 2 \mathbf{j} + 5\mathbf{k}$

$\begin{array}{ll} \mathbf{2a} - \mathbf{b} &= 2(2\mathbf{i} - \mathbf{j}) - (\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) \\ &= 4\mathbf{i} - 2\mathbf{j} - \mathbf{i} - 3\mathbf{j} - 5\mathbf{k} \\ & = 3\mathbf{i} - 5 \mathbf{j} - 5\mathbf{k} \end{array}$

$\begin{array}{ll} \mathbf{c} + \mathbf{2b} - 3\mathbf{a} &= (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) + 2(\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) - 3(2\mathbf{i} - \mathbf{j})\\ &= \mathbf{i} + \mathbf{j} - 2\mathbf{k} + 2\mathbf{i} + 6\mathbf{j} + 10\mathbf{k} - 6\mathbf{i} + 3\mathbf{j}\\ &= -3\mathbf{i} + 10\mathbf{j} + 8\mathbf{k} \end{array}$

3. Let $$\mathbf{p} = 2\mathbf{j} + 3\mathbf{k}$$. Then

$\mathbf{p} + 2\mathbf{a} - \mathbf{b} = 2\mathbf{j} + 3\mathbf{k} + 3\mathbf{i} - 5\mathbf{j} - 5\mathbf{k} = 3\mathbf{i} - 3\mathbf{j} - 2\mathbf{k}$

So $$Q = \left(3, -3, -2\right)$$.

Exercise 13
1. From the given equation, we have $x = 3s - 1$ $y = 8 - 2s$ $z = 1 - s$

Which specifies the line in parametric form.

In vector form, this is:

$\begin{array}{ll} \mathbf{r} &= (3s - 1)\mathbf{i} + (8 - 2s)\mathbf{j} + (1 - s)\mathbf{k} \\ &= (-\mathbf{i} + 8\mathbf{j} + \mathbf{k}) + s(3\mathbf{i} - 2\mathbf{j} - \mathbf{k}) \end{array}$

Therefore, $$\mathbf{a} = -\mathbf{i} + 8\mathbf{j} + \mathbf{k}$$ and $$\mathbf{b} = 4\mathbf{i} + 6\mathbf{j}$$

From the vector form of the line, we have $$\vec{AB} = 3\mathbf{i} - 2\mathbf{j} - \mathbf{k}$$, with $$\lvert \vec{AB} \rvert = \sqrt{3^2 + (-2)^2 + (-1)^2} = \sqrt{14}$$

1. $\begin{array}{ll} M &= (-\mathbf{i} + 8\mathbf{j} + \mathbf{k}) + \frac{1}{2}(3\mathbf{i} - 2\mathbf{j} - \mathbf{k}) \\ &= \frac{1}{2}\mathbf{i} + 7 \mathbf{j} + \frac{1}{2} \mathbf{k} \end{array}$
1. $\begin{array}{ll} N &= (-\mathbf{i} + 8\mathbf{j} + \mathbf{k}) + 0.6(3\mathbf{i} - 2\mathbf{j} - \mathbf{k}) \\ &= 0.8\mathbf{i} + 6.8 \mathbf{j} + 0.4\mathbf{k} \end{array}$

2. $\begin{array}{ll} \mathbf{r} &= (1 - t)(-\mathbf{i} - 2 \mathbf{j}) + t (2\mathbf{i} + \mathbf{j} - \mathbf{k}) \\ &= (-\mathbf{i} - 2\mathbf{j}) + t (3\mathbf{i} + 3\mathbf{j} - \mathbf{k}) \\ &= (3t - 1) \mathbf{i} + (3t - 2) \mathbf{j} - t\mathbf{k} \end{array}$

3. Setting the equations equal, we have

 $\begin{cases} 3s - 1 = 3t - 1 \\ 8 - 2s = 3t - 2 \\ 1 - s = -t \end{cases}$

As this system of simultaneous equations has no solution, the lines do not intersect.

## The Dot Product

The dot product of two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is

$\mathbf{a} \cdot \mathbf{b} = \lvert \mathbf{a} \rvert \lvert \mathbf{b} \rvert \cos \theta$

where $$\theta$$ is the angle between the two vectors.

As the dot product produces a scalar value, it is also known as the scalar product.

Exercise 14

$\mathbf{a} \cdot \mathbf{b} = 2 \times 4 \times \cos \frac{\pi}{3} = 4$ $\mathbf{b} \cdot \mathbf{c} = 4 \times \cos \frac{\pi}{6} = 2 \sqrt{3}$ $\mathbf{a} \cdot \mathbf{c} = 2 \times \cos \frac{\pi}{4} = 0$ $\mathbf{b} \cdot \mathbf{b} = 4 \times 4 \times \cos 0 = 16$

If two vectors are perpendicular, then their dot product is 0.

The dot product of a vector against itself is the square of its magnitude:

$\mathbf{a} \cdot \mathbf{a} = \lvert \mathbf{a} \rvert ^2$

Exercise 15
1. $\begin{array}{ll} (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) &= \mathbf{a} \cdot (\mathbf{a} - \mathbf{b}) + \mathbf{b} \cdot (\mathbf{a} - \mathbf{b})\\ &= \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \\ &= \mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \\ &= \lvert \mathbf{a} \rvert ^ 2 - \lvert \mathbf{b} \rvert ^2 \end{array}$

2. $\begin{array}{ll} \lvert a + b \rvert ^ 2 &= (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) \\ &= \mathbf{a} \cdot (\mathbf{a} + \mathbf{b}) + \mathbf{b} \cdot (\mathbf{a} + \mathbf{b}) \\ &= \mathbf{a} \cdot \mathbf{a} + \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} \\ &= \mathbf{a} \cdot \mathbf{a} + 2 (\mathbf{a} \cdot \mathbf{b}) + \mathbf{b} \cdot \mathbf{b} \\ &= \lvert \mathbf{a} \rvert^2 + 2 (\mathbf{a} \cdot \mathbf{b}) + \lvert \mathbf{b} \rvert ^ 2 \\ \end{array}$

Exercise 16
1. We know that $$\lvert \mathbf{a} \rvert = \lvert \mathbf{b} \rvert = 1$$ and $$\mathbf{a} \cdot \mathbf{b} = 0$$. If the two vectors are perpendicular we must have

$(2\mathbf{a} + 3\mathbf{b}) \cdot (m\mathbf{a} + \mathbf{b}) = 0$

We can rewrite this as

$\begin{array}{ll} 0 &= (2\mathbf{a} + 3\mathbf{b}) \cdot (m\mathbf{a} + \mathbf{b}) \\ &= 2\mathbf{a} \cdot (m\mathbf{a} + \mathbf{b}) + 3\mathbf{b} \cdot (m\mathbf{a} + \mathbf{b}) \\ &= 2\mathbf{a} \cdot m\mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + 3\mathbf{b} \cdot m\mathbf{a} + 3\mathbf{b} \cdot \mathbf{b} \\ &= \lvert 2\mathbf{a} \rvert \lvert m\mathbf{a} \rvert + \lvert 3\mathbf{b} \rvert \lvert \mathbf{b} \rvert \\ &= 2m + 3 \end{array}$

Therefore $$m = -\frac{3}{2}$$

2. $\begin{array}{ll} \lvert \mathbf{c} \rvert &= \sqrt{\lvert \mathbf{c} \rvert ^ 2} \\ &= \sqrt{\lvert 3\mathbf{a} + 5 \mathbf{b} \rvert ^ 2} \\ &= \sqrt{\lvert 3\mathbf{a} \rvert ^ 2 + 2 (3\mathbf{a} \cdot 5 \mathbf{b}) + \lvert 5 \mathbf{b} \rvert ^2} \\ &= \sqrt{\lvert 3\mathbf{a} \rvert ^ 2 + \lvert 5 \mathbf{b} \rvert ^2} \\ &= \sqrt{3^2 + 5^2} \\ &= \sqrt{34} \\ \end{array}$

### Component Form of the Dot Product

If $$\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$$ and $$\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}$$, then $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b _2 + a_3 b_3$

Exercise 17

$\begin{array}{rl} \mathbf{a} \cdot \mathbf{b} &= 4 \times 1 + 1 \times (-3) + (-5) \times 1 \\ &= 4 - 3 - 5\\ &= -4 \end{array}$

As this dot product is negative, the angle between $$\mathbf{a}$$ and $$\mathbf{b}$$ is obtuse.

### Angle Between Vectors

The angle $$\theta$$ between any two non-zero vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is given by

$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\lvert \mathbf{a} \rvert \lvert \mathbf{b} \rvert}$

Exercise 18

$\lvert \mathbf{a} \rvert = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}$ $\lvert \mathbf{b} \rvert = \sqrt{(-1)^2 + 2^2 + 4^2} = \sqrt{21}$ $\begin{array}{ll} \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\sqrt{14} \sqrt{21}} = \frac{- 2 - 6 + 4}{\sqrt{14}\sqrt{21}} = -\frac{4}{7\sqrt{6}} \end{array}$

Therefore $\theta = 1.806$.

### Resolving a Vector into Components

Exercise 19

$\mathbf{a} \cdot \mathbf{i} = a_1$ $\mathbf{a} \cdot \mathbf{j} = a_2$ $\mathbf{a} \cdot \mathbf{k} = a_3$

The component of a vector $$\mathbf{a}$$ in the direction of the unit vector $$\mathbf{\hat{u}}$$ is $$\mathbf{a} \cdot \mathbf{\hat{u}}$$.

Exercise 20
1. $\mathbf{a} \cdot \mathbf{c} = - 2 - 3 + 3 = -2$ $\mathbf{a} \cdot \mathbf{d} = - 4 + 1 = -3$ $\mathbf{a} \cdot \mathbf{e} = -2 + 3 -1 = 0$

Therefore only $$\mathbf{a}$$ and $$\mathbf{e}$$ are perpendicular.

2. Let $$\mathbf{u} = \mathbf{i} + \mathbf{j} + \mathbf{k}$$. Then $$\mathbf{\hat{u}} = \frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k})$$. The component of $$\mathbf{a} + 2\mathbf{b}$$ along $$\mathbf{\hat{u}}$$ is $\begin{array}{ll} (\mathbf{a} + 2 \mathbf{b}) \cdot \mathbf{\hat{u}} &= (2\mathbf{i} − 3 \mathbf{j} + \mathbf{k} + 2(-\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})) \cdot (\frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k})) \\ &= (2\mathbf{i} − 3 \mathbf{j} + \mathbf{k} - 2\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}) \cdot (\frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k})) \\ &= (\mathbf{j} + 9\mathbf{k}) \cdot (\frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k})) \\ &= \frac{1}{\sqrt{3}} + \frac{9}{\sqrt{3}} \\ &= \frac{10}{\sqrt3} \end{array}$

Exercise 21

$$\mathbf{p}$$ makes the angle $$\pi - \alpha$$ with $$\mathbf{i}$$ and $$\frac{\pi}{2} - \alpha$$ with $$\mathbf{j}$$. Therefore the $$\mathbf{i}$$-component of $$\mathbf{p}$$ is $$-\lvert\mathbf{p}\rvert \cos \alpha = - 2.5 \cos \alpha$$, and the $$\mathbf{j}$$-component is $$2.5 \sin \alpha$$.

Hence $$\mathbf{p} = -2.5 \cos \alpha \mathbf{i} + 2.5 \sin \alpha \mathbf{j}$$.

$$\mathbf{q}$$ makes the angle $$\beta$$ with $$-\mathbf{i}$$ and $$\frac{\pi}{2} + \beta$$ with $$\mathbf{j}$$. Therefore, the $$\mathbf{i}$$-component of $$\mathbf{q}$$ is $$-\lvert \mathbf{q} \rvert \cos \beta = -3\cos\beta$$ and the $$\mathbf{j}$$-component is $$-\lvert \mathbf{q} \rvert \sin \beta = -3\sin\beta$$.

Hence $$\mathbf{q} = -3 \cos \beta \mathbf{i} - 3 \sin \beta \mathbf{j}$$.

$$\mathbf{r}$$ makes the angle $$-\gamma$$ with $$\mathbf{i}$$ and $$\frac{\pi}{2} + \gamma$$ with $$\mathbf{j}$$. Therefore, the $$\mathbf{i}$$-component of $$\mathbf{r}$$ is $$\lvert \mathbf{r} \rvert \cos \gamma = 2.5\cos\gamma$$ and the $$\mathbf{j}$$-component is $$-\lvert \mathbf{r} \rvert \sin \gamma = -2.5\sin\gamma$$.

Hence $$\mathbf{r} = 2.5 \cos \gamma \mathbf{i} - 2.5 \sin \gamma \mathbf{j}$$.

## The Cross Product

The cross product of two vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ is given by

$\mathbf{a} \times \mathbf{b} = (\lvert \mathbf{a} \rvert \lvert \mathbf{b} \rvert \sin \theta) \mathbf{\hat{n}}$

where $$\theta$$ is the angle between $$\mathbf{a}$$ and $$\mathbf{b}$$, and $$\mathbf{\hat{n}}$$ is a unit vector perpendicular to both $$\mathbf{a}$$ and $$\mathbf{b}$$.

The direction of $$\mathbf{\hat{n}}$$ can be seen using the right-hand rule - align $$\mathbf{a}$$ with the index finger and $$\mathbf{b}$$ with the ring finger, then $$\mathbf{\hat{n}}$$ is in the direction of the thumb.

Note that the cross product is not commutative. The order in which the vectors are written determines the direction $$\mathbf{\hat{n}}$$.

Exercise 22

$\begin{array}{ll} \mathbf{u} \times \mathbf{v} &= (\lvert \mathbf{u} \rvert \lvert \mathbf{v} \rvert \sin \frac{\pi}{6}) \mathbf{k}\\ &= (2 \cdot 3 \cdot \frac{1}{2})\mathbf{k} \\ &= 3\mathbf{k} \end{array}$

$\begin{array}{ll} \mathbf{u} \times \mathbf{w} &= (\lvert \mathbf{u} \rvert \lvert \mathbf{w} \rvert \sin 0) \mathbf{k}\\ &= \mathbf{0} \end{array}$

$\begin{array}{ll} \mathbf{v} \times \mathbf{w} &= (\lvert \mathbf{v} \rvert \lvert \mathbf{w} \rvert \sin \frac{\pi}{6}) (-\mathbf{k})\\ &= (3 \cdot 4 \cdot \frac{1}{2}) (-\mathbf{k}) \\ &= -6 \mathbf{k} \end{array}$

Exercise 23
1. $\mathbf{i} \times \mathbf{j} = (1 \cdot 1 \cdot 1) \mathbf{k} = \mathbf{k}$ $\mathbf{j} \times \mathbf{k} = (1 \cdot 1 \cdot 1) \mathbf{i} = \mathbf{i}$ $\mathbf{k} \times \mathbf{i} = (1 \cdot 1 \cdot 1) \mathbf{j} = \mathbf{j}$

2. $\mathbf{j} \times \mathbf{i} = -\mathbf{k}$ $\mathbf{k} \times \mathbf{j} = -\mathbf{i}$ $\mathbf{i} \times \mathbf{k} = -\mathbf{j}$

3. $\mathbf{i} \times \mathbf{i} = \mathbf{0}$ $\mathbf{j} \times \mathbf{j} = \mathbf{0}$ $\mathbf{k} \times \mathbf{k} = \mathbf{0}$

4. $\begin{array}{ll} (\mathbf{i} + \mathbf{k}) \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) &= \mathbf{i} \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mathbf{k} \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) \\ &= \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{j} + \mathbf{i} \times \mathbf{k} + \mathbf{k} \times \mathbf{i} + \mathbf{k}\times\mathbf{j} + \mathbf{k}\times\mathbf{k} \\ &= \mathbf{0} + \mathbf{k} - \mathbf{j} + \mathbf{j} - \mathbf{i} + \mathbf{0} \\ &= -\mathbf{i} + \mathbf{k} \end{array}$

$\begin{array}{ll} \left( \mathbf{i} \times \left( \mathbf{i} + \mathbf{k} \right) \right) - \left(\left(\mathbf{i} + \mathbf{j}\right)\times\mathbf{k}\right) \ &= \left( \mathbf{i} \times \mathbf{i} + \mathbf{i}\times \mathbf{k} \right) - \left(\mathbf{i} \times \mathbf{j} + \mathbf{j} \times \mathbf{k}\right) \\ &= \left( \mathbf{0} - \mathbf{j} \right) - \left(- \mathbf{j} + \mathbf{i}\right) \\ &= -\mathbf{i} \end{array}$

5. $\begin{array}{ll} \left( \mathbf{a} + \mathbf{b} \right) \times \left( \mathbf{a} + 2\mathbf{b} \right) &= \mathbf{a} \times \left( \mathbf{a} + 2\mathbf{b} \right) + \mathbf{b} \times \left( \mathbf{a} + 2\mathbf{b} \right) \\ &= \mathbf{a} \times \mathbf{a} + \mathbf{a} \times 2\mathbf{b} + \mathbf{b} \times \mathbf{a} + \mathbf{b} \times 2\mathbf{b} \\ &= \mathbf{0} + 2 \left( \mathbf{a} \times \mathbf{b} \right) - (\mathbf{a} \times \mathbf{b}) + \mathbf{0} \\ &= \mathbf{a} \times \mathbf{b} \end{array}$

$\begin{array}{ll} \left( \mathbf{a} + \mathbf{b} \right) \times \left( \mathbf{a} - \mathbf{b} \right) &= \mathbf{a} \times \mathbf{a} - \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} - \mathbf{b} \times \mathbf{b} \\ &= \mathbf{0} - \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{a} - \mathbf{0} \\ &= \mathbf{b} \times \mathbf{a} - \mathbf{a} \times \mathbf{b} \\ &= \mathbf{0} \end{array}$

if $$\mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$$ and $$\mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k}$$, then

### Component Form of the Cross Product

$\mathbf{a} \times \mathbf{b} = (a_2 b_3 - a_3 b_2) \mathbf{i} + (a_3 b_1 - a_1 b _3) \mathbf{j} + (a_1 b_2 - a_2 b_1) \mathbf{k}$

Exercise 24
1. $\begin{array}{ll} \mathbf{a} \times \mathbf{b} &= (2\mathbf{i} − 3\mathbf{j} + \mathbf{k}) \times (−\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) \\ &= (-3 \cdot 4 - 1 \cdot 2) \mathbf{i} + (1 \cdot -1 - 2 \cdot 4) \mathbf{j} + (2 \cdot 2 - -3 \cdot -1) \mathbf{k} \\ &= -14\mathbf{i} - 9\mathbf{j} + \mathbf{k} \end{array}$

2. $\begin{array}{ll} \mathbf{a} \times \mathbf{c} &= (2\mathbf{i} − 3\mathbf{j} + \mathbf{k}) \times (-4\mathbf{i} + 6 \mathbf{j} - 2 \mathbf{k}) \\ &= \mathbf{0} \end{array}$

3. $\begin{array}{ll} \mathbf{b} \times \mathbf{c} &= (−\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) \times (-4\mathbf{i} + 6 \mathbf{j} - 2 \mathbf{k}) \\ &= -28 \mathbf{i} - 18 \mathbf{j} - 18 \mathbf{k} \end{array}$

From these results, we can see that $$\mathbf{a}$$ and $$\mathbf{b}$$ are parallel, as their cross product is the zero vector.

Exercise 25

A vector perpendicular to these vectors is given as $$\mathbf{a} \times \mathbf{b}$$:

$\begin{array}{ll} \mathbf{a} \times \mathbf{b} &= \left(2\mathbf{i} + 2\mathbf{j} + \mathbf{k}\right) \times \left(4 \mathbf{i} + 4 \mathbf{j} − 7 \mathbf{k} \right) \\ &= -18 \mathbf{i} + 18 \mathbf{j} \end{array}$

Now $\lvert \mathbf{a} \times \mathbf{b} \rvert = \sqrt{(-18)^2 + 18^2} = 18 \sqrt {2}$ so a unit vector perpendicular to $$\mathbf{a}$$ and $$\mathbf{b}$$ is

$\frac{1}{\lvert \mathbf{a} \times \mathbf{b} \rvert} \mathbf{a} \times \mathbf{b} = -\frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j}$

### Geometric Applications

The area of parallelogram with sides $$\mathbf{a}$$ and $$\mathbf{b}$$ is $$\lvert \mathbf{a} \times \mathbf{b}\rvert$$.

The area of triangle with sides $$\mathbf{a}$$ and $$\mathbf{b}$$ is $$\frac{1}{2} \lvert \mathbf{a} \times \mathbf{b}\rvert$$.

The volume of a parellelpiped with edges $$\mathbf{a}$$, $$\mathbf{b}$$ and $$\mathbf{c}$$ is given by  $The quantity$ $$\mathbf{b} \cdot \mathbf{c}$$ is a *scalar triple product$. # Modelling Forces This section looks at modelling four types of forces: gravity, a force on an object, tension from string and the friction between surfaces. ## Particles A particle is a material object whose size and internal structure may be neglected. It has mass but no size, thus occupies a single point in space. Each force on an object can be seen as a force acting on a particular point - this is the point of action of the force. In the particle model, all forces are modelling as acting at the point of the particle. In this model, the overall outcome can be seen as a single vector that is the sum of all forces. If this sum is zero then object doesn’t move, and thus is in a state of equilibrium. A particle subjected to forces $\mathbf{F}_1, \mathbf{F}_2, \ldots, \mathbf{F}_n$ is in equilibrium if $\sum\limits_{i=1}^n \mathbf{F}_i = \mathbf{0}$ ## Weight An object of mass$m has weight $$\mathbf{W}$$ of magnitude $$\lvert\mathbf{W}\rvert = mg$$, where $$g$$ is the magnitude of acceleration due to gravity, with direction towards the centre of the Earth.

Exercise 28

$$\mathbf{W} = 3g\mathbf{k}$$

Exercise 29

We have $$15g\mathbf{W} \cdot \mathbf{i} = 15g \cos \frac{2 \pi}{3} = -\frac{15g}{2}$$, and $$15g\mathbf{W} \cdot \mathbf{j} = 15g\mathbf \cos \frac{\pi}{6} = \frac{15g\sqrt{3}}{2}$$

Therefore, $$\mathbf{W} = \frac{15g}{2} \left( -\mathbf{i} + \sqrt{3} \mathbf{j} \right)$$.

## Normal Reaction

The force exerted by any surface of an object on an object in contact with it is called the normal reaction. We typically write the normal reaction force as $$\mathbf{N}$$. The normal reaction acts normally - that is, at a right angle - with the surface itself.

## Tension

The force exerted on an object hanging from a cable is the tension force, $$\mathbf{T}$$.

A model string is an object with a fixed finite length and no area, volume or mass, that exerts a force at the point of attachement with an object.

The tension force due to a string is a vector quantity that acts along the length of the string away from the point of attachment.

## Friction

The friction force $$\mathbf{F}$$ operates perpendicular to the normal reaction, and opposes any force of motion. The friction force is not unlimited - it can be overcome, at which point motion occurs. The equation $$\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert$$ allows us to approximate when motion will occur, as this equation describes when an object is on the point of slipping. The constant $$\mu$$ is known as the *coefficient of static friction. Exercise 31 We choose the axis to be $$\mathbf{i}$$ and $$\mathbf{j}$$, where these vectors are the standard unit vectors. The block has mass $$2$$kg, and so $$\mathbf{W} = -2g\mathbf{j}$$, and as the block rests on a solid surface, $$\mathbf{N} = 2g\mathbf{j}$$. The block will be on the point of slipping when $$\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert$$, that is, when $$\lvert \mathbf{F} \rvert = 0.74 * 2g \approx 7.26$$. As $$\lvert \mathbf{F} \rvert = 2 \lt 7.26$$, the block is stationary. The magnitude of the friction force is 2N, to balance the force being applied to the block. Exercise 32 When modelling the box as a particle, the surface area is not taken in to account. Thus both of these orientations will behave the same. Exercise 33 We choose the axis $$\mathbf{i}$$ to be down the ramp, and $$\mathbf{j}$$ to be perpendicular to the ramp. To be in equilibrium, we must have $\mathbf{F} + \mathbf{T} + \mathbf{N} + \mathbf{W} = \mathbf{0}$ Resolving each of these forces into components, we have $\mathbf{W} = \lvert \mathbf{W} \rvert \cos \frac{\pi}{3} \mathbf{i} - \lvert \mathbf{W} \rvert \sin \frac{\pi}{3} \mathbf{j} = \frac{1}{2} \lvert \mathbf{W} \rvert \mathbf{i} - \frac{\sqrt{3}}{2} \lvert \mathbf{W} \rvert \mathbf{j}$ $\mathbf{T} = \lvert \mathbf{T} \rvert (- \mathbf{i})$ $\mathbf{F} = \lvert \mathbf{F} \lvert (- \mathbf{i})$ $\mathbf{N} = \lvert \mathbf{N} \lvert \mathbf{j}$ The equilibrium equation can now be re-written in the $$\mathbf{i}$$-component: $\frac{1}{2} \lvert \mathbf{W} \rvert - \lvert \mathbf{T} \rvert - \lvert \mathbf{F} \rvert = 0$ The magnitude of the force from friction when the block is on the point of moving is given by $\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert = 0.2 \frac{\sqrt{3}}{2} \lvert \mathbf{W} \rvert$ Therefore $\lvert T \rvert = \frac{1 - 0.2 \sqrt{3}}{2} \lvert \mathbf{W} \rvert \approx 192$ Exercise 34 1. We choose the axis to be $$\mathbf{i}$$ aligned down the ramp, and $$\mathbf{j}$$ aligned perpendicular away from the ramp. This scenario involves three forces, the weight of the box $$\mathbf{W}$$, the normal reaction from the ramp $$\mathbf{N}$$, and the force of friction preventing the box from moving $$\mathbf{F}$$. As the box is stationary, we have the equation $\mathbf{F} + \mathbf{N} + \mathbf{W} = \mathbf{0}$ Resolving these forces into components, we have $\mathbf{N} = \lvert \mathbf{N} \rvert \mathbf{j}$ $\mathbf{F} = \lvert \mathbf{F} \rvert (- \mathbf{i})$ $\mathbf{W} = \lvert \mathbf{W} \rvert \cos (\frac{\pi}{2} - \alpha) \mathbf{i} - \lvert \mathbf{W} \rvert (\frac{\pi}{2} \sin \alpha) = \lvert \mathbf{W} \rvert \sin \alpha \mathbf{i} - \lvert \mathbf{W} \rvert \cos \alpha \maAthbf{j}$ $\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert$ Resolving the equilibrium equation in $$\mathbf{i}$$-direction gives $-\lvert \mathbf{F} \rvert + 0 + \lvert \mathbf{W} \rvert \sin \alpha = 0$ Hence $\lvert \mathbf{F} \rvert = m g \sin \alpha$ Resolving in the $$\mathbf{j}$$-direction, we have $0 + \lvert \mathbf{N} \rvert - mg \cos \alpha = 0$ And so $$\lvert \mathbf{N} \rvert = mg \cos \alpha$$

Therefore, as

$\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert$

We have

$\mu = \frac{\lvert \mathbf{F} \rvert}{\lvert \mathbf{N} \rvert} = \frac{mg \sin \alpha}{mg \cos\alpha} = \tan \alpha$

2. Part (a) tells us that the coefficient of friction depends only on the angle of the plane, and not the weight of the objects on it. Therefore both mugs will begin to slip at the same angle.

# Two-Particle Systems

In a statics problem with two-particles, we have a system where each particle has forces applied, but is otherwise unaware of the other particle.

## Pulleys

A model pulley is an object with no mass or size, over which a model string may pass without resistance. The magnitude of the tension in the string passing over the pulley is the same on both side of the pulley.

Exercise 35
1. Axis have already been chosen, so next we draw the force diagrams for the sack of flour and the stone.

The equilibrium equations are

$\mathbf{W_1} + \mathbf{N} + \mathbf{T_1} + \mathbf{F} = \mathbf{0}$

$\mathbf{T_2} + \mathbf{W_2} = \mathbf{0}$

Resolving into components of the given axis, we have

$\mathbf{W_1} = 50g(- \mathbf{j})$

$\mathbf{N} = \lvert \mathbf{N} \rvert \mathbf{j}$

$\mathbf{F} = \lvert \mathbf{F} \rvert (- \mathbf{i})$

$\mathbf{T_1} = \lvert \mathbf{T_1} \rvert \cos \frac{\pi}{4} \mathbf{i} + \lvert \mathbf{T_1} \rvert \sin \frac{\pi}{4} \mathbf{j} = \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \mathbf{i} + \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \mathbf{j}$

$\mathbf{T_2} = \lvert \mathbf{T_2} \rvert \mathbf{j}$

$\mathbf{W_2} = 15g (- \mathbf{j})$

Considering the equalibrium equations in the $$\mathbf{i}$$-direction, we have

$0 + 0 + \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} - \lvert \mathbf{F} \rvert = 0$

In the $$\mathbf{j}$$-direction, we have

$-50g + \lvert \mathbf{N} \rvert + \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} = 0$

$\lvert \mathbf{T_2} \rvert - 15g = 0$

Also, as we are working with a model pulley, we know that

$\lvert \mathbf{T_1} \rvert = \lvert \mathbf{T_2} \rvert$

2. We have

$\begin{array}{ll} \lvert \mathbf{N} \rvert &= 50g - \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \\ &= 50g - \frac{15g}{\sqrt{2}} \approx 386 \end{array}$

So the magnitude of the normal reaction on the sack is approximately 386N.

3. From the equilibrium equation, we have

$\begin{array}{ll} \lvert \mathbf{F} \rvert &= \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \\ &= \lvert \mathbf{T_2} \rvert \frac{1}{\sqrt{2}} \\ &= 15g \frac{1}{\sqrt{2}} \approx 104 \end{array}$

So the magnitude of the force of friction on the sack is approximately 104N

4. The system is in equilibrium if $\lvert \mathbf{F} \rvert \le \mu \lvert \mathbf{N} \rvert$

Rearranging this, we have

$\mu \ge \frac{\lvert \mathbf{F} \rvert}{\lvert \mathbf{N} \rvert} \approx \frac{104.05}{386} \approx 0.27$

Thus for the system to remain in equilibrium the coefficient of static friction must be at least 0.27.

5. If the sack is only just in contact with the floor, it is on the point of no longer being on the floor. At this point, $$\mathbf{N} = 0$$.

In this scenario, the forces (resolved into component form) are:

$\mathbf{W_1} = 50g(- \mathbf{j})$

$\mathbf{F} = \lvert \mathbf{F} \rvert (- \mathbf{i})$

$\mathbf{T_1} = \lvert \mathbf{T_1} \rvert \cos \frac{\pi}{4} \mathbf{i} + \lvert \mathbf{T_1} \rvert \sin \frac{\pi}{4} \mathbf{j} = \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \mathbf{i} + \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} \mathbf{j}$

$\mathbf{T_2} = \lvert \mathbf{T_2} \rvert \mathbf{j}$

$\mathbf{W_2} = m g (- \mathbf{j})$

In the $$\mathbf{j}$$-direction we have

$-50g + \lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} = 0$

$\lvert \mathbf{T_2} \rvert = mg$

As $\lvert \mathbf{T_1} \rvert = \lvert \mathbf{T_2} \rvert$ we have

$\lvert \mathbf{T_1} \rvert \frac{1}{\sqrt{2}} = \lvert \mathbf{T_2} \rvert \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}} = 50g$

Therefore $$m = 50\sqrt{2}$$, so the mass of the stone is approximately 71kg.

## Slipping

Exercise 36

The man in the well can be modelled as a particle, which is acted on by two forces: the man’s weight and the tension in the rope

The men pulled the man can be modelled a single particle which is acted on by four forces: the men’s weight, the normal reaction of the groud, friction from the ground and the tension in the rope.

As the man in the well is stationary, we can apply equilibrium equations to both particles. We have

$\mathbf{W_1} + \mathbf{T_1} = \mathbf{0}$

$\mathbf{W_2} + \mathbf{N} + \mathbf{T_2} + \mathbf{F} = \mathbf{0}$

The axel is modelled as a model pulley, so $\lvert \mathbf{T_1} \rvert = \lvert \mathbf{T_2} \rvert$

Resolving these forces into component form, we have

$\mathbf{W_1} = \lvert \mathbf{W_1} \rvert (- \mathbf{j})$ $\mathbf{W_2} = \lvert \mathbf{W_2} \rvert (- \mathbf{j})$ $\mathbf{N} = \lvert \mathbf{N} \rvert (\mathbf{j})$ $\mathbf{T_1} = \lvert \mathbf{T_1} \rvert \mathbf{j}$ $\mathbf{T_2} = \lvert \mathbf{T_2} \rvert (- \mathbf{i})$ $\mathbf{F} = \lvert \mathbf{F} \rvert \mathbf{i}$

In the $$\mathbf{i}$$-direction, the equilibrium equations are

$\lvert \mathbf{F} \rvert - \lvert \mathbf{T_2} \rvert = 0$

While in the $$\mathbf{j}$$-direction, the equilibrium equations are

$\lvert \mathbf{T_1} \rvert - \lvert \mathbf{W_1} \rvert = 0$

$\lvert \mathbf{N} \rvert - \lvert \mathbf{W_2} \rvert = 0$

Therefore $\lvert \mathbf{T_1} \rvert = \lvert \mathbf{W_1} \rvert = \lvert \mathbf{T_2} \rvert$

Also, as $\lvert \mathbf{F} \rvert = \mu \lvert \mathbf{N} \rvert$ we have

$\lvert \mathbf{N} \rvert = \frac{\lvert \mathbf{F} \rvert}{\mu}$

But $\lvert \mathbf{F} \rvert = - \lvert \mathbf{T_2} \rvert$ so

$\lvert \mathbf{N} \rvert = \frac{\lvert \mathbf{W_1} \rvert}{\mu}$

Therefore

$\lvert \mathbf{N} \rvert - \lvert \mathbf{W_2} \rvert = \frac{\lvert W_1 \rvert}{\mu} - \lvert \mathbf{W_2} \rvert = 0$

Now $$\lvert \mathbf{W_1} \rvert = 80g$$, $$\lvert \mathbf{W_2} \rvert = 80ng$$ and $$\mu = 0.45$$. Thus

$\frac{80g}{0.45} = 80ng$

Therefore $$n \approx 2.2$$. Therefore it will require at least three men to maintain equilibrium.

# Torques

This section considers the turning effect of forces. This doesn’t apply to particles, as they have an inifinitely small size.

## Extended and Rigid Bodies

An extended body is a material object with a mass, along with one more of length, breadth and depth.

A rigid body is an extended body that does not change its shape.

The point at which a force acts on an extended body is significant, and changes the results. The weight always acts on the objects centre of mass.

## Turning Effect of a Force

A horizontal rod is a rigid body with length but no breadth of depth. Such a rod pivoted at its centre will remain horizontal under the action of two forces $$\mathbf{F_1}$$ and $$\mathbf{F_2}$$ acting vertical downwards at distances $$l_1$$ and $$l_2$$ either side of the pivot provided that

$\lvert \mathbf{F_1} \rvert l_1 = \lvert \mathbf{F_2} \rvert l_2$

Exercise 38

If the seesaw is at rest and horizontal, then we must have

$60g \cdot 1 = mg \cdot 1.2$

Therefore $$m = 50$$, so Jill must weigh 50kg.

The line of action of a force is a straight line in the direction of the force and through the point of action of the force.

The torque $$\mathbf{\Gamma}$$ of a force $$\mathbf{F}$$ about a fixed point $$O$$ is the cross product

$\mathbf{\Gamma} = \mathbf{r} \times \mathbf{F}$

where $$\mathbf{r}$$ is the position vector relative to the origin $$O$$ of a point on the line of action of the force.

Exercise 39

A position vector on the line of action of $$\mathbf{F_1}$$ is $$2\mathbf{i}$$. For $$\mathbf{F_2}$$ we use $$-2\mathbf{i}$$, and for $$\mathbf{F_3} = 2 \mathbf{k}$$.

Therefore, the respective torques are:

$\mathbf{\Gamma_1} = 2\mathbf{i} \times 3 \mathbf{j} = 6\mathbf{k}$ $\mathbf{\Gamma_2} = -2\mathbf{i} \times 3 \mathbf{j} = -6\mathbf{k}$ $\mathbf{\Gamma_3} = 2\mathbf{k} \times 3 \mathbf{j} = -6\mathbf{i}$

Exercise 40
1. If $$\mathbf{r}$$ is a position vector on the point of action relative to $$O$$, also on the line of action, then the angle between $$\mathbf{r}$$ and $$\mathbf{F}$$ is 0. Therefore

$\mathbf{r} \times \mathbf{F} = (\lvert \mathbf{r} \rvert \lvert \mathbf{F} \rvert \sin 0) \mathbf{\hat{n}} = \mathbf{0}$

Therefore, as the torque of a force $$\mathbf{F}$$ with position vector $$\mathbf{r}$$ relative to $$O$$ on the force’s line of action, in this case we see that the torque is zero.

# Appyling the Principles

When a rod is fixed to a surface at a hinge, we introduce an extra force - the *reaction force $$\mathbf{R}$$.

Exercise 41

In this scenario the rod is acted on by three forces - $$\mathbf{T}$$, the of the rope; $$\mathbf{W}$$, the weight of the rod; and $$\mathbf{R}$$, the reaction at the hinge. These forces act around the origin $$O$$, which is positioned at the hinge. All forces act in the vertical plane, so we choose the axis $$\mathbf{i}$$ and $$\mathbf{j}$$.

The equilibrium equations are

$\mathbf{W} + \mathbf{R} + \mathbf{T} = \mathbf{0}$

$\mathbf{\Gamma_1} + \mathbf{\Gamma_2} + \mathbf{\Gamma_3} = \mathbf{0}$

Resolving these forces into component form, we have

$\mathbf{W} = mg(-\mathbf{j})$

$\mathbf{T} = - \lvert \mathbf{T} \rvert \cos \alpha \mathbf{i} + \lvert \mathbf{T} \rvert \sin \alpha \mathbf{j}$

$\mathbf{R} = \lvert \mathbf{R} \rvert \cos \beta \mathbf{i} + \lvert \mathbf{R} \rvert \sin \beta \mathbf{j}$

where $$\beta$$ is an unknown quantity.

We have position vectors relative to the origin for each of these forces:

$\mathbf{r_W} = \frac{l}{2} \mathbf{i}$

$\mathbf{r_R} = \mathbf{0}$

$\mathbf{r_T} = l\mathbf{i}$

Thus the resulting torques are

$\mathbf{\Gamma_w} = \mathbf{r_W} \times \mathbf{W} = \frac{l}{2} \mathbf{i} \times -mg \mathbf{j} = -\frac{mgl}{2}\mathbf{k}$

$\mathbf{\Gamma_R} = \mathbf{0}$

$\mathbf{\Gamma_T} = l\mathbf{i} \times \lvert \mathbf{T} \rvert (- \cos \theta \mathbf{i} + \sin \theta \mathbf{j}) = l \lvert \mathbf{T} \rvert \sin \theta \mathbf{k}$

Resolving the equilibrium equation for torque in the $$mathbf{k}$$-direction, we have

$l\lvert \mathbf{T} \rvert \sin \theta = \frac{mgl}{2}$

Thus $\lvert \mathbf{T} \rvert = \frac{1}{2} mg \mathrm{cosec} \theta$

Resolving the first equilibrium equation in the $$\mathbf{i}$$-direction, we have

$0 - \lvert \mathbf{T} \rvert \cos \alpha + \lvert \mathbf{R} \rvert \cos \beta = 0$

That is,

$R_i = \frac{1}{2} mg \mathrm{cosec} \theta \cos \theta = \frac{1}{2}mg \cot \theta$

Resolving the same equation in the $$\mathbf{j}$$-direction, we have

$-mg + \lvert \mathbf{T} \rvert \sin \alpha + R_j = 0$

That is,

$R_j = mg - \lvert \mathbf{T} \rvert \sin \alpha = \frac{1}{2}mg$

Therefore, the reaction force at the hinge is

$\mathbf{R} = \frac{1}{2}mg \left(\cot \theta \mathbf{i} + \mathbf{j} \right)$

Rewriting the equilibrium equation in terms of $$\mathbf{T}$$, we have

$\mathbf{T} = -\mathbf{R} - \mathbf{W} = -\frac{1}{2}mgcot\theta \mathbf{i} + mg \mathbf{j}$

Therefore, as the $$\mathbf{j}$$ component of $$\mathbf{T}$$ and $$\mathbf{R}$$ is equal, we conclude that the angle is the same for $$\mathbf{R}$$ and $$\mathbf{T}$$