### Vector Geometry

Many ways problem

## Solution

In each situation below, three forces are acting on a particle. Given two of the forces and the fact the particle is in equilibrium, find the third force.

$\mathbf{a} = \quantity{-3\mathbf{i} + 2\mathbf{j}}{N}$

$\mathbf{b} = \quantity{4\mathbf{i} + \mathbf{j}}{N}$

$\mathbf{a}$ has magnitude $\quantity{10}{N}$ and direction $45^{\circ}$ anticlockwise from the positive $x$ axis

$\mathbf{b}= \quantity{6\mathbf{i} - 3\mathbf{j}}{N}$

Is it possible to use a different approach for each situation?

In each case the particle is in equilibrium, which means the forces are balanced in all directions and the net force acting on the particle is zero. Below we discuss three possible approaches.

#### Triangle of forces

Since the particle is in equilibrium we can draw a triangle of forces and see approximately what the third force might be.

Why must the third force join up the other two forces to make a triangle?

Using the geometry of the triangle, we can find the magnitude of the third force with the cosine rule.

\begin{align*} \mathbf{|c|}^2 &= 20^2 + 35^2 - 2 \times 20 \times 35 \times \cos60 \\ \mathbf{|c|} &= 5\sqrt{37} \\ &\approx \quantity{30.4}{N} \end{align*}

To describe the third force fully we need the direction it is acting in. Many methods could be used, but we created two right angled triangles that share a side.

We could have also used Lami’s Theorem, which states that if three forces, acting from a point, are in equilibrium, then the magnitude of each force is proportional to the sine of the angle between the other two forces.

$\dfrac{\mathbf{|a|}}{\sin \alpha} = \dfrac{\mathbf{|b|}}{\sin \beta} = \dfrac{\mathbf{|c|}}{\sin \gamma}$

#### Sum of forces

$\mathbf{a} = \quantity{-3\mathbf{i} + 2\mathbf{j}}{N}$

$\mathbf{b} = \quantity{4\mathbf{i} + \mathbf{j}}{N}$

It is always useful to draw a diagram. It can help to check whether answers seem sensible.

Where would you estimate the third force to be?

Since the particle is in equilibrium the sum of all the forces acting on it must be zero, i.e. $\mathbf{a} + \mathbf{b} + \mathbf{c} = 0$. Therefore

$-3\mathbf{i} + 2\mathbf{j} + 4\mathbf{i} + \mathbf{j} + \mathbf{c} = 0,$

so the third force is $\mathbf{c} =\quantity{-\mathbf{i} - 3\mathbf{j}}{N}.$

What are the similarities between this approach and drawing a triangle of forces as used in the first situation?

#### Resolving forces

$\mathbf{a}$ has magnitude $\quantity{10}{N}$ and direction $45^{\circ}$ anticlockwise from the positive $x$ axis

$\mathbf{b}= \quantity{6\mathbf{i} - 3\mathbf{j}}{N}$

Resolving forces means we break forces down into perpendicular components (usually horizontal and vertical). Since it is in equilibrium we know that the sum of the horizontal forces must be zero, as must the vertical forces. If our third force is $\mathbf{c} = p\mathbf{i}+q\mathbf{j}$ then we have the following diagram.

So $\mathbf{c} = \left(-6-5\sqrt{2}\right)\mathbf{i} + \left(3 - 5\sqrt{2}\right)\mathbf{j}.$

What are the similarities and differences between these approaches?

Would each of these approaches work equally well for the different situations? If not, why not?