# A-level Mathematics/OCR/M1/Force as a Vector
## Vector

A vector is a mathematical quantity that has both a magnitude and a direction. The direction is usually expressed as an angle in degrees. For example if it is said that a ball is rolling down a hill at 30 metres per second and the hill has a 30 degree slope then this is a vector quantity. However if it is only said that it is rolling down the hill at 30 metres per second then it is a scalar quantity. A scalar quantity will only give a magnitude and not a direction. In most mechanics problems the direction will have a profound impact on the motion.

### Representation of Vectors

When you write a vector you must add an arrow to the top like so ${\vec {v}}$. When you draw a vector you need to draw as an arrow slanted at approximately the same degree as the direction. The vector described above would be drawn as such: .

## Components of Force

A force is a vector that can be separated into x and y components which are perpendicular to each other. We separate a force into components so that we can add multiple forces together.
If you draw an accurate enough diagram you can find the x and y components. However this can be tedious and is error prone and is unacceptable on the A-level examination. Instead you need to use trigonometry to separate a vector into x and y components. The force is always the hypotenuse of the triangle. The angle, which is the direction that the force is applied, is usually given in degrees. However sometimes you will also find the angle given in the Cardinal directions (east, south, west, north,etc.).
| Value | Formulae | Value | Formulae |
| $\theta \,$ | $\cos \left({\frac {x}{r}}\right)$ | $x\ force\,$ | $F\cos \theta \,$ |
| $\theta \,$ | $\sin \left({\frac {y}{r}}\right)$ | $y\ force\,$ | $F\sin \theta \,$ |
| $\theta \,$ | $\tan \left({\frac {y}{x}}\right)$ | $\left|\mathbf {F} \right|=F\,$ | ${\sqrt {x^{2}+y^{2}}}\,$ |

### Example 1

Find the x and y components of a 40 Newton force acting on a ball at a 30° angle.
* Using the mathematical formulae we get:
** $y\ Force=40N\sin 30=20N\,$
** $x\ Force=40N\cos 30=34.64N\,$

### Example 2

Find the Angle $\theta \,$ if the resultant force is 24N and the y component of the force is 18.11N.
** Using the mathematical formulae we get:
** $\theta =\arcsin \left({\frac {18.11N}{24N}}\right)=49^{\circ }\,$

## Resultant Force

When multiple forces are acting on one object you will get a resultant force. The resultant force is how all the forces combined will act on the object. The procedure for finding the resultant force is as follows:
* Draw the forces and the angles on a Cartesian plane. This is to ensure that you get the right angles.
* Make all angle in terms of the positive x axis.
* Separate each force into its x and y components.
* Add all the x and y components. You can also use the i and j notation, where i is the horizontal component and j is the vertical component.
* Find the resultant force.
* Find the angle. If a force is negative make it positive. Then place the angle in the right quadrant, because the inverse tangent will place all angles in the first quadrant. If necessary review Core 2 Trigonometry.

### Example 1

A group of people are playing football. The ball is kicked simultaneously by one person with a force of 60N at an angle of 73°, another kicks the ball with a force of 40N at a 20 angle west of south and the third kicks with a 100N force at a 124° angle. What is the resultant force and angle?
* Draw the forces and the angles on a Cartesian plane. This is to ensure that you get the correct angles.
* Make all angle in terms of the positive x axis.
    The 60 N is acting at a 73° angle.
    The 40 N is acting at a 250° angle.
    The 100N is acting at a 124° angle.
* Separate each force into its x and y components.
** For the x components
    $x\ Force=60N\cos 73=17.54N\,$
    $x\ Force=40N\cos 250=-13.68N\,$
    $x\ Force=100N\cos 124=-55.92N\,$
** For the y components
    $y\ Force=60N\sin 73=57.38N\,$
    $y\ Force=40N\sin 250=-37.59N\,$
    $y\ Force=100N\sin 124=82.9N\,$
* Add all the x and y components.
** x Force = 17.54N - 13.68N - 55.92N = -52.06
** y Force = 57.38N - 37.59N + 82.9N = 102.69
* Find the resultant force.
    $F_{resultant}={\sqrt {}-52.06^{2}+102.69^{2})=115.13N$
* Find the angle.
    $\theta =\arctan \left({\frac {102.69}{52.06}}\right)=63.11^{\circ }$
    Since the cosine is negative and the sine is positive the angle is in the second quadrant. 180° - 63.11° = 116.88°.
The resultant force will be 115.13N and will act at a 116.88° angle.