Interactive Lecture
Forces as Vectors in 2D
by Dr Arnaldo Delli Carri
Forces have an intensity (magnitude), a direction (line of action) and a verse (sense), and a single number is not sufficient to represent these aspects, so they need a special mathematical object to represent them: this object is known as vector.
Vector forces have a Cartesian representation: \[\mathbf{F} = F_x \, \mathbf{i} + F_y \, \mathbf{j} \] where
or a polar representation: \[\mathbf{F} = \left(F,\theta \right) = F \,e^{i\varphi} \] where
Graphically, a vector can be represented with an arrow. For example, the force \(\mathbf{F}=3\mathbf{i}+4\mathbf{j}\) is the vector represented below (click the "Toggle Cartesian" button to verify). You can read this as "3 units in the x-direction (horizontal) and 4 units in the y-direction (vertical)".
The polar representation of the same vector force is \(\mathbf{F}=\left(5,53°\right) = 5\, e^{i 0.93}\) (click the "Toggle Polar" button to verify) and can be read as "magnitude 5 units at an angle 53 degrees from the horizontal".
Drag the arrowhead directly to change the vector and see how its Cartesian and polar representations update.
You can also convert between representations using basic trigonometry and Pythagoras' Theorem:
The left column converts Cartesian to Polar; the right converts Polar to Cartesian.
Multiplying a vector \(\mathbf{F}\) by a real number \(\alpha\) scales it — this is why \(\alpha\) is called a "scalar" (\(\alpha \in \mathbb{R}\)).
\[ \mathbf{R}=\alpha \mathbf{F} \]
It is easy to see, using Cartesian representation, that the scale factor affects both components in the same way \[\mathbf{R} = \alpha\,\mathbf{F}=\left(\alpha F_x\right) \, \mathbf{i} + \left(\alpha F_y\right) \, \mathbf{j}\]
And as a result, it follows that the magnitude (length) is also scaled by the same amount \[\|\mathbf{F}\|=F=\sqrt{F_x^2+F_y^2}\] \[\|\mathbf{R}\|=R=\sqrt{\alpha^2F_x^2+\alpha^2F_y^2}=|\alpha|\sqrt{F_x^2+F_y^2}=|\alpha| F\]
Feel free to play with the applet above and note the following facts:
Vectors can be easily added together using their Cartesian representation: \[ \mathbf{A}+\mathbf{B}=\left(A_x\,\mathbf{i}+A_y\,\mathbf{j}\right)+\left(B_x\,\mathbf{i}+B_y\, \mathbf{j} \right) = \left(A_x+B_x\right)\,\mathbf{i}+\left(A_y+B_y\right)\,\mathbf{j} \] Graphically, you can resort to the parallelogram rule that states that the resultant vector is always the diagonal of the parallelogram formed by the two addends.
Another way of graphically sum multiple vectors is the so-called triangle rule, that states that the resultant vector is a vector connecting the start and end points of the polygon formed by all addends, placed consecutively.
Vector summation is commutative: the order of the addends does not matter.
Force, velocity, acceleration — any quantity with both magnitude and direction is a vector. Here are some areas where vector algebra is central:
| Concept | Equation | Description |
|---|---|---|
| Vector Magnitude | \(|\mathbf{V}| = \sqrt{V_x^2 + V_y^2}\) | Length of vector in 2D |
| Unit Vector | \(\hat{u} = \frac{\mathbf{V}}{|\mathbf{V}|}\) | Vector with magnitude 1 |
| Vector Addition | \(\mathbf{R} = \mathbf{A} + \mathbf{B}\) | Component-wise addition |
| Scalar Multiplication | \(\mathbf{V}' = k\mathbf{V}\) | Scales magnitude by k |
| Cartesian Form | \(\mathbf{V} = V_x\mathbf{i} + V_y\mathbf{j}\) | Component representation |
Convert the following vector from Cartesian to polar form: \(\mathbf{V} = -3\mathbf{i} + 4\mathbf{j}\)
Show Solution
The magnitude of the vector is:
\(|\mathbf{V}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
The angle can be found using:
\(\theta = \tan^{-1}(V_y/V_x) = \tan^{-1}(4/(-3)) = \tan^{-1}(-1.333) = -53.13°\)
However, since the x-component is negative and the y-component is positive, the vector is in the second quadrant, so we need to adjust:
\(\theta = 180° - 53.13° = 126.87°\)
Answer: \(\mathbf{V} = (5, 126.87°)\)
Vector \(\mathbf{A}\) has a magnitude of 10 units and points in the direction of 30° from the positive x-axis. Vector \(\mathbf{B}\) has a magnitude of 15 units and points in the direction of 135° from the positive x-axis. Find the resultant vector \(\mathbf{R} = \mathbf{A} + \mathbf{B}\) in both Cartesian and polar form.
Show Solution
First, convert both vectors to Cartesian form:
\(\mathbf{A} = 10\cos(30°)\mathbf{i} + 10\sin(30°)\mathbf{j} = 8.66\mathbf{i} + 5\mathbf{j}\)
\(\mathbf{B} = 15\cos(135°)\mathbf{i} + 15\sin(135°)\mathbf{j} = -10.61\mathbf{i} + 10.61\mathbf{j}\)
Adding the vectors:
\(\mathbf{R} = \mathbf{A} + \mathbf{B} = (8.66 - 10.61)\mathbf{i} + (5 + 10.61)\mathbf{j} = -1.95\mathbf{i} + 15.61\mathbf{j}\)
Converting back to polar form:
Magnitude: \(|\mathbf{R}| = \sqrt{(-1.95)^2 + (15.61)^2} = \sqrt{3.80 + 243.67} = \sqrt{247.47} \approx 15.73\)
Direction: \(\theta = \tan^{-1}(15.61/(-1.95)) = \tan^{-1}(-8.00) = -82.9°\)
Adjusting for the quadrant (negative x, positive y): \(\theta = 180° - 82.9° = 97.1°\)
Answer: \(\mathbf{R} = -1.95\mathbf{i} + 15.61\mathbf{j} = (15.73, 97.1°)\)
Given vector \(\mathbf{V} = 4\mathbf{i} - 3\mathbf{j}\), find:
Show Solution
a) Magnitude: \(|\mathbf{V}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
Direction: \(\theta = \tan^{-1}((-3)/4) = \tan^{-1}(-0.75) = -36.9°\)
Since the x-component is positive and the y-component is negative, the vector is in the fourth quadrant, so the angle is correct as is: \(\theta = -36.9° = 323.1°\) (if expressing in the 0-360° range)
b) \(2\mathbf{V} = 2(4\mathbf{i} - 3\mathbf{j}) = 8\mathbf{i} - 6\mathbf{j}\)
This has magnitude \(|2\mathbf{V}| = 2|\mathbf{V}| = 2 \times 5 = 10\) and the same direction as \(\mathbf{V}\), which is \(-36.9°\)
c) \(-0.5\mathbf{V} = -0.5(4\mathbf{i} - 3\mathbf{j}) = -2\mathbf{i} + 1.5\mathbf{j}\)
This has magnitude \(|-0.5\mathbf{V}| = |-0.5||\mathbf{V}| = 0.5 \times 5 = 2.5\) and direction \(\theta + 180° = -36.9° + 180° = 143.1°\) (because of the negative scalar multiplier)
d) The unit vector is \(\hat{\mathbf{V}} = \frac{\mathbf{V}}{|\mathbf{V}|} = \frac{4\mathbf{i} - 3\mathbf{j}}{5} = 0.8\mathbf{i} - 0.6\mathbf{j}\)
This has magnitude 1 and the same direction as \(\mathbf{V}\), which is \(-36.9°\)
Answer: a) \(|\mathbf{V}| = 5\), \(\theta = -36.9°\); b) \(2\mathbf{V} = 8\mathbf{i} - 6\mathbf{j}\); c) \(-0.5\mathbf{V} = -2\mathbf{i} + 1.5\mathbf{j}\); d) \(\hat{\mathbf{V}} = 0.8\mathbf{i} - 0.6\mathbf{j}\)