# Forces in Cartesian Vector Form 13

There are usually three ways a force is shown. It’s important to know how we can express these forces in Cartesian vector form as it helps us solve three dimensional problems. The following video goes through each example to show you how you can express each force in Cartesian vector form. If you do not want to watch the video, you can read the steps below.

In this example, the force is shown with coordinate direction angles. When we have a force shown with coordinate direction angles, all we need to do is multiply the force by the cosine of each angle to get the component. Let’s start with force $F_1$.

First, let’s figure out the x-component of force $F_1$.

Looking at the diagram, we can see that there is a 60o angle between the positive x-axis and force $F_1$. Therefore, the x-component of force $F_1$ is:

$F_{1x}\,=\,5\text{cos}\,(60^0)\,=\,2.5$ kN

Now, let’s figure out the y-component of force $F_1$.

Looking at the diagram, we see that there is a 450 angle between the positive y-axis and force $F_1$. As before, we can then write the y-component of force $F_1$ as:

$F_{1y}\,=\,5\text{cos}\,(45^0)\,=\,3.5$ kN

Lastly, let’s figure out the z-component of force $F_1$.

From the diagram, we see that there is a 600 angle between the positive z-axis and force $F_1$. Therefore the z-component of force $F_1$ is:

$F_{1z}\,=\,5\text{cos}\,(60^0)\,=\,2.5$ kN

Let’s now write force $F_1$ in Cartesian vector form like so:

$F_1\,=\,\left\{2.5i+3.5j+2.5k\right\}$ kN

Notice how each value corresponds to each of the components we found.

Now, let’s look at force $F_2$.

Looking at the diagram, you can see that it actually lies on the negative y-axis. That means it doesn’t have a x-component or a z-component. It only has a y-component. When we write it in Cartesian vector notation, we can write the x and z components as 0.

$F_2$ in Cartesian vector form is:

$F_2\,=\,\left\{0i-2j+0k\right\}$ kN

Note how our j component is negative. That’s because force $F_2$ is in the negative y direction.

Let’s now look at our next example and see how we can express these forces in Cartesian vector form.

In this example, the angles you see are NOT coordinate direction angles. Let’s start off by looking at force $F_1$.

Notice how force $F_1$ is in the xz plane. That means force $F_1$ does not have a y-component.

We can also see that force $F_1$ forms a right angle triangle with the x-axis. Using our trig functions, we can solve for $F_{1x}$ and $F_{1z}$.

$\text{cos}\,(30^0)\,=\,\dfrac{F_{1x}}{300}$

(remember cosine is adjacent over hypotenuse)

$F_{1x}\,=\,300\text{cos}\,(30^0)\,=\,260$ N

$\text{sin}\,(30^0)\,=\,\dfrac{F_{1z}}{300}$

(remember sine is opposite over hypotenuse)

$F_{1z}\,=\,300\text{sin}\,(30^0)\,=\,150$ N

Now we can write force $F_1$ in Cartesian vector form like so:

$F_1\,=\,\left\{260i+0j-150k\right\}$ kN

Notice how our k value is negative. That’s because the z-component of force $F_1$ is in the negative z-direction.

Let’s now focus on force $F_2$.

Force $F_2$ has all three components. These are labeled on the diagram. To find these forces, we will have to find F’.

F’ is the force that lies on the xy plane. It can also be found using trigonometry.

As before, we can use our basic trig functions on this right angle triangle to figure out F’ and the z-component.

$F'\,=\,500\text{cos}\,(45^0)\,=\,353.5$ N

$\text{z-component}\,=\,500\text{sin}\,(45^0)\,=\,354$ N

We can now use the value of F’ we found to figure out the x and y components.

Notice how F’ is now the hypotenuse of the new right angle triangle we formed. Let’s calculate the x and y components using trigonometry as before.

$\text{x-component}\,=\,353.5\text{sin}\,(30^0)\,=\,177$ N

$\text{y-component}\,=\,353.5\text{cos}\,(30^0)\,=\,306$ N

Let us now write force $F_2$ in Cartesian vector form:

$F_2\,=\,\left\{177i+306j-354k\right\}$ kN

Notice that our k term is negative. This is because the z-component of force $F_2$ is in the negative direction.

Let’s look at our final example and see how we can express force vectors directed along a line as forces in Cartesian vector form.

We will first figure out where point A is with respect to the origin.

We can write where A is in Cartesian vector notation like so:

A: (0.5i-1.5j+0k) m

Notice that our j value is negative. That’s because point A is on the negative y-axis side. Let us now figure out where point B is.

We see that point B is at:

B:(-1.5i-2.5j+2k) m

Now that we know where point A and B is with respect to the origin, we can now find the position vector, denoted $r_{AB}$.

$r_{AB}$ is found by subtracting the corresponding components of B from A.

$r_{AB}\,=\,\left\{(-1.5-0.5)i+(-2.5-(-1.5))j+(2-0)k\right\}$

$r_{AB}\,=\,\left\{-2i-1j+2k\right\}$ m

The next step is to find the magnitude of $r_{AB}$ which can be found by finding the square root of each component squared and added together.

Magnitude of $r_{AB}$ = $\sqrt{(-2)^2+(-1)^2+(2)^2}$

Magnitude of $r_{AB}$ = $3$ m

Now, we can find the unit vector, denoted $u$. The unit vector is simply each term of $r_{AB}$ divided by the magnitude.

$u\,=\,-\dfrac{2}{3}i\,-\,\dfrac{1}{3}j\,+\,\dfrac{2}{3}k$

We can now write force $F_B$ in Cartesian vector notation. To do so, we multiply the force by each component in our unit vector.

$F_B\,=\,600\left(-\dfrac{2}{3}i\,-\,\dfrac{1}{3}j\,+\,\dfrac{2}{3}k\right)$

$F_B\,=\,-400i\,-\,200j\,+\,400k$ N

Let us now express force $F_C$ in Cartesian vector form. We already know where point A is, since we figured it out, so we only need to figure out where point C is, with respect to the origin.

We can write the location of point C in Cartesian vector notation like so:

C:(-1.5i+0.5j+3.5k) m

We already know where point A is, which was:

A: (0.5i-1.5j+0k) m

Again, we will find the position vector, $r_{AC}$ by subtracting each corresponding component of C from A.

$r_{AC}\,=\,\left\{(-1.5-0.5)i+(0.5-(-1.5))j+(3.5-0)k\right\}$

$r_{AC}\,=\,\left\{-2i+2j+3.5k\right\}$ m

As before, the next step is to figure out the magnitude of $r_{AC}$ by finding the square root value of each component in $r_{AC}$ squared and added together.

Magnitude of $r_{AC}$ = $\sqrt{(-2)^2+(2)^2+(3.5)^2}$

Magnitude of $r_{AC}$ = $4.5$ m

Now, we will find the unit vector, $u$. Remember, we find the unit vector by dividing each term in $r_{AC}$ by it’s magnitude.

$u\,=\,-\dfrac{2}{4.5}i\,+\,\dfrac{2}{4.5}j\,+\,\dfrac{3.5}{4.5}k$

We can now write force $F_C$ in Cartesian vector form. To do so, we multiply each component of our unit vector by the value of the force.

$F_B\,=\,450\left(-\dfrac{2}{4.5}i\,+\,\dfrac{2}{4.5}j\,+\,\dfrac{3.5}{4.5}k\right)$

$F_B\,=\,-200i\,+\,200j\,+\,350k$ N

That is how we can express forces in Cartesian vector form. Look through more examples to better your understanding.

The images in this post are taken from:  Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.