How to find magnitude of y component?

Sometimes we have the x and y components of a force, and we want to find the magnitude and direction of the force.

Let's see how we can do this.

There are three possible cases to consider:

If a force F has the x and y components both different from zero, in order to find F we start by roughly representing the components on an xy-plane (based on the magnitude of the components and their sign).

If, for instance, both the components are positive, with the x component slightly larger in magnitude, we would represent them something like this:

Then we draw the rectangle with Fx and Fy as two of the sides:

The diagonal of the rectangle that goes from the origin is the force F:

We can find the magnitude of F, by applying Pythagoras' Theorem:

And what about the direction of F?

The direction is often expressed by the direction angle, i.e. the counterclockwise angle that F makes with the positive x axis.

Let's see how we can find it:

First we find θ, the angle F makes with its component Fx:

According to trigonometry:

After we have found θ, we can easily determine the direction angle.

Sometimes θ will already be the direction angle, other times you will need to add θ to 180° or subtract it from 180° etc., it depends in what quadrant your force is.

Check out the exercises below to see some examples.

Often a force has either the x or y component equal to zero and the other component different from zero.

In that case, the magnitude and direction of the force is equal to the magnitude and direction of the non-zero component:

For example let's assume that a force F has y component zero, and x component > 0:

If we represent the two components graphically, we should see something like this:

Fy is zero, so we can't actually see it.

It is clear that F will be in the direction of the positive x axis and have the same magnitude as Fx:

On the other hand, if the x component of F is negative,

F will be in the negative direction of the x axis, and the magnitude will be the same as that of Fx.

And since Fx is negative, the magnitude will be −Fx (remember a magnitude is always positive), therefore:

So if Fx is −10 N, then F has magnitude 10 N.

The same can be shown for a force that has the x component equal to zero, and the y component different from zero.

If both the components are equal to zero, then the force is also equal to zero:

To test your understanding, make sure to do the exercises below.

The x component of a force is −7.0 N, the y component is 0 N. Find the magnitude and direction of the force.

Find a force knowing that its x and y components are 50.0 N and 21.2 N respectively.

Assuming that a force has the x component −387 N and the y component −532 N, find magnitude and direction of the force.

Find F knowing that Fx is −9.48 N and Fy 5.67 N.

F has the following components: 0 N in the x direction, and 8.3 × 102 N in the y direction. Determine magnitude and direction of F.

Fx is 0.41 N, Fy is −0.80 N. Find F.

{eq}A=\sqrt{A_{x}^{2}+A_{y}^{2}} {/eq}

where {eq}A {/eq} is the magnitude of the vector, {eq}A_{x} {/eq} is the magnitude of the x-component, and {eq}A_{y} {/eq} is the magnitude of the y-component. All three variables must be in the same unit.

After calculating the magnitude of the vector we can use right triangle trigonometry to determine the angle the vector makes with the x-axis.

So, let's try using these steps to calculate the magnitude and direction of a vector in the following two examples!

Determine the magnitude and direction of vector {eq}A {/eq}.

Step 1: Use the equation {eq}A=\sqrt{A_{x}^{2}+A_{y}^{2}} {/eq} to calculate the magnitude of the vector.

From the figure, we can see that the x-component of the vector is {eq}85.8 {/eq} and the y-component of the vector is {eq}34.7 {/eq}. So we can say that {eq}A_x = 85.8 {/eq} and {eq}A_y = 34.7 {/eq}. Plugging these into the formula, we will have

$$\begin{align*} A&=\sqrt{85.8^{2}+34.7^{2}}\\ A &= 92.6 \end{align*} $$

Step 2: Use the equation {eq}\Theta =\tan^{-1}\left(\frac{A_{y}}{A_{x}} \right) {/eq} to calculate the direction of the vector.

Again, plugging in the values of {eq}A_x {/eq} and {eq}A_y {/eq} into the formula, we will have

$$\begin{align*} \Theta &=\tan^{-1}\left(\frac{34.7}{85.8} \right)\\ &= 22^{\circ} \end{align*} $$

Therefore, the magnitude of vector {eq}A {/eq} is {eq}92.6 {/eq} and its direction is {eq}\Theta =22^{\circ} {/eq}.