# How to decompose a force into x and y components

It is often useful to decompose a force into x and y components, i.e. **find two forces such that one is in the x direction, the other is in the y direction, and the vector sum of the two forces is equal to the original force.**

Let's see how we can do this.

Suppose we have a force F that makes an angle of 30° with the *positive* x axis, as shown below:

And we want to decompose F into x and y components.

The first thing we need to do is to represent the two components on the **xy-plane**. We do this by dropping two perpendiculars from the head of F: one to the x axis, the other to the y axis.

Like this:

And we join the **origin** of the xy-plane with the **x-intercept** to represent the x component of F:

And again, we join the **origin** with the **y-intercept** to represent the y component of F:

F_{x} and F_{y} are two vectors, i.e. they both have *a magnitude and a direction*. However, since F_{x} and F_{y} are in the directions of the x and y axes, they are commonly expressed by the *magnitude alone*, preceded by a positive or negative sign: positive when they point in the positive directions, and negative when they point in the negative directions of the x and y axes.

In our example F_{x} and F_{y} are positive because both point in the positive directions of the x and y axes.

The positive values of F_{x} and F_{y} can be found using trigonometry:

_{x}= F cos30°

_{y}= F sin30°

To keep it simple, just remember that **if a component is adjacent to the angle, then it is cos, otherwise it is sin.**

Often F_{x} will be the component adjacent to the angle, so it will be cos, and F_{y} will be sin.

Let's now consider a force that has one of its components negative:

In this case F_{x} is negative because it points in the negative direction of the x axis.

Therefore:

_{x}= −F cos15°;

_{y}= F sin15°

Notice the minus sign before F cos15° which we have added to make F_{x} negative.

**You have to be very careful if your angle is not between 0° and 90°, because the sin or(and) cos of that angle may be already negative, so the product is also negative and you don't need to add a minus sign.**

To be on the safe side, we recommend to always work with angles between 0° and 90°, so that the sin and cos are always positive, and therefore the product is also always positive.

## The bottom line

We can summarize the process of decomposing a force F, as follows:

**1. Represent the x and y components of the force on the xy-plane** by dropping perpendiculars from the head of the force to the x and y axes, and then joining the origin of the xy-plane with the two intercepts (*the goal of graphically representing the components is to help you see which component is adjacent to the angle and what the signs of the two components are*).

**2. Find the values of the x and y components**: the component adjacent to the angle will be F cos θ and the other will be F sin θ. Components that point in the negative directions of the x and y axes are negative, therefore you will need to add a minus sign (given that you are working with θ between 0° and 90°, so that F cos θ and F sin θ are always positive).

## Forces with tail not in the origin

Sometimes a force does not have the tale in the origin of the xy-plane.

For example:

In cases like this, we draw two straight-lines parallel to the x and y axis that pass through the tail of the force, and then we drop two perpendiculars from the head of the force to the straight-lines:

_{x}= F cos30°

_{y}= F sin30°

## Forces that are already in the x or y direction

Often we deal with forces that are already in the x or y direction. In that case we can determine the x and y components in a simpler and more intuitive way, without using trigonometry.

If, for example, we have a force F that is in the direction of the positive x axis:

It is obvious that the y component of F is 0, and the x component is *positive* with magnitude equal to the magnitude of F:

_{x}= F

_{y}= 0

On the other hand, if we have a force F in the direction of the negative x axis:

Then the y component is again 0, and the x component is *negative* (because it points in the negative direction of the x axis) and has the same magnitude as F:

_{x}= −F

_{y}= 0

The same can be shown for forces in the y direction: They will always have x component 0, and y component either positive or negative with magnitude equal to the magnitude of the force.

To verify your understanding of the concept, do the exercises below. Then, if you want to see how useful force decomposition is in practice for solving real problems, head to the step-by-step guide for solving force problems.

## Exercises

### #1

A force of *19N* is in the direction of the negative x axis. Find the x and y components of the force.

### Solution

F_{x} is negative, and has the same magnitude as the force (19N). F_{y} is zero.

_{x}= −19N

_{y}= 0N

### #2

A force of *114N* makes an angle of *67°* with the positive x axis. Decompose the force into x and y components.

### Solution

Both the components are positive.

_{x}= F cos67° = 44.5N

_{y}= F sin67° = 105N

### #3

A force makes an angle of *221°* with the positive x axis. Assuming the force has magnitude *3.1×10 ^{3}N*, find the x and y components.

### Solution

Instead of dealing with the 221° angle, we want to deal with the 41° angle (221° − 180°) that the force makes with the negative x axis:

As you can see, both F_{x} and F_{y} are negative:

_{x}= −F cos41° = −2.3×10

^{3}N

_{y}= −F sin41° = −2.0×10

^{3}N

### #4

A force of *4.5×10 ^{5}N* has the direction of the positive y axis. Determine its components.

### Solution

This one is simple. F_{x} is zero. F_{y} is positive and has the same magnitude as the force.

_{x}= 0N

_{y}= F = 4.5×10

^{5}N

### #5

A *90.0N* force makes an angle of *33°* with the positive y direction. Calculate the x and y components of the force.

*Tip:* Since the angle is *+33°*, it goes counterclockwise from the positive y axis.

### Solution

In this case F_{y} is adjacent to the angle, therefore its magnitude is the force times the cos of the angle, while the magnitude of F_{x} is the force times the sin.

Also notice that F_{x} is negative:

_{x}= −F sin33° = −49.0N

_{y}= F cos33° = 75.5N

Alternatively you could have considered the 57° angle (90° − 33°) which F makes with F_{x} . That way F_{x} would have been adjacent to the angle.

### #6

A force that has magnitude *3.21×10 ^{4}N* makes an angle of

*−50°*with the positive x axis. Determine the components.

*Tip:* The angle is negative, meaning it goes clockwise from the positive x axis.

### Solution

_{x}= F cos50° = 2.06×10

^{4}N

_{y}= −F sin50° = −2.46×10

^{4}N