# How to find the magnitude and direction of a force given the x and y components

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:

## • The two components are both different from zero

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 F_{x} and F_{y} 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**:

_{x}

^{2}+ F

_{y}

^{2}

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 F_{x}:

According to trigonometry:

θ = tan^{-1} | F_{y} |

F_{x} |

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.*

## • One of the two components is equal to zero

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*:

_{x}> 0

_{y}= 0

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

F_{y} 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 F_{x}:

_{x}

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

_{x}< 0

_{y}= 0

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

And since F_{x} is negative, the magnitude will be −F_{x} (remember a magnitude is *always positive*), therefore:

_{x}

So if F_{x} is *−10N*, then F has magnitude *10N*.

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

## • The two components are both equal to zero

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

_{x}= 0; F

_{y}= 0

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

## Exercises

### #1

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

*Solution*

_{x}= −7.0N

_{y}= 0N

F will be in the negative x direction, and have the same magnitude as the x component:

_{x}= 7.0N

It is −F_{x} because F_{x} is negative, and the magnitude must be positive.

### #2

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

*Solution*

We first roughly represent F_{x} and F_{y} on an xy-plane, and from that we draw the rectangle and F.

_{x}= 50.0N

_{y}= 21.2N

Let's find the magnitude of F applying **Pythagoras' Theorem:**

_{x}

^{2}+ F

_{y}

^{2}

^{2}+ 21.2

^{2}N

Next we find θ:

θ = tan^{-1} | F_{y} |

F_{x} |

θ = tan^{-1} | 21.2N |

50.0N |

^{-1}0.424

In this case θ is already the direction angle of F. Indeed θ is the counterclockwise angle that F makes with the positive x axis.

Therefore the force has magnitude *54.3N* and the direction angle is *23.0°*.

### #3

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

*Solution*

_{x}= −387N

_{y}= −532N

Let's determine the magnitude of F:

_{x}

^{2}+ F

_{y}

^{2}

^{2}+ (−532)

^{2}N

And then θ:

θ = tan^{-1} | F_{y} |

F_{x} |

θ = tan^{-1} | −532N |

−387N |

^{-1}1.37

We need the direction angle of F, i.e. the counterclockwise angle F makes with the positive x axis.

Looking at the xy-plane above, we see the we just need to add *180°* to θ. Therefore the direction angle of the force will be *53.9° + 180° = 233.9°* :

Hence, the magnitude is *658N* and the direction angle is *233.9°*.

### #4

Find F knowing that F_{x} is *−9.48N* and F_{y} *5.67N*.

*Solution*

_{x}= −9.48N

_{y}= 5.67N

The magnitude of F will be:

_{x}

^{2}+ F

_{y}

^{2}

^{2}+ (5.67)

^{2}N

And θ:

θ = tan^{-1} | F_{y} |

F_{x} |

θ = tan^{-1} | 5.67N |

−9.48N |

^{-1}(−0.598)

Since the tangent is negative, θ came out negative. But we are just interested in the magnitude, so we ignore the minus sign:

The direction angle of F will be *180°* − θ (look at the figure above), i.e. *180° − 30.9° = 149.1°*:

### #5

F has the following components: *0N* in the x direction, and *8.3×10 ^{2}N* in the y direction. Determine magnitude and direction of F.

*Solution*

_{x}= 0N

_{y}= 8.3×10

^{2}N

F will be in the direction of the positive y axis, and have the same magnitude as F_{y}:

_{y}= 8.3×10

^{2}N

### #6

F_{x} is *0.41N*, F_{y} is *−0.80N*. Find F.

*Solution*

_{x}= 0.41N

_{y}= −0.80N

Let's first determine the magnitude of F:

_{x}

^{2}+ F

_{y}

^{2}

^{2}+ (−0.80)

^{2}N

Next let's find θ:

θ = tan^{-1} | F_{y} |

F_{x} |

θ = tan^{-1} | −0.80N |

0.41N |

^{-1}(−1.95)

θ is negative because the tangent is negative. But we are just interested in the magnitude of the angle, so we can ignore the minus sign:

Looking carefully at the xy-plane above, we can see that the direction angle of F is θ subtracted from 360°, i.e. 360° − 63° = 297° :