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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 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**:

F = âˆšF_{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 Î¸, the angle F makes with its component F_{x}:

According to trigonometry:

Î¸ = tan^{-1} | F_{y} |

F_{x} |

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

Sometimes Î¸ will already be the direction angle, other times you will need to add Î¸ to 180Â° or subtract it from 180Â° 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:

F_{x} > 0

F_{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}:

F = F_{x}

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

F_{x} < 0

F_{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 âˆ’F_{x} (remember a magnitude is *always positive*), therefore:

F = âˆ’F_{x}

So if F_{x} is âˆ’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:

F_{x} = 0

F_{y} = 0

â†“

F = 0

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

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

F_{x} = âˆ’7.0 N

F_{y} = 0 N

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

F = âˆ’F_{x} = 7.0 N

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

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

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

F_{x} = 50.0 N

F_{y} = 21.2 N

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

F = âˆšF_{x}^{2} + F_{y}^{2}

F = âˆš50.0^{2} + 21.2^{2} N

F = âˆš2949 N

F = 54.3 N

Next we find Î¸:

Î¸ = tan^{-1} | F_{y} |

F_{x} |

Î¸ = tan^{-1} | 21.2 N |

50.0 N |

Î¸ = tan^{-1} 0.424

Î¸ = 23.0Â°

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

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

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

F_{x} = âˆ’387 N

F_{y} = âˆ’532 N

Let's determine the magnitude of F:

F = âˆšF_{x}^{2} + F_{y}^{2}

F = âˆš(âˆ’387)^{2} + (âˆ’532)^{2} N

F = 658 N

And then Î¸:

Î¸ = tan^{-1} | F_{y} |

F_{x} |

Î¸ = tan^{-1} | âˆ’532 N |

âˆ’387 N |

Î¸ = tan^{-1} 1.37

Î¸ = 53.9Â°

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 that we just need to add 180Â° to Î¸. Therefore the direction angle of the force will be 53.9Â° + 180Â° = 233.9Â°:

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

Find F knowing that F_{x} is âˆ’9.48 N and F_{y} 5.67 N.

F_{x} = âˆ’9.48 N

F_{y} = 5.67 N

The magnitude of F will be:

F = âˆšF_{x}^{2} + F_{y}^{2}

F = âˆš(âˆ’9.48)^{2} + (5.67)^{2} N

F = âˆš122 N

F = 11.0 N

And Î¸:

Î¸ = tan^{-1} | F_{y} |

F_{x} |

Î¸ = tan^{-1} | 5.67 N |

âˆ’9.48 N |

Î¸ = tan^{-1} (âˆ’0.598)

Î¸ = âˆ’30.9Â°

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

Î¸ = 30.9Â°

The direction angle of F will be 180Â° âˆ’ Î¸ (look at the figure above), i.e. 180Â° âˆ’ 30.9Â° = 149.1Â°:

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

F_{x} = 0 N

F_{y} = 8.3 Ã— 10^{2} N

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

F = F_{y} = 8.3 Ã— 10^{2} N

F_{x} is 0.41 N, F_{y} is âˆ’0.80 N. Find F.

F_{x} = 0.41 N

F_{y} = âˆ’0.80 N

Let's first determine the magnitude of F:

F = âˆšF_{x}^{2} + F_{y}^{2}

F = âˆš(0.41)^{2} + (âˆ’0.80)^{2} N

F = âˆš0.808 N

F = 0.90 N

Next let's find Î¸:

Î¸ = tan^{-1} | F_{y} |

F_{x} |

Î¸ = tan^{-1} | âˆ’0.80 N |

0.41 N |

Î¸ = tan^{-1} (âˆ’1.95)

Î¸ = âˆ’63Â°

Î¸ 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:

Î¸ = 63Â°

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