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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:

The x and y components of a force, both different from zeroxyxFyF

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

The rectangle that has Fx and Fy as two of its sidesxyxFyF

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

The force F is the diagonal of the rectangleFxyxFyF

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

F = Fx2 + Fy2

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:

Angle theta formed by F and its x componentFxθyxFyF

According to trigonometry:

θ = tan-1 Fy
Fx

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:

Fx > 0
Fy = 0

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

The x component of the force F is greater than 0xFxy

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:

The force F is in the direction of the positive x axisFxy
F = Fx

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

Fx < 0
Fy = 0
The x component of the force F is lesser than 0xFxy

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:

The force F is in the direction of the negative x axisFxy
F = −Fx

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.

• The two components are both equal to zero

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

Fx = 0
Fy = 0
F = 0

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

Exercises

#1

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

Solution

Fx = −7.0 N
Fy = 0 N
The x component of the force points in negative direction, y component is zeroxyxF

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

F points in the negative direction of xFxy
F = −Fx = 7.0 N

It is Fx because Fx is negative, and the magnitude must be positive.

#2

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

Solution

We first roughly represent Fx and Fy on an xy-plane, and from that we draw the rectangle and F.

Fx = 50.0 N
Fy = 21.2 N
Force with Fx 50.0N and Fy 21.2NFxθyxFyF

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

F = Fx2 + Fy2
F = 50.02 + 21.22 N
F = 2949 N
F = 54.3 N

Next we find θ:

θ = tan-1 Fy
Fx
θ = tan-1 21.2 N
50.0 N
θ = tan-1 0.424
θ = 23.0°

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 54.3 N and the direction angle is 23.0°.

#3

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

Solution

Fx = −387 N
Fy = −532 N
Force with components: x -387N and y -532NFxθyxFyF

Let's determine the magnitude of F:

F = Fx2 + Fy2
F = (−387)2 + (−532)2 N
F = 658 N

And then θ:

θ = tan-1 Fy
Fx
θ = tan-1 −532 N
−387 N
θ = tan-1 1.37
θ = 53.9°

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 180° to θ. Therefore the direction angle of the force will be 53.9° + 180° = 233.9°:

The force makes a counterclockwise angle of 233.9 deg with the positive x axisFxyxFyF233.9°

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

#4

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

Solution

Fx = −9.48 N
Fy = 5.67 N
Force with components: Fx -9.48N and Fy -5.67NFxθyxFyF

The magnitude of F will be:

F = Fx2 + Fy2
F = (−9.48)2 + (5.67)2 N
F = 122 N
F = 11.0 N

And θ:

θ = tan-1 Fy
Fx
θ = tan-1 5.67 N
−9.48 N
θ = tan-1 (−0.598)
θ = −30.9°

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

θ = 30.9°

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

The force makes an angle of 149.1 deg with the positive xFxyxFyF149.1°

#5

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.

Solution

Fx = 0 N
Fy = 8.3 × 102 N
The y component points in positive direction of the y axis, the x component is zeroyFxy

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

F points in the positive direction of yFxy
F = Fy = 8.3 × 102 N

#6

Fx is 0.41 N, Fy is −0.80 N. Find F.

Solution

Fx = 0.41 N
Fy = −0.80 N
Force with the components: 0.41N and -0.80NFxθyxFyF

Let's first determine the magnitude of F:

F = Fx2 + Fy2
F = (0.41)2 + (−0.80)2 N
F = 0.808 N
F = 0.90 N

Next let's find θ:

θ = tan-1 Fy
Fx
θ = tan-1 −0.80 N
0.41 N
θ = tan-1 (−1.95)
θ = −63°

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:

θ = 63°

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°:

The force makes a counterclockwise angle of 297 deg with the positive x axisFxyxFyF297°

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