How To Calculate The Magnitude Of A Force In Physics

Calculating magnitudes for forces is an important part of physics. When you're working in one dimension, the magnitude of the force isn't something you have to consider. Calculating magnitude is more of a challenge in two or more dimensions because the force will have "components" along both the ​x-​ and y-axes and possibly the z-axis if it's a three-dimensional force. Learning to do this with a single force and with the resultant force from two or more individual forces is an important skill for any budding physicist or anyone working on classical physics problems for school.

The first step to understanding what it means to calculate the magnitude of a force in physics is to learn what a vector is. A "scalar" is a simple quantity that just has a value, such as temperature or speed. When you read a temperature of 50 degrees F, it tells you everything you need to know about the temperature of the object. If you read that something is traveling at 10 miles per hour, that speed tells you all you need to know about how quickly it's moving.

A vector is different because it has a direction as well as a magnitude. If you watch a weather report, you'll learn just how fast the wind is traveling and in what direction. This is a vector because it gives you that extra bit of information. Velocity is the vector equivalent of speed, where you find out the direction of motion as well as how fast it's moving. So if something is traveling 10 miles per hour toward the northeast, the speed (10 miles per hour) is the magnitude, northeast is the direction, and both parts together make up the vector velocity.

In many cases, vectors are split into "components." Velocity might be given as a combination of speed in the northerly direction and speed in the easterly direction so that the resultant motion would be toward the northeast, but you need both bits of information to work out how quickly it's moving and where it's going. In physics problems, east and north are usually replaced with ​x​ and ​y​ coordinates, respectively.

To calculate the magnitude of force vectors, you use the components along with Pythagoras' theorem. Think of the ​x​ coordinate of the force as the base of a triangle, the ​y​ component as the height of the triangle, and the hypotenuse as the resultant force from both components. Extending the link, the angle the hypotenuse makes with the base is the direction of the force.

If a force pushes 4 Newtons (N) in the x-direction and 3 N in the y-direction, Pythagoras' theorem and the triangle explanation show what you need to do when calculating magnitude. Using ​x​ for the ​x​-coordinate, ​y​ for the ​y​-coordinate and ​F​ for the magnitude of the force, this can be expressed as:

\(F=\sqrt{x^2+y^2}\)

In words, the resultant force is the square root of ​x​² plus ​y​². Using the example above:

\(\begin{aligned} F&=\sqrt{4^2+3^2}\&=\sqrt{16+9}\&=\sqrt{25}\&=5\text{ N}\end{aligned}\)

So, 5 N is the magnitude of force.

Note that for three-component forces, you add the ​z​ component to the same formula. So:

\(F=\sqrt{x^2+y^2+z^2}\)

The direction of the force isn't the focus of this question, but it's easy to work out based on the triangle of components and the resultant force from the last section. You can work out the direction using trigonometry. The identity best-suited to the task for most problems is:

\(\tan{\theta}=\frac{y}{x}\)

Here ​_θ_​ is standing in for the angle between the vector and the ​x​-axis. This means you can use the components of the force to work it out. You can use the magnitude and the definition of either cos or sin if you prefer. The direction is given by:

\(\theta=\tan^{-1}(y/x)\)

Using the same example as above:

\(\theta=\tan^{-1}(3/4)=36.9\text{ degrees}\)

So, the vector makes about a 37-degree angle with the x-axis.

If you have two or more forces, work out the resultant force magnitude by first finding the resultant vector and then applying the same approach as above. The only extra skill you need is finding the resultant vector, and this is fairly straightforward. The trick is that you add the corresponding ​x​ and ​y​ components together. Using an example should make this clear.

Imagine a sailboat on the water, moving along with the force from the wind and the current of the water. The water imparts a force of 4 N in the x-direction and 1 N in the y-direction, and the wind adds a force of 5 N in the x-direction and 3 N in the y-direction. The resultant vector is the ​x​ components added together (4 + 5 = 9 N) and the ​y​ components added together (3 + 1 = 4 N). So you end up with 9 N in the x-direction and 4 N in the y-direction. Find the magnitude of the resultant force using the same approach as above:

\(\begin{aligned} F&=\sqrt{9^2+4^2}\&=\sqrt{81+16}\&=\sqrt{97}\&=9.85\text{ N}\end{aligned}\)