What is Torque? – A Short and Simple Explanation

Often times when doing physics we simply say “a force is acting on a body” without specifying which point of the body it is acting on. This is basically point-mass physics. We ignore the fact that the object has a complex three-dimensional shape and assume it to be a single point having a certain mass. Sometimes this is sufficient, other times we need to go beyond that. And this is where the concept of torque comes in.

Let’s define what is meant by torque. Assume a force F (in N) is acting on a body at a distance r (in m) from the axis of rotation. This distance is called the lever arm. Take a look at the image below for an example of such a set up.

Great Formulas II_html_m7f65e4ec

(Taken from sdsu-physics.org)

Relevant for the rotation of the body is only the force component perpendicular to the lever arm, which we will denote by F’. If given the angle Φ between the force and the lever arm (as shown in the image), we can easily compute the relevant force component by:

F’ = F · sin(Φ)

For example, if the total force is F = 50 N and it acts at an angle of Φ = 45° to the lever arm, only the the component F’ = 50 N · sin(45°) ≈ 35 N will work to rotate the body. So you can see that sometimes it makes sense to break a force down into its components. But this shouldn’t be cause for any worries, with the above formula it can be done quickly and painlessly.

With this out of the way, we can define what torque is in one simple sentence: Torque T (in Nm) is the product of the lever arm r and the force F’ acting perpendicular to it. In form of an equation the definition looks like this:

T = r · F’

In quantitative terms we can interpret torque as a measure of rotational push. If there’s a force acting at a large distance from the axis of rotation, the rotational push will be strong. However, if one and the same force is acting very close to said axis, we will see hardly any rotation. So when it comes to rotation, force is just one part of the picture. We also need to take into consideration where the force is applied.

Let’s compute a few values before going to the extremely useful law of the lever.


We’ll have a look at the wrench from the image. Suppose the wrench is r = 0.2 m long. What’s the resulting torque when applying a force of F = 80 N at an angle of Φ = 70° relative to the lever arm?

To answer the question, we first need to find the component of the force perpendicular to the lever arm.

F’ = 80 N · sin(70°) ≈ 75.18 N

Now onto the torque:

T = 0.2 m · 75.18 N ≈ 15.04 Nm


If this amount of torque is not sufficient to turn the nut, how could we increase that? Well, we could increase the force F and at the same time make sure that it is applied at a 90° angle to the wrench. Let’s assume that as a measure of last resort, you apply the force by standing on the wrench. Then the force perpendicular to the lever arm is just your gravitational pull:

F’ = F = m · g

Assuming a mass of m = 75 kg, we get:

F’ = 75 kg · 9.81 m/s² = 735.75 N

With this not very elegant, but certainly effective technique, we are able to increase the torque to:

T = 0.2 m · 735.75 N = 147.15 Nm

That should do the trick. If it doesn’t, there’s still one option left and that is using a longer wrench. With a longer wrench you can apply the force at a greater distance to the axis of rotation. And with r increased, the torque T is increased by the same factor.


This was an excerpt from my Kindle ebook More Great Formulas Explained.

Check out my BEST OF for more interesting physics articles.



Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s