The concept of torque appears when we start studying rigid bodies instead of particles. When we studied Newton’s laws, we assumed that the system size is very small, and it can be considered as a particle. A particle has no shape, no volume, no size. The mechanics of rigid bodies need other concepts, in addition to those used in Newton's Laws. One of these concepts is Torque.

Torque is a common experience. Try to open a door pushing in the middle of it instead doing in the edge.

You need much more force to open it if you push it in the middle than in the edge (probably you need double amount of force). The key concept is that the effects of a force in a rigid body do not only depends on the magnitude of the force, but also of the point where this force is applied. First of all you need to determine the axis of rotation. In the example described before, (the door one) the axis is easy to find.

The effect of the force is higher when the door is pushed far from the axis. Now, we define a new magnitude, that describes this effect. It is called TORQUE, and the symbol for it is \(\tau\)


But torque does not only depends on the force (\(F\)) and the distance (\(d\)). The force direction is also important. Let’s analyze this.

Notice (and you can also do it at home as well) when you push in the door direction, like \((F_2)\), there is no effect on the movement of the door. Although, if you push perpendicularly, the effect is maximum.  

Let’s summarize: In order to study the rotation of a rigid body, a new magnitude is required, instead of Force. It is called Torque. Torque depends on the magnitude of the force (\(F\)), the distance to the axis (\(d\)) and the direction in which the force is applied (\(\sin \alpha\)). Look at the figure to understand that \(\alpha\) is the angle between \(F\) and \(d\).

Example 1.

In each case scketched, all forces are equal in magnitude and all the rods have the same length. Rank these situations from the greatest to least on the basis of the magnitude of torque.








Torque depends on three magnitudes, force, distance and direction. The forces are equal in all cases, so we have to study distance and directions.

The maximum torque is in case b, because the force is applied at the most distant point from the axis and the angle between force and distance is 90º. (and \(\sin 90º=1\))

The minimum torque is in case d, because the force is applied at the middle of the rod, and the angle between force and distance is 45º, and \(\sin 45º = 0.707\)

What about cases a and c? We have to analize the product \(d\sin (\alpha)\)

For case a:  \({d\over 2} (\sin 90º) = 0.5 d\)

For case c: \(d\sin 60º = 0.87d\)

So, the torque for case c is higher than case a. Finally, ranking the torque from greatest to least we have \(b > c > a > d\).

Example 2.

The rod is being pushed by two forces, as shown in the figure below. Calculate the magnitude of F2 in order to keep the rod in equilibrium.

If the rod must be kept in equilibrium, the net torque must be 0. There are two forces that generates torque. If you look closely, you can conclude that each force tends to rotate the rod for different sides. So, we can assume that \(\tau_1=\tau_2\).


If we apply the torque’s definition, we have

\(F_1d_1\sin\alpha_1= F_2d_2\sin\alpha_2\)

Now we substitute the values of forces, distances and the magnitudes of the angles

\(10N \times2.8\;m\times\sin90º = F_2\times1.2\;m\times\sin 45º\)

And solving the equation, we have

\({F_2} = {28 \over 0.8484}\)

\(F_2 = 33\;N\).