Calculating Torque as a Cross Product

Torque is the rotational effect of force. For moving, a body from rest, a force is required similar to set up a body in rotation from rest a torque is required. The following section will deal with details about how large or small is torque? In which direction the torque acts.

Consider a screw shown in the diagram. If forces are applied at different locations then how the rotating effects are created. The axis of rotation passes through the center of the screw and perpendicular to the plane of the diagram.
First, consider only \({F_1}\) force, then will it create any rotational effect? The screw will not rotate, thus the force \({F_1}\) will not create any torque.
Now, if only force \({F_2}\) is applied, then again the screw will not rotate.
Forces \({F_3}\) and \({F_4}\) will rotate the screw, thus they will create some torque or turning effect. However, which one will easily rotate the screw? The force \({F_3}\) is far off the axis of rotation and easily rotate the screw as compared to \({F_4}\).

\({F_1}\) and \({F_3}\) are acting at the same point, but their angle is different. \({F_1}\) and \({F_3}\) have different turning effects. Thus, on what factors does the turning effect depend upon?
As \({F_3}\) and \({F_4}\) are producing torque and they are at different distances from the axis of rotation, thus torque must depend on the distance of the force from the axis of rotation. This is the reason why the handles are made at the end of doors, so that the distance of force increases from the axis of rotation and more turning effect can be created by applying lesser force.
As \({F_1}\) and \({F_3}\) are at the same distance from the axis of rotation, but at different angles, thus the torque depends on the orientation of force.
Torque also depends on the magnitude of the force. The greater the magnitude of force greater is the turning effect. Suppose you want to open a jammed screw, then you have to apply greater force to produce greater torque.

On accumulating all the above mentioned factors, it can be concluded that about any point
\[\vec \tau = \vec r \times \vec F\] Here \(\vec r\) is the position vector of the point of application of force with respect to the point about which torque is to be calculated,
\(\vec F\) is the force applied,
\(\vec \tau \) is the torque.
Direction of torque can be calculated by the rules of cross product.

Consider the above diagram in which the angle between \(\vec r\) and \(\vec F\) is \(\theta\). In this case if the line of action of the force is extended and a perpendicular is dropped on it from the point of calculation of torque then this perpendicular is called as moment arm.
The moment arm equals \(r sin \theta\),
Magnitude of the torque about point 'O' equals \(r F sin \theta \),
Therefore, torque can also be written as the product of force and moment arm.

\(\large \color {blue} {NOTE} \)

If torque is to be calculated about a point on the line of action of force, then the torque comes out to be zero. This is because in this case the angle between the position vector 'r' and force 'F' will be zero.

When torque is calculated about a point, it automatically points along an axis which can be found by using right hand thumb rule. If torque is to be calculated about any different axis, then the following steps are needed to be taken,
1) Calculate torque about any point on the axis
2) Calculate the component of torque about the specified axis.

Consider the diagram shown above, in which force 'F' is acting on a body at point 'P', perpendicular to the plane of the figure. Thus 'r' is perpendicular to the force and torque about point 'O' is in x-y plane at an angle \(\theta \) to y-axis. Torque about 'O' equals \({\tau_O} = r F \)
To calculate the torque about the y-axis, take the component of torque about y-axis.
Thus,
\({\tau_{yaxis}} = r F cos \theta \)
The perpendicular (AF) drawn on y-axis from the line of action of the force is equal to \(AF=r cos \theta \), Thus, the torque about an axis can be calculated as the product of force and the perpendicular distance between the line of action of the force and axis about which torque is to be calculated.

\(\large \color{blue} NOTE \) For the above formula to be correct the line of action of the force and the axis about which torque is to be calculated has to be skew symmetric line. Skew symmetric lines are those which are neither parallel nor intersecting.
If the line of action of the force and axis about which torque is to be calculated are parallel, then the torque about the axis will be zero.