Physics Bootcamp

Subsection 9.8.1 Torque of a Force that is in a Plane Perpendicular to the Axis

Consider a force on the wrench as in Figure 9.8.2. The axis is perpendicular to the page and the forces lie in the page.

The axis of rotation intersects this plane at point O while the force acts at point P. We denote the displacement vector from O to P by \(\vec r \text{.}\) Then, we define the torque of force \(\vec F \) about axis through O by the cross product of vectors \(\vec r\) and \(\vec F\text{.}\)

Magnitude and Direction of Torque:

The definition of the cross product gives us the rules for the magnitude and the direction of the torque vector \(\vec{\mathcal{T}}\text{.}\) You will find those rules explained in the chapter on Vectors.

To apply the Right Hand Rule here, hold your right hand such that the thumb points in the direction of \(\vec r \) and any of the other fingers points in the direction of \(\vec F\text{,}\) then the torque \(\vec{\mathcal{T}}\) will be perpendicular to the two vectors and will be coming out of the palm.

There are other ways of doing the right-hand rule also. For instance, if you point \(\vec r \) in the direction of \(x\) axis, and \(\vec F\) in the \(xy \) plane, then \(\vec{\mathcal{T}}\) will be in the direction of positive or negative \(z \) axis of a right-handed Cartesian coordinate system.

Components of Torque:

The definition of the cross product also gives us the rules for the components of the torque vector \(\vec{\mathcal{T}}\text{.}\) Let components of \(\vec r\) be \((x,\, y\, z)\) and that of \(\vec F\) be \((F_x,\, F_y\, F_z)\text{.}\) Then the components \((\mathcal{T}_x,\, \mathcal{T}_y\, \mathcal{T}_z)\) of \(\vec{\mathcal{T}}\) will be

If \(\vec r \) and \(\vec F \) are in the \(xy\) plane, then, this shows that torque will have only the \(z\) component non-zero.

In this case, we can speak of the direction of the torque as towards positive or negative \(z \) axis, or equivalently, in the clockwise or counterclockwise sense when we look at the force from the direction of positive \(z \) axis.

Subsection 9.8.2 Zero Torque Situations

Figure 9.8.3 has four situations where torque for each force about the axis shown is zero. In (a), torque is zero because force is parallel or antiparallel to the axis. In (b), torque is zero because lever arm of each force is zero.

We can take advantage of these zero torque directions to simplify calculations when we encounter a force in a completely arbitrary direction. Thus, we would first project a force on a plane perpendicular to the axis as shown in Figure 9.8.4, and work with \(\vec F_{\perp}\) component.

In our treatment in this book, if we do not explicitly acknowledge this aspect of a force, we will assume that the force is in the plane perpendicular to the axis, and \(\vec F_{\perp}\) is same as \(\vec F\text{.}\)

Subsection 9.8.3 The Lever Arm Picture

The lever arm, denoted by symbol \(r_{\perp}\text{,}\) is the distance from the pivot point to the line of action of the force, where the line of action of the force stands for the infinite line that is along the force vector through the point where force acts.

Figure 9.8.5 shows the geometric construction that helps compute the lever arm. We find that lever arm \(r_{\perp}\) is

Substituting \(r_{\perp}\) for \(r\,\sin\,\theta\) in Eq. (9.8.2), we can write the magnitude of the torque vector \(\vec{\mathcal{T}}\) more simply as the product of the magnitude of the force and the lever arm.

The direction of the torque is still obtained the same way as before, i.e., using the right hand rule.

Subsection 9.8.4 The Sense of Rotation and Torque

Torque is nonzero if displacement \(\vec r \) and force \(\vec F \) are not parallel or antiparallel. These two vectors fall in a plane. The torque \(\vec{\mathcal{T}} \text{,}\) which is their cross product, must be perpendicular to this plane.

Let \(\vec r \) and \(\vec F \) are in the \(xy \) plane as shown in the two figures below, then, \(\vec{\mathcal{T}} \) will either be in the positive \(z\) axis direction or in the negative \(z \) axis direction.

In Figure 9.8.6 on the right, \(\vec{\mathcal{T}} \) is in the positive \(z \) axis direction. Now, when you look at the body from the positive \(z \) axis direction, i.e., from above the page, it will appear that the force is “trying” to rotate the body in the counterclockwise sense.

In Figure 9.8.7 on the right, \(\vec{\mathcal{T}} \) is in the negative \(z \) axis direction. Now, when you look at the body from the positive \(z \) axis direction, i.e., from above the page, it will appear that the force is “trying” to rotate the body in the clockwise sense.

Thus, we have two ways of indicating the direction of a torque: (1) towards positive or negative \(z \) axis, or, (2) counterclockwise or clockwise tendency.

Subsection 9.8.5 Net Torque

How do you add influences of two or more torques? Since torque is a vector, the sum of torques will be a vector sum.

For a rotation about \(z\)-axis, we will be working with only the \(z\)-components, which will just add algebraically.

If we are thinking in terms of clockwise and countercockwise, all torques that have counterclockwise sense add together and all torques that have clockwise sense add together, and then we subtract the clockwise torque from the counterclockwise.

A simple example: Suppose we have four torques, \(\vec{\mathcal{T}}_1 \) of magnitude \(12\text{ N.m}\) in the counterclockwise sense, \(\vec{\mathcal{T}}_2 \) of magnitude \(30\text{ N.m}\) in the clockwise sense, \(\vec{\mathcal{T}}_3 \) of magnitude \(25\text{ N.m}\) in the counterclockwise sense, and \(\vec{\mathcal{T}}_4 \) of magnitude \(15\text{ N.m}\) in the clockwise sense. We can add all the counterclockwise together and separately add all the clockwise together. Then, subtract the clockwise from the counterclockwise. If the result is positive, the net torque will be counterclockwise and if negative, the net torque will be clockwise. What we are doing here is summing the \(z \) components of these vectors.

Therefore, in this example, \(\vec{\mathcal{T}}_{\text{net}} \) has magnitude \(8 \text{ N.m}\text{,}\) in the negative \(z \) axis, which is also stated as clockwise sense.