### Video Transcript

In this video, weโre going to learn about torque. Weโll see what torque is, how to calculate it as a vector and a scalar, and how to solve for the direction in which torque acts.

To get started, imagine that you are on an expedition hiking through a range of mountains. You and your fellow hikers are expecting to cross this range of mountains through a well-known path youโve hiked before. But as you get nearer to the path, you see that a gigantic boulder has fallen across the trail blocking the way. Looking around you at what materials might be at hand for moving the boulder, the only things you see are some small stones lying near the path and a few tall trees off to the side. You wonder if, using just these materials, it would be possible to move the boulder out of the way. To figure out how this might be, it will be helpful to know something about torque.

Torque, which we often symbolize using the Greek letter ๐, is a force that tends to make an object rotate. If we had, for example, a massive object whose center of mass was at a particular location, then a force applied to this object at some other point than the center of mass would create a torque, that is it would tend to make the mass rotate.

In order to develop some intuition about torque, imagine that we were looking downward on a door that rotated on a hinge. And the door would rotate in as we push on it. What if we put our hand on the far-right side of the door and push, then the door would rotate easily we know. If we brought our hand closer to the hinge though, in order to make the door move the same way we would have to push with a greater force. And if we moved our hand yet further in, we would have to push with yet more force in order to make the door move the same way as it did when we pushed on its outer edge with less force.

The secret here, or the reason we had to push with increasingly greater force to have the door move the same way, was that our lever arm, which weโve marked out here in blue, got less and less as we moved closer to the hinge. So, torque then depends on the force being applied. And it also depends on where we apply that force relative to the axis of rotation, in this case, the door hinge.

If we go back to our diagram of our generic mass, we can draw a vector from the center of mass to where our forces applied and call that ๐, which is the distance vector from our center of mass to where we apply our force. Noting that the force ๐น as well as the distance ๐ are both vectors, weโre ready for a mathematical definition of torque.

Torque, which is itself a vector, is equal to the cross-product of our distance vector ๐ multiplied by the applied force ๐น. In this example, weโve shown the distance vector ๐ originating at the center of mass of our mass of interest. But we couldโve defined the origin of our distance vector wherever we like. For example, if we had a set of coordinate axes, itโs common to have our origin of those axes be where the distance vector ๐ begins.

When we talk about torque as a vector, itโs important to realize that because itโs a cross-product of two other vectors, that means that torque is perpendicular to both the vectors that make it up, ๐ and ๐น. A helpful tool for quickly figuring out the direction of torque knowing ๐ and ๐น is called the right-hand rule. The rule has this name because we use our right hand to find this direction.

First, we orient our hand so that are four fingers point in the direction of the distance vector ๐. We then rotate our fingers to point in the direction of the force ๐น. In this case, out of the page. Having done this, our thumb then points in the direction of the torque being exerted by this force on this moment arm. Thatโs the convention called the right-hand rule for finding the direction in which torque acts.

Notice that if we accidentally used our left hand instead of our right hand, then we would find torque acting in exactly the opposite direction. Though torque is a vector, sometimes weโre only concerned with the magnitude of that vector. To get a sense for the magnitude of torque, letโs reconsider this scenario with our force and our distance vector ๐.

Since both the force ๐น and the distance ๐ are three-dimensional vectors, in three-dimensional space, there is some angle between these two vectors that weโve called ๐ณ. The magnitude of torque is equal to the magnitude of the force times the magnitude of the distance multiplied by the sine of that angle between them, the sine of ๐ณ. With what we know so far about torque, we can test and see if this equation makes sense.

Letโs consider a first case, where the distance vector ๐ and the force vector ๐น are at 90 degrees to one another. Well, we know that the sine of 90 is one. So, in that case, our torque is simply equal to the magnitude of ๐น times the magnitude of ๐. This is the angle between ๐ and ๐น at which torque is maximum. But what if we had a case, where ๐ and ๐น were in line with one another, that is the angle between them is zero? We know that the sine of zero degrees is zero. So, zero times any magnitude is still zero. And the total torque in this case is zero.

And this makes sense. Notice that the applied force in this case is much more like a linear force that would cause translation rather than rotation. Indeed, we see because torque is zero, it cannot cause an object to rotate. Speaking of translation and rotation, letโs consider scenarios where the net force on an object and the net torque on an object are zero.

We know from Newtonโs second law of motion that the net force on an object is equal to the objectโs mass times its acceleration. We could call this the linear version of the second law. Imagine that we had a mass being acted on by a set of forces which all cancel one another out so that the net force on this mass is zero. When this condition is met, that means that the mass is in equilibrium. Itโs not accelerating.

Using what weโve learned about torque, we can write a rotational version of Newtonโs second law, which is similar to the linear version. It goes like this. Torque is equal to the moment of inertia ๐ผ of an object multiplied by its angular acceleration.

Say that we had a long massive plank being balanced on a fulcrum at the center of the plank. Imagine further that we start applying forces at different points on this plank perpendicular to the axis of the plank. But weโre careful to do it in such a way that the sum of the torques around the rotation point at the fulcrum is zero. This means that the net torque acting on the plank is zero, and, therefore, that the plank is in equilibrium, that is it wonโt rotate clockwise or counterclockwise about its fulcrum.

So, just like we can apply the linear version of Newtonโs second law of motion to understand object equilibrium, so we can apply the rotational version of the second law to understand whether an object will rotate or not. Knowing all this, letโs get some practice with torque using an example.

A uniform plank of length 8.0 meters rests on a level surface with 1.2 meters of its lengths overhanging the end of the surface, as shown in the diagram. The plank has a mass of 40.0 kilograms. What is the maximum mass that can be supported at the overhanging end of the plank that maintains the equilibrium of the plank?

Weโll call this maximum supportable mass capital ๐. Weโre told in this statement that the overall mass of the plank is 40.0 kilograms. We can depict this scenario as hanging a mass of value capital ๐ from the end of the overhanging plank. If this mass was heavy enough to cause the plank to rotate, that rotation would happen at the point where the tableโs edge is. So, that is effectively our axis of rotation. And we want to calculate the torques acting on the plank with respect to that point.

In addition to the torque created by our hanging mass, there is also a torque created by the mass of the plank itself. That torque is equal to the force of gravity acting on the plank equal to ๐ sub ๐ times ๐ multiplied by the distance between the center of mass of the plank and our fulcrum point. We know that the condition of our mass capital ๐ is that it be the largest mass possible without upsetting the plankโs equilibrium. This means we can set the magnitude of the torque created by the mass capital ๐ equal to the magnitude of the torque created by the plank under its own weight.

Recalling that torque magnitude is equal to force magnitude times the distance magnitude times the sine of the angle between these vectors. And recognizing that in our case, ๐ is always 90 degrees. Since the forces being exerted are always perpendicular to the lever arm directions, we can write that capital ๐ times ๐ times 1.2 meters, the distance from the axis of rotation to the point where the force of capital ๐ is applied, is equal to the mass of the plank times ๐ times ๐.

And we see that the acceleration due to gravity cancels from both sides. We can solve for ๐ by knowing the overall length of the plank as well as the length of the plank on the surface. Based on our diagram, ๐ is equal to 6.8 meters minus 40.0 meters, or 2.8 meters. Now that we know ๐ and since weโre given the mass of the plank ๐ sub ๐, we can rearrange and solve for capital ๐.

Capital ๐ is equal to ๐ sub ๐ times 2.8 meters divided by 1.2 meters. Plugging in for ๐ sub ๐ 40.0 kilograms, when we calculate capital ๐ to two significant figures, we find itโs 93 kilograms. Thatโs the maximum amount of mass we can hang from the end of the plank and still maintain its equilibrium.

Letโs summarize what weโve learnt so far about torque. Weโve seen that torque, often symbolized with the Greek letter ๐, is a measure of a force that tends to make an object rotate. Written as an equation, the vector torque is equal to ๐, the distance vector from some reference point to where the force ๐น is applied, crossed with that force ๐น.

And when weโre interested only in the magnitude of torque, as an equation, thatโs equal to the magnitude of the force times the magnitude of the distance ๐ multiplied by the sine of the angle between these two vectors. And finally, weโve seen that the direction of torque is perpendicular to the force and the direction vector ๐ and that we use the right-hand rule to solve for that direction.