### Video Transcript

In this video, we’re going to learn
about torque. We’ll 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 you’ve 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 𝜏, 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 we’ve 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 𝑟, which is the distance vector from our center of mass to
where we apply our force. Noting that the force 𝐹 as well as
the distance 𝑟 are both vectors, we’re ready for a mathematical definition of
torque.

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

When we talk about torque as a
vector, it’s important to realize that because it’s a cross-product of two other
vectors, that means that torque is perpendicular to both the vectors that make it
up, 𝑟 and 𝐹. A helpful tool for quickly figuring
out the direction of torque knowing 𝑟 and 𝐹 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 𝑟. We then rotate our fingers to point
in the direction of the force 𝐹. 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. That’s 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 we’re only concerned with the magnitude of that vector. To get a sense for the magnitude of
torque, let’s reconsider this scenario with our force and our distance vector
𝑟.

Since both the force 𝐹 and the
distance 𝑟 are three-dimensional vectors, in three-dimensional space, there is some
angle between these two vectors that we’ve called 𝛳. 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 𝛳. With what we know so far about
torque, we can test and see if this equation makes sense.

Let’s consider a first case, where
the distance vector 𝑟 and the force vector 𝐹 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 𝐹 times the magnitude of 𝑟. This is the angle between 𝑟 and 𝐹
at which torque is maximum. But what if we had a case, where 𝑟
and 𝐹 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, let’s consider scenarios where the net force on an object and the net
torque on an object are zero.

We know from Newton’s second law of
motion that the net force on an object is equal to the object’s 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. It’s not accelerating.

Using what we’ve learned about
torque, we can write a rotational version of Newton’s second law, which is similar
to the linear version. It goes like this. Torque is equal to the moment of
inertia 𝐼 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 we’re 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 won’t rotate clockwise or counterclockwise about its fulcrum.

So, just like we can apply the
linear version of Newton’s 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, let’s 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?

We’ll call this maximum supportable
mass capital 𝑀. We’re 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 𝑀 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 table’s
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 𝑚 sub 𝑝 times 𝑔 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 𝑀 is that it be the largest mass possible without upsetting the
plank’s equilibrium. This means we can set the magnitude
of the torque created by the mass capital 𝑀 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,
𝜃 is always 90 degrees. Since the forces being exerted are
always perpendicular to the lever arm directions, we can write that capital 𝑀 times
𝑔 times 1.2 meters, the distance from the axis of rotation to the point where the
force of capital 𝑀 is applied, is equal to the mass of the plank times 𝑔 times
𝑑.

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

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

Let’s summarize what we’ve learnt
so far about torque. We’ve seen that torque, often
symbolized with the Greek letter 𝜏, is a measure of a force that tends to make an
object rotate. Written as an equation, the vector
torque is equal to 𝑟, the distance vector from some reference point to where the
force 𝐹 is applied, crossed with that force 𝐹.

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