Hi. It’s Mr. Andersen and this AP Physics essentials video 52. It is on torque. Torque
is simply the product of the force, which is perpendicular to the lever arm. And this
only works in rotational motion. So what is rotational motion? Imagine that we have a
wrench that is sitting here. And I have marked the center of gravity. And so let’s say
it is just sitting in space to make this easier. And let’s say I apply a force right at that
center of gravity. What we are getting is linear, or we call that translational motion.
We apply a force in one direction and we are seeing an acceleration in that direction.
Let’s say I do the same thing but now I am going to do it off center. So I am applying
a force not at the center of gravity. And watch what we get. We get two forms of motion.
We are getting that translational motion but we are also getting rotational motion. And
it is hard to separate those two. And so an easy way to look at rotational motion is to
pin it down. So now I have this same thing. We have the wrench. But now it is centered
around this nut on one side. And so this is not moving. So it is going to rotate around
that point. Now let me apply that same force on one direction. And now we are just getting
that rotational motion. Let me return the wrench for a second. And we can start to talk
about what torque is. And so this is going to be the lever arm. The lever arm is going
to go from this point in the middle, that axis of rotation. It is going to go out from
that. It is perpendicular to that axis of rotation. And it is just like a lever that
you would use as a simple machine. And then we are applying a force perpendicular to that
lever arm. And so what is torque? We are multiplying these two values. We are multiplying the force
that we apply, perpendicular to the lever times the lever arm itself. And so it is two
things and the units are not just going to be newtons, it is going to be newton meters.
And so in a rotational system like this, where we have an axis of rotation, what is torque?
It is simply the product of that lever arm which is going to be a vector, it has to be
perpendicular to that axis of rotation. Again the axis of rotation is coming straight at
us through the middle and we are going to multiply that times the force. The force has
to be perpendicular to that lever arm. And so if we multiply those two values then we
are going to have ourselves a torque. And this is our equation right here. Tau, which
stands for torque, is equal to that lever arm, perpendicular and then that force. And
so if we have a system that is not moving, like our rotational system is not moving anymore
we know that all of those forces must be balanced. In other words all the torques must be balanced.
And so the net torque on a balanced system is equal to 0. Have you ever noticed that
a door will always have the door knob on the outside or far away from the hinge? Well the
reason that is just deals with torque. And so what I have here is the door itself. This
is the door hinge right here. And here is the door knob on the outside. And so if I
apply a force out here what am I going to get? Torque. We have the lever arm and then
we are applying a force perpendicular. And so we are going to get an acceleration or
rotational acceleration like that. What happens if I apply a greater force? What am I going
to have? I am going to have a greater torque. And so we are going to see a faster acceleration
like that. What happens if I take that smaller force however and move it to the inside? What
happens if I put the doorknob on the inside of the door? If I try to pull on it with that
small force nothing will happen. In other words I have decreased that torque so much,
since I have decreased that lever arm, that I would not have enough torque to open that
door. And so that lever arm distance is incredibly important. Just like a lever in a lever system,
it is giving us a mechanical advantage to open up that door. So in torque we are applying
that lever arm times the force. That is how we calculate the torque. And the equation
looks like this. Tau is equal to r perpendicular F. And so let’s add some numbers in here
and I will show you how to calculate that. Let’s say the doorknob is 75 centimeters
from the hinge and we apply a 6.7 newton force. What is going to be our torque in this situation?
Well I am going to plug in those values. Again I had to convert the centimeters into meters.
We have to use SI units. And now I simply multiply those values. What are my units?
It is now in newton meters. You can see I have too many significant digits. And so we
would get a 5.0 newton meter torque if the doorknob is at the outside. Now watch what
happens if we move it 15 centimeters from the hinge. Watch what happens to our torque.
Again the force is going to be the same. But now we are going to get a 1.0 newton meter
torque. It is 1/5 of what it was before, which is not surprising, because the lever arm is
now 1/5 of what it was before. Let’s take a second to look at a balanced system. And
a see-saw or a teeter-totter is a great example of that. We have the axis of rotation right
here in the middle. And we can have a lever arm on this side and a lever arm on the other
side. So what we can do is we can apply a torque on one side and a torque on the other
side. And if those torques are equal, then we are going to have a balanced system where
you should not see any movement. And so watch what happens when I remove the supports. Since
the torque is the same on either side, nothing happens. Let me apply a 5 kilogram weight
on either side, if I remove the supports, it is totally balanced. Torque is equal on
either side. Let’s say I add a 10 kilogram weight to the right side. And now remove the
supports. What happens? Well we have greater torque on the right side and so we are getting
rotational motion in that direction. Now a good question I might ask you is where could
I move that 5 kilogram mass on the left side so the torque on the left equals the torque
on the right? How could I balance this system out? Well if we look at our equations again,
it is simply the product of the lever arm times the force. And so if I throw in some
values here and set them equal to each other, on the left side we have the unknown lever
distance times that 50 newton force. Where do I get the 50 newton force? I am simply
taking 5 kilograms times about 10 for the acceleration due to gravity. So it is a 50
newton force down on that side. On the other side it is going to be 100 newton force. So
I could solve for this. We would have a torque of 400 newton meters on the right side. And
so to solve for that we should have a distance on the left side equal to 8 meters. And so
if I take that 5 kilogram mass, move it out to 8 meters, what are we going to get? A balanced
system. The torque on either side is going to be exactly the same. And so did you learn
the relationship between a force and a torque? Again we have to multiply the force times
that lever arm. Do you see what happens when we apply different forces? We increase the
torque. Or what happens if we move that force that in? We are decreasing the torque because
we are decreasing the lever arm. Can you design an experiment that would allow you to kind
of manipulate these balanced forces? Again we used a teeter-totter to do that. And finally
can you calculate torque in a two-dimensional system like this?
I hope so. And I hope thatwas helpful.