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.