WEBVTT physics/ap-physics-1-2/fullerton
00:00:00.000 --> 00:00:03.000
Hi folks and welcome back to Educator.com.
00:00:03.000 --> 00:00:06.000
This lesson is on torque.
00:00:06.000 --> 00:00:14.000
Our objectives are to calculate the torque on a rigid object and apply conditions of equilibrium to analyze a rigid object under the influence of a variety of forces.
00:00:14.000 --> 00:00:18.000
As we do this, let us start by defining torque.
00:00:18.000 --> 00:00:24.000
Torque is a vector -- τ is a force that causes an object to turn.
00:00:24.000 --> 00:00:29.000
Now in order for it to cause a rotation, torque must be perpendicular to the displacement.
00:00:29.000 --> 00:00:37.000
So if we look at this diagram of a wrench, we are trying to turn it around this point here.
00:00:37.000 --> 00:00:44.000
What we need to do is we need to have a force that is perpendicular to this line of action.
00:00:44.000 --> 00:00:52.000
The stronger the force, the more torque and the further away you are from that point, the more torque.
00:00:52.000 --> 00:01:00.000
Because of that, the further away you are, you obtain more leverage, so that line, that distance is called the lever arm.
00:01:00.000 --> 00:01:13.000
Now, officially, torque is a cross-product; it is a vector product -- a vector multiplication of the r-vector, which is the vector from the point of rotation to where the force is applied, and the force vector.
00:01:13.000 --> 00:01:20.000
Now for our purposes, instead of getting into detail around cross-products, let us focus on the magnitude of the force (F).
00:01:20.000 --> 00:01:29.000
The magnitude is going to be rF sin θ and the reason is since this force -- only the portion that is perpendicular to this line of action counts. . . .
00:01:29.000 --> 00:01:42.000
...if we were to draw the component of the force here that is perpendicular to the line of action -- if that is angle θ, that then must be the opposite side, so that is where we get the sin of(θ).
00:01:42.000 --> 00:01:53.000
So the magnitude of the torque vector is the distance (r) × the force (F) × the sin of the angle between the line of action and the force.
00:01:53.000 --> 00:01:59.000
The direction of the torque vector again is a little tricky -- kind of like the angular velocity angular acceleration vectors.
00:01:59.000 --> 00:02:09.000
The direction of the torque vector is perpendicular to both the position vector and the force vector and again we figure it out using the Right Hand Rule.
00:02:09.000 --> 00:02:20.000
For example, if we have a position vector (r) from the point of rotation to where the force is applied, let us call that (r) and then we have a force (F).
00:02:20.000 --> 00:02:32.000
The way we find the direction of the torque vector is we take the fingers of our right hand, point those in the direction of (r), bend the fingers toward (F) and your thumb will point in the direction of the positive torque vector.
00:02:32.000 --> 00:02:36.000
It is always perpendicular to both (r) and (F).
00:02:36.000 --> 00:02:47.000
Now, positive torques cause counterclockwise rotations while negative torques typically cause clockwise rotations.
00:02:47.000 --> 00:02:55.000
What is really nice here is -- once again -- just like when we were talking about rotational kinematics, we have the same sort of parallels as we talk about torque.
00:02:55.000 --> 00:03:00.000
The net force on an object in a linear sense was the mass times linear acceleration.
00:03:00.000 --> 00:03:12.000
The net torque on an object is equal to (I), the moment of inertia, or also known as the rotational inertia times the angular acceleration α.
00:03:12.000 --> 00:03:15.000
So we start to see all these parallels again.
00:03:15.000 --> 00:03:21.000
In the linear world, we have force (F); in the rotational world, we have torque.
00:03:21.000 --> 00:03:28.000
In the linear world, we have mass -- a measure of inertia -- in the rotational world, we have moment of inertia or rotational inertia.
00:03:28.000 --> 00:03:36.000
In the linear world, we have acceleration; in the rotational world, we have angular rotation.
00:03:36.000 --> 00:03:42.000
In the linear world, we have velocity (V); in the angular world, angular velocity.
00:03:42.000 --> 00:03:53.000
We also have displacement, linear displacement, Δ x; -- in the angular world, angular displacement Δ θ.
00:03:53.000 --> 00:03:57.000
All of these parallels just keep coming up.
00:03:57.000 --> 00:04:00.000
Let us talk for a minute about types of equilibrium.
00:04:00.000 --> 00:04:08.000
Static equilibrium -- that implies that the net force and the net torque on an object are 0 and the system is at rest.
00:04:08.000 --> 00:04:18.000
Dynamic equilibrium implies that the net force and net torque again are 0, but the system is moving at constant translational and rotational velocity.
00:04:18.000 --> 00:04:22.000
No linear acceleration. No angular acceleration.
00:04:22.000 --> 00:04:31.000
Rotational equilibrium implies that the net torque on an object is 0, therefore, no angular acceleration.
00:04:31.000 --> 00:04:33.000
Let us take a look at a couple of examples.
00:04:33.000 --> 00:04:40.000
A pirate captain takes the helm and turns the wheel of his ship by applying a force of 20N to a wheel spoke.
00:04:40.000 --> 00:04:50.000
If he applies the force at a radius of 0.2 m from the access of rotation and at an angle of 80 degrees to the line of action, what torque does he apply to the wheel?
00:04:50.000 --> 00:05:12.000
Well, that is a straightforward calculation where the magnitude of the torque vector is rF sin θ, where our (r) -- radius 0.2 m -- the distance from the point of rotation to where the force is applied -- × the force itself (20N) × the sin of our angle (80 degrees).
00:05:12.000 --> 00:05:22.000
So he applies a torque of 3.94Nm -- units of torque -- N x m.
00:05:22.000 --> 00:05:26.000
Let us take a look at another one -- our auto mechanic.
00:05:26.000 --> 00:05:34.000
A mechanic tightens the lugs on a tire by applying a torque of 100Nm at an angle of 90 degrees to the line of action.
00:05:34.000 --> 00:05:39.000
What force is applied if the wrench is 0.4 m long?
00:05:39.000 --> 00:05:55.000
Well again, magnitude of the torque vector is rF sin θ, therefore, if we want the force -- that is just going to be the torque/r sin θ.
00:05:55.000 --> 00:06:11.000
If our torque is 100Nm, our (r) is 0.4 m × the sin of our angle (90 degrees) -- sin 90 = 1, so 100/0.4 = 250N.
00:06:11.000 --> 00:06:16.000
How long must the wrench be if the mechanic is only capable of applying a force of 200N?
00:06:16.000 --> 00:06:44.000
Well if we want the length there, the torque = Fr sin θ again, therefore, r = the torque/F sin θ, which is 100Nm/F (200N) × sin 90 degrees, which is 0.5 m.
00:06:44.000 --> 00:06:54.000
Example 3 -- We have a 3 kg cafe sign hung from a 1 kg horizontal pole, as shown in the diagram.
00:06:54.000 --> 00:07:00.000
We have attached a guy-wire to prevent the sign from rotating. Find the tension in the wire.
00:07:00.000 --> 00:07:04.000
Well to start off with -- let us draw a diagram of our situation.
00:07:04.000 --> 00:07:09.000
There is our pole. It is attached over here to the pivot.
00:07:09.000 --> 00:07:15.000
We have the weight of the pole down; its mass is 1 kg.
00:07:15.000 --> 00:07:22.000
So the force on it -- down is 1 mg, so 1 × g.
00:07:22.000 --> 00:07:30.000
We also have the 3 kg sign which is over here at a distance of 3 m from the pole, so that is 3g.
00:07:30.000 --> 00:07:37.000
We also have the tension in our guy-wire here at this angle of 30 degrees.
00:07:37.000 --> 00:07:43.000
How could I find the tension in the wire? I am going to use Newton's second law for rotation.
00:07:43.000 --> 00:07:51.000
The net torque is going to be equal to -- well, counterclockwise, we will call positive.
00:07:51.000 --> 00:08:00.000
We have t sin 30 degrees × the distance from the point -- our reference -- 4.
00:08:00.000 --> 00:08:13.000
Now going the other direction -- going in the clockwise direction or the negative direction, we also have -3g F × its distance from our point -- 3.
00:08:13.000 --> 00:08:19.000
We also have this 1g F - 1g, at a position of 2.
00:08:19.000 --> 00:08:24.000
All of that must equal 0 since it is in rotational equilibrium.
00:08:24.000 --> 00:08:57.000
So I could solve for (t) to say that (t) must be -- we have 9g + 2g -- 11g/4 sin 30 degrees, which is going to be 11 × 10 m/s²/4 sin 30 degrees or 110/sin 30 1/2, 2, which is going to give us 55N as the tension in that wire.
00:08:57.000 --> 00:09:02.000
Let us take a look at one more -- the seesaw problem.
00:09:02.000 --> 00:09:06.000
A 10 kg tortoise sits on a seesaw 1 m from the fulcrum.
00:09:06.000 --> 00:09:13.000
Where must a 2 kg hare sit in order to maintain static equilibrium and what is the force on the fulcrum?
00:09:13.000 --> 00:09:17.000
First off, we are going to assume it is a massless seesaw.
00:09:17.000 --> 00:09:23.000
It does not tell us anything about the mass, so let us just assume the mass of the seesaw does not come into play.
00:09:23.000 --> 00:09:28.000
Let us draw what we have here. We have a seesaw. We have a fulcrum.
00:09:28.000 --> 00:09:42.000
Over here on one side, we have a 10 kg tortoise, so its weight -- the force on it -- is 10g and that is 1 m from the fulcrum -- 1 m.
00:09:42.000 --> 00:09:55.000
On the other side, we have our hare -- 2 kg, so its force is 2g and it is some unknown distance from our fulcrum.
00:09:55.000 --> 00:10:02.000
If it is in static equilibrium though, we know that the net torque must equal 0.
00:10:02.000 --> 00:10:10.000
Looking at our torques, the 10g over here -- that force times the distance (1 m) -- 10g × 1.
00:10:10.000 --> 00:10:30.000
Now, the negative direction, -2g times whatever that distance happens to be from the fulcrum (x) must equal 0, so 10g - 2g(x) = 0, 10g = 2g(x) or x = 10g/2g, which must be 5 m.
00:10:30.000 --> 00:10:36.000
The hare must sit 5 m from the fulcrum.
00:10:36.000 --> 00:10:38.000
What is the force on the fulcrum?
00:10:38.000 --> 00:10:46.000
To find the force on the fulcrum, all we have to do now is look at all of the different forces that we have here.
00:10:46.000 --> 00:10:51.000
Newton's Second law says that we must have some force from the fulcrum pointing back up.
00:10:51.000 --> 00:11:05.000
If we write Newton's Second Law for that object, F^net in the y direction, we have Fp - 10g - 2g = 0.
00:11:05.000 --> 00:11:13.000
Therefore, the force of that pivot point, the fulcrum, must be 12 g or 120N.
00:11:13.000 --> 00:11:16.000
Hopefully, that gets you a good start on torque.
00:11:16.000 --> 00:11:17.000
Thanks for watching Educator.com.
00:11:17.000 --> 00:11:21.000
Looking forward to seeing you next time. Make it a great day!