AP Physics 1 & 2 Torque

Hi folks and welcome back to Educator.com.0000

This lesson is on torque.0003

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.0006

As we do this, let us start by defining torque.0014

Torque is a vector -- τ is a force that causes an object to turn.0018

Now in order for it to cause a rotation, torque must be perpendicular to the displacement.0024

So if we look at this diagram of a wrench, we are trying to turn it around this point here.0029

What we need to do is we need to have a force that is perpendicular to this line of action.0037

The stronger the force, the more torque and the further away you are from that point, the more torque.0044

Because of that, the further away you are, you obtain more leverage, so that line, that distance is called the lever arm.0052

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.0060

Now for our purposes, instead of getting into detail around cross-products, let us focus on the magnitude of the force (F).0073

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. . . .0080

...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).0089

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.0102

The direction of the torque vector again is a little tricky -- kind of like the angular velocity angular acceleration vectors.0113

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.0119

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).0129

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.0140

It is always perpendicular to both (r) and (F).0152

Now, positive torques cause counterclockwise rotations while negative torques typically cause clockwise rotations.0156

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.0167

The net force on an object in a linear sense was the mass times linear acceleration.0175

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.0180

So we start to see all these parallels again.0192

In the linear world, we have force (F); in the rotational world, we have torque.0195

In the linear world, we have mass -- a measure of inertia -- in the rotational world, we have moment of inertia or rotational inertia.0201

In the linear world, we have acceleration; in the rotational world, we have angular rotation.0208

In the linear world, we have velocity (V); in the angular world, angular velocity.0216

We also have displacement, linear displacement, δ x; -- in the angular world, angular displacement δ θ.0222

All of these parallels just keep coming up.0233

Let us talk for a minute about types of equilibrium.0237

Static equilibrium -- that implies that the net force and the net torque on an object are 0 and the system is at rest.0240

Dynamic equilibrium implies that the net force and net torque again are 0, but the system is moving at constant translational and rotational velocity.0248

No linear acceleration. No angular acceleration.0258

Rotational equilibrium implies that the net torque on an object is 0, therefore, no angular acceleration.0262

Let us take a look at a couple of examples.0271

A pirate captain takes the helm and turns the wheel of his ship by applying a force of 20N to a wheel spoke.0273

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?0280

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).0290

So he applies a torque of 3.94Nm -- units of torque -- N x m.0312

Let us take a look at another one -- our auto mechanic.0322

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.0326

What force is applied if the wrench is 0.4 m long?0334

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.0339

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.0355

How long must the wrench be if the mechanic is only capable of applying a force of 200N?0371

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.0376

Example 3 -- We have a 3 kg cafe sign hung from a 1 kg horizontal pole, as shown in the diagram.0404

We have attached a guy-wire to prevent the sign from rotating. Find the tension in the wire.0414

Well to start off with -- let us draw a diagram of our situation.0420

There is our pole. It is attached over here to the pivot.0424

We have the weight of the pole down; its mass is 1 kg.0429

So the force on it -- down is 1 mg, so 1 × g.0435

We also have the 3 kg sign which is over here at a distance of 3 m from the pole, so that is 3g.0442

We also have the tension in our guy-wire here at this angle of 30 degrees.0450

How could I find the tension in the wire? I am going to use Newton's second law for rotation.0457

The net torque is going to be equal to -- well, counterclockwise, we will call positive.0463

We have t sin 30 degrees × the distance from the point -- our reference -- 4.0471

Now going the other direction -- going in the clockwise direction or the negative direction, we also have -3g F × its distance from our point -- 3.0480

We also have this 1g F - 1g, at a position of 2.0493

All of that must equal 0 since it is in rotational equilibrium.0499

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.0504

Let us take a look at one more -- the seesaw problem.0537

A 10 kg tortoise sits on a seesaw 1 m from the fulcrum.0542

Where must a 2 kg hare sit in order to maintain static equilibrium and what is the force on the fulcrum?0546

First off, we are going to assume it is a massless seesaw.0553

It does not tell us anything about the mass, so let us just assume the mass of the seesaw does not come into play.0557

Let us draw what we have here. We have a seesaw. We have a fulcrum.0563

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.0568

On the other side, we have our hare -- 2 kg, so its force is 2g and it is some unknown distance from our fulcrum.0582

If it is in static equilibrium though, we know that the net torque must equal 0.0595

Looking at our torques, the 10g over here -- that force times the distance (1 m) -- 10g × 1.0602

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.0610

The hare must sit 5 m from the fulcrum.0630

What is the force on the fulcrum?0636

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.0638

Newton's Second law says that we must have some force from the fulcrum pointing back up.0646

If we write Newton's Second Law for that object, F^net in the y direction, we have Fp - 10g - 2g = 0.0651

Therefore, the force of that pivot point, the fulcrum, must be 12 g or 120N.0665

Hopefully, that gets you a good start on torque.0673

Thanks for watching Educator.com.0676

Looking forward to seeing you next time. Make it a great day!0677