Torque Basics | Simple harmonic motion and rotational motion | AP Physics 1 | Khan Academy

Imagine you've got a door here with a blue doorknob. Any one of these 10-newton forces will cause the door to rotate around the hinge, or the axis, or sometimes this is called the pivot point. Any one of these forces will cause the door to rotate.

My question is, if you could exert one of these in one of these locations, which one of these forces, if any, would cause the most angular acceleration of this door? You might think, "Oh, well, 10 newtons is 10 newtons; they'll all cause the same amount," but that's not true.

It turns out we put doorknobs at the end of doors for a reason. This red 10 newtons at the outside edge will cause the most angular acceleration; that will cause this door to speed up most rapidly. This used to bother me; I was like, "How come this is getting an advantage?"

I think the best way to think about it is this: even though these forces are all going through the same angle—so they've gone through 20 degrees, now they've all gone through 30 degrees, now they've all gone through 45, 60, and 90—even though these forces have all gone through the same angle, they have not gone through the same distance.

Some of these forces have been exerted through a larger distance. So, just look at it. If you imagine rotating this thing, that red force, this outside pink force here, goes through a much larger distance than that inner yellow force. This force has gone through very little distance whatsoever.

And you might think, "Well, why does that matter?" Well, it matters because, if you remember, work done is proportional to the amount of force. But these are all 10 newtons, so that doesn't really matter here. It's also proportional to the amount of distance through which that force is applied.

Because this outside force has gone through so much more distance than these inner forces, it's done more work over the same angle. If you do more work, you input more kinetic energy into the door; it's going to be moving faster for the same angle compared to what's caused by these other forces here.

This is why, in angular mechanics, you can't just think about forces; you have to think about something called torque. The symbol for torque is this fancy T; it's the Greek letter tau. The amount of torque caused by a force—so you need a force to cause a torque—but it's more than just force; you have to multiply the distance from the axis to the force by the amount of force in order to find how much torque is being exerted by a given force.

The more torque that's exerted, the more angular acceleration you'd get, the faster you'd get something to speed up. Now, you might wonder, "Okay, I get that more force gives me more torque; how come this is just r and not like r squared?" It seems kind of random; maybe it's like the square root of r.

Well, if you remember arc lengths from back in the day, arc length is r times theta. So, if I'm twice as far away from an axis, I get twice the arc length. If I get twice the arc length, I get twice the work done. You get twice the work done; you get twice the input kinetic energy. It turns out twice the kinetic energy will give you twice the angular acceleration.

This is why everything's just proportional to r in terms of torque; it's not like r squared here. So, for example, let's just say the distance from the axis—because that's what matters—to this 10 newtons here was one meter, and from the axis to the purple force was two meters, and from the axis to this doorknob force was three meters.

What this torque formula means is that, even though these are all 10 newtons, they'd all be exerting different amounts of torque. I'd have to take the 1 meter times 10 newtons, which would give me 10 newton meters. So the unit for torque is meters times newtons, but we usually write it as newton meters.

If you buy a torque wrench, you could set it in newton meters or in foot-pounds if you're doing the U.S. system. And then this purple force, even though it's 10 newtons, you'd have to take 2 meters times 10 newtons. This would exert a torque of 20 newton meters, and the doorknob force wins the battle because it would have 3 times 10, which would exert a torque of 30 newton meters.

So the same size force can exert a different amount of torque depending on how far away it is from the axis. One area you have to be careful: this torque is technically a vector. It has a direction; it could be positive or negative. If you're doing full-blown engineering 3D physics, technically these torques would point out of the screen here, out of the page.

But for intro algebra-based physics, and for most problems, you could usually get away with just considering counter-clockwise or clockwise as being the direction of the torque. That is to say, these forces were making this object rotate in the counter-clockwise direction, so they would all have the same sign.

The convention is to call counter-clockwise positive, so we'd call these all positive. If there were any forces that tried to rotate the system clockwise, you'd call those torques negative. You could do it either way, as long as you're consistent. Most books pick this as the convention, though, so you should be aware of that.

And then the last little bit to be careful about: I'm drawing all these forces nice and perpendicular to the r, and if that's the case, you just do r times f. If your force has different components, you need to make sure that the only component you plug in here is the perpendicular piece. So, if this had some weird angle here, you'd only want the piece that was directly into this perpendicular lever arm here at a perpendicular angle.

We'll talk about that more later. For now, let's just try some problems to kick the tires and get used to this formula. So, imagine this example here where you've got the fancy door, you know, with a fancy hotel or restaurant; that's a rotating circle, and you can go in from either direction. This would be a bird's-eye view.

Now, imagine you go into the hotel; you're pushing over here. You took physics; you know what to do, so you exert those 20 newtons over here. Let's say someone else comes in from the other edge—it's all awkward—and they're trying to go in the other way, and it's a stalemate. You're both pushing with forces, but nothing's happening.

And that doesn't mean the two forces are equal. If you're in a stalemate here in terms of angular motion, that means your torques are equal and opposite. They're opposed; they have the same magnitude but will have opposite directions of torque.

So, if you're locked in a stalemate here, that means the torque that you exert with your 20 newtons has to be equal to the torque from the other person. So, let's try to figure out how much force would this person have to exert. It's not going to be 20 newtons; they're pushing closer to the axis here, so they're going to have to push with more force.

How much more force? Well, we can use the formula for torque to find it. The torques have to be equal; if there's no rotation here, you're balanced out. If your force is 20 newtons, you're exerting a force 3 meters away from the axis; that's your r.

Would be 3 meters times 20 newtons, which means you're exerting 20 times 3, so 60 newton meters of torque. That means the other person has to be exerting 60 newton meters of torque. But there isn't two—be careful here—you always have to measure from the axis, the point where you're rotating about.

That'd be 1 meter; this door is all symmetric here, so it'd be 1 meter times f. If you take this 60 newton meters and you divide by 1 meter, you're going to get that this force here is going to have to be 60 newtons. So this person is going to have to exert more force. In fact, they pushed three times closer to the axis.

So they're going to have to exert three times the force that you do. You have a three times advantage here in holding this door compared to the other person. Alright, let's try one more just to make sure we understand it. Let's say it's now rush hour; you know, bird's eye view here, same circular door.

Three people are trying to go through it once; it's going to be a madhouse. This time, I want to know: it's not going to be a stalemate; this door is going to rotate in some direction. I want to know what the net torque is.

So just like you can find that force, you can find the net torque. But you've got to be careful; these might have different signs. You've got to add or subtract accordingly. So, let's start over here. How much torque would be from this 10 newtons? Well, it's exerted 3 meters away from the axis, so its r is 3 meters.

So, the torque from that force would be 3 meters times 10 newtons. Since this is directed counterclockwise, I'm just going to call that positive, and I'll have to be consistent with that choice. So now let's consider this 8 newtons. You might think it would have an oppositely directed sign of torque from this 10 newtons, because the 8 is down and the 10 is up.

But it's also trying to rotate this door in the counterclockwise direction, so in terms of forces, this 10 newton and 8 newton are oppositely directed. But in terms of torques, they're the same direction. They're both causing rotation counterclockwise.

So if I call this torque from the 10 newtons positive, I've got to call the torque from this 8 newtons positive, because it's trying to exert a torque in the same direction. So I'd have 1 meter as the r for the 8 times 8 newtons would be the torque from the 8 newtons.

And then I have one more force here. This 5 newtons is trying to rotate clockwise. Since I called counterclockwise positive, I'm going to have to make this a negative torque. So minus 3 meters—the r from the axis to this 5 newtons is 3 meters—multiplied by 5 newtons.

If you take 30 plus 8 minus 15, you're going to get a total of positive 23 newton meters of torque. So this is not a stalemate; there will be an amount of angular acceleration caused by this net torque.

So to recap, just like net forces can cause regular acceleration, net torques can cause angular acceleration. If there is no net torque, that means there is no angular acceleration. The way you find the torque from a given force is you take r, the distance from the axis to where that force is applied, and you multiply it by the amount of force, as long as it's that amount of force that runs perpendicular to this lever arm or this r direction.

Be careful that torque is a vector. We typically count counterclockwise as positive and clockwise as negative, but if you're consistent, you can call whichever one of these you want to be positive as long as you call the other one negative.