The Simplest Explanation of How Torque Works in Physics

Added: 2026-05-07

[00:00:00]Well, now we're going to talk about the concept of torque and the basic idea behind torque. One thing that you should remember is it's kind of force along a circle, okay? Angular force, sort of, okay? When you open a door that's hinged, okay? You're pulling on it and that door is swinging through an angle, well, you're kind of pulling around some angular motion there. You're exerting a torque on the door. When you take a wrench and you're trying to loosen a nut or something like this, you're trying to kind of exert a force through an angle or something like this, trying to exert and kind of create an angular motion there, you're pushing kind of around something, you're exerting a torque, okay?

[00:00:37]So, let's kind of talk about that mathematically.

[00:00:41]Topic of torque.

[00:00:44]Okay?

[00:00:46]Imagine that you have some kind of a wrench, okay? And I'm going to draw it like this.

[00:00:52]This is a wrench off like this, okay? The wrench has some distance, some length. Now, I'm instead of using L, I'm going to use R and you'll understand why in a second, but anyway, it's just the it's just the the length of the wrench here.

[00:01:07]And let's say I've got some nut or something in here and I'm trying to loosen it. Well, you all know from experience that it makes most sense to kind of push on this guy kind of perpendicular, kind of push kind of straight up as I've had it as I have it drawn here on the board. So, I'm going to exert some some force.

[00:01:24]Okay, like this.

[00:01:27]Perpendicular to the wrench and that's going to create what we call a torque around this pivot point here, okay? The torque is going to be a a T, but it's going to be a fancy T with a kind of a curvy top and that's called tau. That's a a Greek letter tau, okay? Tau, t a u, and it's just a letter, it just means torque. And it's kind of like a T, so it's kind of kind of easy to remember. The torque in this case as as as I have it drawn here is just equal to the distance R, the length here times the force I exert, okay?

[00:01:59]Now, I want to tell you though that the main thing you need to be aware of here is that this force that you put into this equation has got to be the force perpendicular to the object that you're actually pushing, okay?

[00:02:14]In order to calculate the proper torque.

[00:02:15]So, this force is the force that's perpendicular, this is my symbol for perpendicular to R.

[00:02:25]And R is just the object that you're dealing with here, okay? So, it kind of makes sense. I mean, torque is equal to R times F. So, it kind of makes sense.

[00:02:33]The harder you push, the more torque you exert. Makes sense, right? The the longer something is, the more torque you exert. Now, here's where you start to get a little bit into leverage. If you have a broomstick that's really long, okay? And you're trying to loosen something up, well, you all know from experience that the longer something is, the more leverage you kind of have. It's precisely because the longer something is, as you push up on the end of it is going to generate a higher torque down at the end over there. So, a longer wrench is going to give you more leverage is what we really talk about usually, right? But what it's really doing for you is it's generating more torque for you, which is really the force that's loosening that nut down there. So, the longer something is or the more you push is going to generate more torque. The only caveat to really be aware of is that this force has got to be and to use in this equation has got to be the force perpendicular here.

[00:03:24]Now, let's take a little bit a more complicated case. Let's say I have a wrench like this.

[00:03:36]Nothing different, okay? It's also the same length R long, okay? But let's say I don't push on it vertically like this. Let's say I push on it or I pull on it or whatever you want to say off at an angle like this.

[00:03:50]Now, you should know because I've already cautioned cautioned you, you know, 17 times that you can't just take this force and put it into this equation and expect to get the proper torque because this force has got to be the force that's in the perpendicular direction to the actual object that you're moving. And then clearly here, I'm pulling it off to the side.

[00:04:09]But what you can do is you can use the magic of vectors and decompose this guy into a component of force this way and a component of force this this way. And then by using this component of force, then I can use the equation just fine.

[00:04:23]So, and draw my little dotted lines here, there's going to have to be some angle involved anytime you're moving something off at an angle. And I submit to you that this force here, which I'm just going to call um the force here that's the vertical component of the force is just going to equal F times the sine of theta. And you should know that and be able to look at the triangle and figure that out.

[00:04:51]Okay?

[00:04:53]Once you have this, then you can put that force in here and you'll get the right torque. So, over here, the torque is just going to be R times F times sine of theta. Okay? It's not too hard, you see?

[00:05:08]Because uh basically all you're doing is you're decomposing the force and you're getting the component that matters, which is the component that's acting perpendicular here. And you can kind of convince yourself that any force acting in this direction, okay? So, let's say you move this force down here and you started pulling on on it like this, well, that's not going to generate any torque because it's just going to want to pull the wrench out. You're you're not going to be able to exert any angular force if you're pulling on it like this. You need to be pulling on it up like this and that's why you use the F sine theta. This is the general form for the equation of torque.

[00:05:40]Okay? Anytime there's an angle involved that's not 90° here, perpendicular, then you'll have to use this guy. And then if you just happen to know that 90° is the angle, you can use this one. And never forget that the sine of 90° is just simply equal to one. So, this this equation and this equation are exactly the same. In the case over here where 90° is the angle that you have, well, then this just goes off to one. So, it doesn't even matter.

[00:06:06]So, it's not really and truly two different equations. It's just that this is kind of a special case of this equation, okay?

[00:06:13]The unit of torque is Newton meters, Newton times meter and that's just because you have Newton here and a meter here. So, anytime you write the answer to your problem down, always remember that you're going to be dealing with Newton meters.

[00:06:26]Let's go ahead and work a couple quick problems and see what we can learn.

[00:06:33]First problem is the torque required to loosen a nut it's holding a uh uh flat tire in place on a car and you're trying to loosen this nut to get the tire off has a magnitude of 40 Newton meters. That's the torque.

[00:06:46]That's how much torque you you need to exert. What minimum force must be exerted by the mechanic at the end of a 30 cm wrench to accomplish this task?

[00:06:57]So, what you have here again is you have a wrench.

[00:07:02]Okay?

[00:07:03]And this wrench is 30 cm long, okay?

[00:07:09]And I'm trying to exert a force like this that's going to generate some torque down here, okay?

[00:07:16]And I want to figure out uh what force I need to exert to generate a torque of 40 Newton meters. So, over here, I'm trying to generate a torque of 40 Newton meters.

[00:07:31]Okay?

[00:07:33]40 Newton meters. First thing you know, okay? Is that the torque is just going to be equal to the the distance of the the length of the wrench R times the perpendicular force F exerted.

[00:07:46]Now, in this case with a wrench, you're going to be pushing perpendicular to it.

[00:07:49]That's our assumption here, okay? What you want to do is find this force, okay?

[00:07:53]So, the torque is 40 Newton meters.

[00:07:57]The length of the wrench is 30 cm, but never forget you got to be dealing in the proper units. This is the base unit of torque. The base unit of centimeters is meters. So, I mean, the base unit of length is meters. So, 30 cm you should be able to divide there and figure out is 0.3 m, okay? So, we want to use that number. 0.3 goes in here and then the perpendicular force required if you just divide 40 by 0.3 is just going to be equal to 133 Newtons. This is what you would actually feel yourself having to exert exert out here in order to generate a torque uh that that large. And of course, if you have a longer wrench, then you you this number would get bigger and then you would not have to exert as much force in order to give you the same torque. And that's why longer wrenches are are easier to work with.

[00:08:57]Second problem is a little more interesting. One of my favorite things to do is about fishing. You're out there fishing and you've got a fishing pole and uh the fishing pole is 20° angle, 20° from the ground and a fish is tugging on the fishing pole because you caught a fish.

[00:09:14]Um and how much torque the the question is how much torque does this fish exert on your hand because it's obviously pulling on the line here.

[00:09:24]Okay, so let's take a look at the drawing here. What you have is a fishing pole that's 2 m long inclined at an angle of 20°. You have a fishing line that's at an angle of 37° to the horizontal going into the water with a fish on it and the fish is pulling with 100 Newtons. Question is, you know, this fish is pulling, it's going to cause the pole to bend, it's going to generate a torque around this pivot point here.

[00:09:46]Question is how much torque is it generating?

[00:09:49]Okay?

[00:09:50]Well, first thing you need to do because you know that this is not going to be perpendicular to the pole here is you need to figure out what the perpendicular force is that's acting kind of down perpendicular here. Okay?

[00:10:02]Easiest way to do that is just to realize that if you were to extend this fishing pole out a little bit, let's do that.

[00:10:11]Like this.

[00:10:12]This angle is going to be 20°.

[00:10:14]Okay?

[00:10:16]So, in order to find this guy, which is going to be which is going to be the the force that's acting perpendicular to to my fishing pole here, all I need to do is figure out what this force is here. Now, I've got 100 N over here. I've drawn it here as well. This is the force along the line. What would be this guy here?

[00:10:38]This would be the force here would be 100 N times the sine of this angle here, and this angle here is just 20 + 37, which is 57 degrees. Okay? So, I've got 57° here, and I've got 100 N here. So, the force here is obviously going to be less, and so you have to take the sine of 57, multiply by 100, and that's going to give you that.

[00:11:01]The torque over here is just going to be equal to R, okay, here, which is the distance of the of the fishing pole times F times sine of theta, which is just telling you that it's the force, okay, times the sine of the angle between them so that you get the perpendicular force to the pole there.

[00:11:21]So, the torque is just going to equal to two times 100 times sine of 57.

[00:11:31]Okay? 2 * 100 * sine of 57. These are all numbers, so this torque is just going to equal 168 N meters, and it's obviously going to be acting clockwise because it's pulling it down. That would cause a moment to be generated this way. It's going to be acting clockwise here. Learn anything at mathandscience.com.