Hey guys. So in this video, we're going to introduce the idea of torque, which is the rotational equivalent of force. Let's check it out. Alright. So you can think of torque as a twist that a force gives an object around an axis of rotation. Here's the most classic example. If you have a door that is fixed around an axis here, this is the hinge of the door which is also its axis of rotation, meaning the door is free to rotate around the axis. And if you push this way with a force f, it causes the door to spin this way. I'm going to say that the door accelerates in that direction. It gains an alpha. So when you push on a door, it rotates around its hinges. More generally speaking, when a force acts on an object as it does here, away from its axis, it produces a torque on it. Let's talk about these two parts, away from the axis. If you push on a door here, it doesn't actually cause spinning. I'm going to say alpha equals 0. Right? If you try to open the door by pushing the hinge, it doesn't spin. You have to be away from the axis of rotation. And then it produces a torque. So this is the idea that a force causes a torque which causes an acceleration. So what you're doing, you're not producing a torque, you're producing a force which then results in a torque on that object. So now the other point is that a force may produce a torque. We already talked about how if you push here, it doesn't cause it to rotate so you don't produce a torque. So force may or may not produce a torque and a torque may or may not produce a rotation, an angular acceleration. And that's, let's say if you're pushing this way, but then someone else, F2 is pushing this way and these 2 cancel out. In that case, it wouldn't really produce them. But the most important point I want to make is that you have f produces a t, which produces an alpha. That's the sequence. We just talked about this. Let's fill it in here. Similar to forces cause linear acceleration, remember, sum of all forces equals m a. Sum of all forces, as long as you have a net force, you're going to have an acceleration. It's the same thing with torques. Torques cause angular acceleration. We're not going to talk about that just yet, alpha. We're going to talk about this a little bit later. Now another difference between torque and force is that force is a straightforward number. If I tell you we push with 10, that's the end of it. But for torques, torque depends on how hard you push, it depends on how far you push, and some other stuff. So, we actually have an equation for torque. You don't have an equation for force. You're given a force, but for torque, you have to calculate it. And it is frsinθ, where f is the force you push with, it's a vector. R is a vector. It says here r is a vector from the axis of rotation to the point where the force is applied. Remember, little r in most rotation problems has to do with distance from the center. It's the same thing here. Now theta is the angle between these two vectors. See these two vectors right here, f and r? Theta is simply the angle between those two guys. f and r are meant to put an f and an r. And one thing I like to do is I like to think of like an arrow pointing towards these two guys here. Right? And that's to remind me that in that equation, the theta is the angle between the f and the r. The 2 guys are hanging out next to the theta. Okay? Theta is the angle between these two guys. Now the unit for force is newtons. The newton for the units for distance r is meters. So torque is measured in newton times meter. Okay? That's the units for torque. One last point that I want to make here is that when we need to maximize the torque, to get the most possible torque, another way to think about this is the way to apply the least amount of force and get the most amount of results to be the most efficient with causing something to rotate is to apply the force as far as possible and perpendicular and apply the force perpendicular, perpendicular to the r vector. Perpendicular means 90 degrees. It's got the little perpendicular symbol, to the r vector. So what does that mean? So let's draw the r vector real quick on this picture here. The r vector is from the axis of the rotation, which is here, to the point where the force is applied, which is over here. This is the r vector. You want your force to make this is your force. You want your force to make 90 degrees with the vector, which in this case, it does. That's how you get the maximum torque, the easiest way to rotate. Imagine if you're instead pushing the door this way, right? This would be a little bit weaker. You have to push harder to get the same rotation. So you want to push at an angle of 90 degrees. The other thing is that you want to push as far as possible away from the axis of rotation. So you've all opened doors before. If you try to open the door by pushing on it, let's say right here, I'm going to make a little bit of a mess, but I'm going to erase this. If you push right here, it's much harder to open the door here than to open the door at the end. In fact, that's why the doorknob is on the opposite side of the hinge because that's where you're supposed to push. It's easiest. So, and you can relate that directly to the equation, right? So if you want the torque to be as high as possible, you obviously want to push as much as hard as you can, as far away or as distance from the axis, as far away as you can. And you want to maximize sine of theta. Now let me remind you that sine and cosine fluctuate between negative one and one. Right? So it looks like that. So the greatest possible value of sine you can have is 1. Now where does this happen? This happens when theta is 0 I'm sorry, when theta is 90. Sine of 90 is 1. That's why if you look at the equation, that's why you get the greatest possible value for torque. Okay? So again, your torque is maxed when you push as far away from the edge as possible and when you push perpendicular, make it a 90 degree angle with the r vector. Alright? Now let's do an exam

# Intro to Torque - Online Tutor, Practice Problems & Exam Prep

### Intro to Torque

#### Video transcript

### Torque of fish pulling on pole

#### Video transcript

Hey guys, let's check out this example where we are asked to find the torque produced by a force with 2 different angles. So here we have a fish catching your bait. Your fishing pole is at an angle of 50 degrees above the x-axis. So, let's say you are like this, holding onto your fishing pole that is 3 meters long, weighs 2 kilograms, and it makes an angle of 50 degrees right there. We want to calculate the torque that's produced on your fishing pole about an axis of rotation in your hands. In other words, think that the fishing pole rotates about the hands right there. So that's the axis of rotation, right here. If the fish pulls on it with 40 newtons, that's a force directed at 20 degrees below the x-axis. The line is here. You're pulling from here. The fish is pulling at an angle direct at 20 degrees below the x-axis. So here's the positive x-axis. 20 degrees below the x-axis looks like this. I'm going to do negative 20 over here. And the fish is pulling with a force of 40 newtons. We want to know how much of a torque this force produces on this axis right here. So what we're going to do is to write the torque equation, torque equals \(fr \sin(\theta)\). Now remember the steps. The first thing we're going to do is to write the, draw our \(r\) vector. Then we're going to figure out what \(\theta\) is, and then we're going to plug it into the equation. Okay? So what is our \(r\) vector? \(R\) vector is an arrow. It's a vector from the axis of rotation to the point where the force happens, where the force is applied on the object. The axis of rotation is here. The force pulls right there. We draw an arrow. This is our \(r\) vector. Now, the length of the \(r\) vector is 3 meters. So that's what I'm going to put here. \(40 \times 3 \times \sin(\theta)\). Now, drawing the \(r\) vector is important, actually not so much so that you can figure out how long it is because you could have just looked at this pole and said the pole is 3 meters long. That's the answer. What's really important about drawing the \(r\) vector is so that you can figure out what angle to use, which is the hardest part. You've got to make sure you're using the right angle. So here, should we use 50? Should we use 20? Should we use negative 20? It actually turns out that in this problem, it's none of these options. It's a different angle. It's a combination of the 2. I want to remind you that \(\theta\) is the angle between \(f\) and \(r\). And to figure out which angle goes between them, the technique I like to use is to try to get \(f\) and \(r\) to be pointing from the same common point, something like this. And then it's just a matter of finding this angle right here. To do this, I'm going to shift \(f\) around or shift \(r\) around so that they start from the same point. So what I'm going to do here is I'm going to push, I'm going to shift my \(r\) up so that they both start from this point. So let me draw this again. I have 50. Let me draw this a little lower. I have 50 over here, and then I have 20 here. What I'm going to do is I'm going to get the \(r\) vector and push it over here so that I have \(r\) and \(f\). And then the angle I need is the angle between these two guys. Now, you see here how there is a 50 between the \(r\) and the x-axis, right? Right here, there's a 50 gets transferred over here. And now I hope you see that the total angle between \(r\) and \(f\) is actually 70 degrees. So you're supposed to add up those 2 guys. This the entire purpose of this question was to look into how to solve questions with nontrivial, \(\theta\) values, angle values, and how to figure out the correct angle to use. If you multiply all of this, you get a 113 newtons-meter, and that's your final answer. Cool. That's it for this one. Let me know if you have any questions. Let's keep going.

### Example 3

#### Video transcript

Hey, everyone. So let's check out this practice problem. We're trying to calculate the minimum amount of force necessary to produce 100 newton meters of torque using a 20 centimeter wrench. I've got my torque (τ) equal to 100 and the length of the wrench is 0.2 meters. Right? That's the length of this wrench over here. And I want to calculate the minimum force (f_{min}). Right? That's the minimum amount of force necessary. So how do we do that? Well, I've got forces and torques. This is really just a play on the torque equation, so let's set it up.

So I've got torque equation as τ=fr*sin(θ). Here's what's going on in this problem. If I want to figure out the least amount of force necessary to produce 100 newton meters, really another way of phrasing it is I want to find the maximum torque possible using the least amount of force, right? You want to grab the wrench and using the least amount of force, produce the most efficient torque possible. Okay? That's really what's going on here. So I've got my torque equal to 100 Newton meters. That's what I need in order to tighten this bolt. So this is f_{min} over here.

And then what am I going to plug in for r and then sin of theta? So really, if I want to minimize this number here, then I want these numbers to be the maximum they possibly can be. If these values are the greatest they possibly can be, then this number can be the smallest number and still get to 100 Newton meters. So, I want these two things to be max. Alright, so what do we plug in for r and sin(θ)? If you grab this wrench here, you imagine that there's a bolt right here. So, what is the distance in which you can maximize your torque? Well, it's basically the farthest you can possibly go from the axis of rotation. In other words, you want your r to be as far as possible, so you want basically to pull right here at the end of the wrench. If you grab in the middle, it's not going to turn as much, but you grab all the way at the end, that's where you get the most amount of torque. It's kind of like if you push a door. You don't want to push the door in the center; you want to push it all the way at the end to get the most amount of push.

So, this 'r' really just becomes 'l', the length of the wrench itself. So r equals l, and that's 0.2. So that's what goes inside of here. Then, how do we maximize sin(θ)? What's the angle at which we push it? Well, hopefully, you guys remember that the sine of an angle ranges between 0 and 1, and the maximum possible value is when you have sin of 90 degrees. So, what this means is that when you're at the end of the wrench here, you don't want to pull off in one direction or another. You basically want to pull off exactly at 90 degrees. That's the angle. If you pull at any other angle here, you're actually going to produce less force or less torque. So, that's what happens is you just have the sin of 90°.

So then, let's go ahead and plug everything in. So we've got fmin*0.2*1=τ, which equals 100. So, 100 divided by 0.2, that's your f_{min}. And if you work this out, what you're going to get is 500 Newtons. Alright, and that's the answer. Hopefully that makes sense. Let me know if you have any questions.

You pull with a 100 N at the edge of a 25 cm long wrench, to tighten a bolt (gold), as shown. The angle shown is 53° . Calculate the torque your force produces on the wrench, about an axis perpendicular to it and through the bolt.

9.9 N•m

990 N•m

15 N•m

1500 N•m

20 N•m

2000 N•m

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