Intro to Torque

Concept: Intro to Torque

16m
Video Transcript

Hey guys! In this video we're going to introduce the idea of torque, which is a rotational equivalent of force. Let's check it out. 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. 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 _. When you push on a door, it rotate 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 doesnÕt actually cause spinning. I'm going to say _ = 0. If you try to open a 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. This is the idea that a force causes a torque which causes an acceleration. What you're doing, you're not producing a torque, you're producing a force which then results in a torque on that object. 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. Force may or may not produce a torque and a torque may or may not produce a rotation, an angular acceleration. Let's say if you're pushing this way but then someone else, F2, is pushing this way and these two cancel out. In that case, you 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 _. 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 ma. 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, _. We're going to talk about this a little bit later. 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. 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. It is Frsin_, 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. _ is the angle between these two vectors. See these two vectors right Ð F and r. _ is simply the angle between those two guys. One thing I like to do is I like to think of like an arrow pointing towards these two guys here. That's to remind me that in that equation the _ is the angle between the F and the r. The two guys are hanging out next to the _. _ is the angle between these two guys. The unit for force is Newtons. The units for distance r is meters so a torque is measured in Newton*meter. That's the units for torque. One last point I want to make here is that 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 to the r vector. Perpendicular means 90 degrees. It's got the little perpendicular symbol to the r vector. What does that mean? Let's draw the r vector real quick on this picture here. The r vector is from the axis of 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. This would be a little bit weaker. You have to push harder to get the same rotation. 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. You've all opened doors before. If you tried to open the door by pushing on it let's say 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 what you're supposed to push. ItÕs easiest. You can relate that directly to the equation. 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. You want to maximize sin_. Let me remind you that sin and cos fluctuates between -1 and 1, so it looks like that. The greatest possible value of sin you can have is 1. Where does this happen? This happens when _ theta is 90. Sin90 is 1. That's why if you look at the equation, that's why you get the greatest possible value for torque. Again, your torque is maxed when you push as far away from the edge as possible and when you push perpendicular, make it 90 degree angle with the r vector. Let's do an example. We're going to use three steps to solve all of these torque questions. We're going to draw the r vector. You're going to figure out what your _ is and then you're going to plug numbers into an equation. Here I have five different forces, just to show you all the different variations. You push or pull on a 3-meter wide door so the length of this door here, let me actually just write length = 3 meters, with 10 Newtons in a bunch of different ways. All of these forces are 10. We want to calculate the torque that each force produces on this door and then the rest is just explaining that F1, F4 and F5 all act on the edge of the door right here. In other words they act at a distance of 3 from the axis. This guy is halfway through the door and this guy's at the hinge. I'm going to write in all this information here. I'm going to say this is 1.5 meters. The rest here is 1.5 as well so the whole thing is 3. What else? It says that F5 is directed 60 degrees below the x-axis. If you do this, this guy is 60 degrees right there. I got all the information. Let's do this. Torque1, remember the equation is just F1r1sin_1. The force is 10, but I got to figure out the r and I got to figure out the sin. The first thing you do, here are the three steps: YouÕre going to draw the r vector. The r vector is from the point where the axis is to the point where the force happens. The force happens right here, so the r vector for F1 looks like this. I'm going to say that this is my r1. How long is r1? It's 3 meters. Sin is the angle between r and F. F1 is this way. Let me actually draw this down here. F1 is this way and r1 is this way. The angle between these two guys is simply 90 degrees. I'm going to put it here. Remember, the sine of 90 is 1. I actually need you to remember that. That's gonna make your life easier but if you forget, just check in the calculator real quick. This is gonna be 10*3*1 which is 30 and the unit is Nm. WeÕre going to just four more times. T2 = F2r2sin_2, the force is 10. I like to leave spaces for r in _. What do you think the r is here? All the r vector look like, I want you to draw that real quick and tell me how long that r vector is and I hope you're thinking that it is 1.5 meters because it's just half. r2 looks like this. This is r2. I'm going to draw it down here so I don't make a huge mess. r2 looks like that. The force is F2 and the angle between them is also 90 degrees. We'll talk a little bit more about angles because there's some places where it might get confusing but for now we're just going to keep going. If you multiply everything, the answer is 15 Nm. Right away you see how when you pushed farther which was in the first situation, you got a bigger torque than when you pushed closer to the edge. Let's keep going. What about T3? F3 which is 10 r sin _, what do you think the r vector looks like here and what do you think the length of the vector is? I hope as you're thinking about this, you're thinking well if you push up the hinge, it doesn't really move at all. The torque should be zero. You're not producing anything that causes acceleration, not causing acceleration, and that's correct and that's because the r vector is going to have a length of 0. The r is the distance betweenÉ You can even draw it. It would look like this. This is like your r3, it's a dot because it's from the axis of rotation which is here to where the force acts which is also here so there's no distance because it's the same two points. In that case it doesn't even matter what the sign is. You can't even technically draw a sign because there's no arrows for you to figure out what the angle between them is. It doesn't matter. This is just zero and that's because you act on the axis. Whenever you act on the axis, torque is zero. End of story. What about T4? ItÕs 10rsin_. What do you think the r vector here is and how long is it? The r vector for 4, 4 act over here so the r vector is the same one as this guy here. This is r1 and this is the same as r4. The length of this vector is the entire length of the door which is 3. So far so good. What about the angle? What do you think you put for the angle and do you think there's a torque here? The second question might be easier to answer. If you pull on a door in the direction of F4, the door doesn't open or close. The door doesn't spin. You should expect that the torque is zero and it is actually 0. But the reason why that happens in terms of the equation, you can see this on the equation it's because your r vector looks like this. This is r4 and this is your Force4. The angle between these two guys is zero degrees. They're both going in the same direction. If you got this wrong and you thought that the angle is 180, that's okay. You get the same answer. But the angle is zero degrees and that's because you can put them sort of side by side. You can see how the angle is 0. The angle here is 0 degrees and if you do sin0, the answer is 0. The last one, T5, it's at an angle, the force is 10. How long is the r vector and what is the angle that we use? Again T5 is at a distance of 3 so this vector here, the long arrow is the same for all three of these guys. What about the angle? The angle has to beÉ Be careful here. A lot of torque problems will give you an angle and people will just sort of blindly put a 60 here and sometimes that's wrong. They do that on purpose. You gotta watch out for that. You should actually assume that they're trying to trick you and make sure that you're using the correct angle. How do you know this? I'm going to draw r5 right here. This is where you've slowed down. Make sure you do this correctly. I'm going to draw F5 here and here is the 60. Is that the angle between those two? The technique that I like to use is to make it where the two arrows point from the same dot. I want to have something like this Ð F & r Ð because then it's easy to confirm that this is in fact the angle. To do this, what I do is I shift the arrow around. For example, see this r5 right here? I'm going to move it this way. I haven't changed the direction of the r5. I'm just shifting it from pointing into the dots to coming out of the dot. It's the same thing but the nice thing about this is that it becomes very easy to see that this angle is in fact the one between F5 and r5 because I have the two vectors like this and it's just the angle between them. Sometimes you're going to do a lot of the shifting. Make sure that the angle that they gave you was the correct angle. For example if they had said 30 here, it wouldn't be the correct angle. The correct angle is instead the angle between the r and the F which is 60. That's good. Sin60 goes here and then if you multiply this whole thing you get 26, I'm rounding it, 26 Nm as the answer. It's really important you know how to calculate torques. You should make sure you understood this and you can do all five of these things on your own. This is just the beginning. WeÕre going to a tons more stuff but you gotta make sure you know how to calculate basic torques. Let's keep going.

Example: Torque of fish pulling on pole

5m
Video Transcript

Hey guys! Let's check out this example where weÕre asked to find the torque produced by a force with two different angles here. Here we have a fish catching your bait. Your fishing pole is at an angle of 50 degrees above the x-axis. Let's say you are like this holding on to your fishing pole that has 3 meters long, 2 kilograms and it makes an angle of 50 right there. We want to calculate the torque that's produced on your fishing pole about an axis of rotation on your hands. In other words, think that the fishing pole rotates about the hands right there, so that's the axis of rotation. If the fish pulls on it with 40 N, 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 directed at 20 degrees below the x-axis. Here's the positive x-axis. 20 degrees below the x-axis looks like. I'm going to do -20 over here and the fish is pulling with the force of 40 N. I want to know how much does this force, how much of a torque does this force produce on this axis right here. What weÕre gonna do is write the torque equation, T = Frsin_. Remember the steps. The first thing I'm going to do is draw our r vector then we're going to figure out what _ is and then we're going to plug it into the equation. 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 with the forces applied on the object. The axis of rotation is here, the force pulls right there. We draw an arrow, this is our r vector. The length of the R vector is 3 meters, so that's what I'm going to put here, so 40*3 sin_. 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 says 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 you which is the hardest part. You've got to make sure you're using the right angle. Here, should we use 50, should we use 20, should we use -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 two. I'm going to remind you that _ is the angle between F and r and to figure out what 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. 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. Let me draw this again. 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. You see here how there is a 50 between the r in the x-axis. Right here there's a 50 between r and the x-axis, so this angle here between r and the x-axis is also a 50. This 50 gets transferred over here. I hope you see that the total angle between r and F is actually 70 degrees. You're supposed to add up those two guys. The entire purpose of this question was to look into how to solve questions with non-trivial _ values, angle values, and how to figure out the correct angle to use. If you multiply all this, you get 113 Nm and that's your final answer. That's it for this one. Let me know if you have any questions. Let's keep going.

Problem: A 4 m-long ladder rests horizontally on a flat surface. You try to lift it up by pulling on the left end of the ladder with a force of 50 N that makes an angle of 37° with the vertical axis. Calculate the torque that your force produces, about an axis through the other (right) end of the ladder.

5m

Example: Maximum torque on wrench

3m
Video Transcript

Hey guys! In this torque example, we're trying to figure out what's the minimum force needed to produce a certain torque. Another way to think about this is how to maximize your torque with the least amount of force, how to be the most effective in causing something to spin. It says here you must produce a torque of 100, T = 100, to properly tighten a given bolt using a 20-centimeter wrench. 20 centimeter is the length of the wrench, so L = 0.2 m. I want to know what is the minimum force needed to do this. This is just a play on the torque equation. We're going to write T = Frsin_. The torque value is fixed, it's 100. I want to know what is my force. To do this, I have to figure out which values to plug in for r and _. Remember, if you want the most effectiveness with your torque, with your efforts, you want the greatest possible force. But here we're trying to find the least amount of force which means that I want to have the most effective r and the most effective _. Let me draw a little wrench here. What you want to do to produce a torque is you want to push as far away from the axis. The wrench, the bolt is here right so we're spinning like that. You want to push as far away from the wrench. Think about this. If you got the wrench and you're pulling here, it doesn't do as much. If you pull the wrench all the way at the end, all the way over here, that's why you grab it here you produce the most torque. You want to be as far as possible from the axis of rotation. That's where this 20 centimeter comes in. If the wrench is 20 centimeters long, you want to be as far as possible. We're going to assume that we can push it at the very tip, therefore we're going to use an r of 0.2. Also, not only do you want to push all the way over here at the end, you want to push with an angle of 90 degrees to the r vector. If you push it here, your r vector is axis to point where the force is applied, r and F. You want to make an angle of 90 degrees. In all of these questions where you want to maximize torque or minimize force, you want your r to be as long as possible which usually means the entire length of the thing. You want your _ to be as big as possible which usually means an angle of 90 to give you a sine of 1 and then that's it. This gives us F*.02*1, so force equals 100 divided by 0.2, the force here would be 500 N. That's it for this one. Let me know if you have any questions.

Problem: 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.

4m

Intro to Torque Additional Practice Problems

A wheel is rotating clockwise on a fixed axis perpendicular to the page (x). A torque that causes the wheel to slow down is best represented by the vector

A. 1

B. 2

C. 3

D. 4

E. 5

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A gasoline engine produces power and torque as shown in the following figure. In a direct drive configuration, the wheels of a car spin with the same angular speed as the engine does. If the wheels have a diameter of 45 cm, how fast would the car have to be moving in order for the engine to be producing maximum power? What about maximum torque? 

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In cars, a gear train (shown in the figure) is used in the transmission to allow the wheels of the car to rotate at high angular speeds while at low engine rpm's, allowing for less gas to be used. If the drive gear in a gear train is connected directly the engine, and the driven gear directly to the wheels (meaning both rotate with their respective gears), how much torque would be delivered to the wheels when the torque produced by the engine is 100 Nm and the drive gear is 3 times larger than the driven gear?

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In the figure, a uniform rectangular crate 0.40 m wide and 1.0 m tall rests on a horizontal surface. The crate weighs 930 N, and its center of gravity is at its geometric center. A horizontal force F is applied at a distance h above the floor. If h = 0.61 m, what minimum value of F is required to make the crate start to tip over? Static friction is large enough that the crate does not start to slide.

A) 763 N

B) 688 N

C) 305 N

D) 413 N

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A wrench is acting on a nut. The nut is at the origin and the wrench extends from the nut along the +x-axis. A force of 150 N acts on the wrench at the position x = 15.0 cm at an angle of 30.0°. What is the torque the wrench exerts on the nut?

A) 1949 N•m

B) 11.3 N•m

C) 19.5 N•m

D) 2250 N•m

E) 22.5 N•m

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Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?

A) both exert equal non-zero torques

B) the first at the midpoint

C) both exerts zero torques

D) the second at the doorknob

E) additional information is needed

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A uniform bar (I = 1/3 MLfor an axis at one end) has mass M = 5.00 kg and length L = 6.00 m. The lower end of the bar is attached to the wall by a frictionless hinge. The bar is held stationary at an angle of 60° above the horizontal by a cable that runs from the upper end of the bar to the wall. The cable makes an angle of 37° with the bar. For an axis at the hinge, what is the torque due to the weight of the bar?

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