TXT

# 11.Torque.SMI

By Irene Edwards,2014-05-19 18:22
7 views 0
11.Torque.SMI

00:00:22 problems where τ is not zero.

00:00:25 There is some object on which some torques are acting and in responds 00:00:29 to the torques-- in response to the torques,

00:00:34 the body will accelerate.

00:00:36 Then, we went to another class of problems where there were no 00:00:41 external torques, there are only torques from inside. 00:00:45 Excuse me, will you people who come late use the other door? 00:00:48 Yeah, you're blocking actually the taping right now. 00:00:51 Yes.

00:00:52 All right.

00:00:54 So, today, I'm going to consider an extreme case where there 00:01:01 is no torque at all.

00:01:04 If there's no torque, we know α is zero and the

00:01:08 angular velocity is constant.

00:01:09 But I'm going to take a case where even angular velocity is zero. 00:01:14 There is no motion; there is no torque.

00:01:16 So you can say, "You know, what's there to study?" 00:01:18 Well, sometimes it's of great interest to us that the object has 00:01:22 no angular velocity.

00:01:24 A common example is, you are trying to do something in 00:01:28 your house, you get up on a ladder, that's somewhere there. 00:01:33 This ladder should have no angular velocity.

00:01:36 Right?

00:01:37 That's not a trivial statement to a person who is going up there. 00:01:41 So you want to ask, "What does it take to keep the 00:01:45 ladder from falling over?"

00:01:47 Okay, that's what we're going to discuss.

00:01:52 All right.

00:01:53 So, let me start with a simple problem and take some horizontal 00:01:56 rod like this.

00:01:58 And I say, "What does it take to keep the rod from moving around?" 00:02:03 And the first thing, you know, is the forces on it should 00:02:06 add up to zero because otherwise F = ma tells you this rod's 00:02:10 going to be flying off somewhere.

00:02:11 So, I've taken a body, a rod of length L,

00:02:14 and I don't want it to move at all.

00:02:15 So, total forces adding up to zero we know is not good enough because 00:02:20 here is the problem where I apply two forces equal and opposite, 00:02:25 add up to zero, but that doesn't do it.

00:02:29 This rod will still rotate.

00:02:31 It is still true that when the total forces add up to zero, 00:02:35 the center of mass will not accelerate.

00:02:38 But that doesn't mean the body will not move because keeping the center 00:02:41 of mass fixed, the body can rotate.

00:02:46 All right.

00:02:47 So, you have to ask yourself, "What does it take to keep it from 00:02:50 rotating?"

00:02:51 Then we know that the cause of rotation is the torque. 00:02:54 So, I'm going to write down now the conditions for a body, 00:02:58 a planar body lying in the plane of the blackboard, to stay 00:03:02 still and not move.

00:03:04 They are that all the forces in the x direction. 00:03:10 i is some label, force number 1, force number 2. 00:03:13 That should add up to zero.

00:03:14 All the forces in the y direction;

00:03:17 that should add up to zero.

00:03:19 This will make sure the center of mass as a whole is not 00:03:21 going anywhere.

00:03:22 And finally, to keep the body from moving or spinning, 00:03:26 all the torques should add up to zero.

00:03:29 So, bulk of the lecture today will just be dealing with this condition. 00:03:33 So, what we're going to study today is a collection of objects that are 00:03:39 somehow subject to a variety of forces but still in equilibrium, 00:03:44 in static equilibrium.

00:03:45 That means they are not moving.

00:03:47 So, simplest problem from the family is a seesaw. 00:03:53 Let's say that's a point where the seesaw is supported and let's say we 00:03:58 put one kid here applying some force F1.

00:04:02 So F1 is the mg for the kid.

00:04:07 And let's say this end is a distance x1

00:04:10 from the support.

00:04:12 The question is, "How much of a kid should I put here 00:04:15 to keep this in balance?"

00:04:20 It should not move under the combined weight of these two kids. 00:04:24 So, I take the body and I write down all the forces on it. 00:04:30 So, I have F1 coming down here.

00:04:34 I have F2 there.

00:04:38 Then there is-- Of course, that cannot be the whole 00:04:41 story because the seesaw is not going into the ground. 00:04:44 That is the support and the normal force from the support. 00:04:49 And these three forces together should conspire to keep the 00:04:53 body in equilibrium.

00:04:54 So let's see.

00:04:56 So, here is what I want you to understand.

00:04:58 F1 is known.

00:05:00 Let's give it some name, 100 Newtons.

00:05:02 This distance x1 is also known.

00:05:06 That distance x2 is also known.

00:05:08 I'm trying to find out what force F2 is needed. 00:05:12 So, F2 is what I'm trying to solve for.

00:05:16 So, I write all the equations I know.

00:05:18 So, horizontally, there's nothing going on.

00:05:20 So, it's zero equals zero.

00:05:21 In the vertical direction, I have F1 pushing

00:05:26 down, F2 pushing down equals N.

00:05:31 And that's not going to be enough to solve this problem because 00:05:35 I have two unknowns.

00:05:37 I don't know N and I don't know F2.

00:05:39 I just know this guy.

00:05:41 So, that's where the torque equation's going to come in. 00:05:46 Now, here is the subtlety when you do the torque in this problem. 00:05:51 So, how do you find the torque due to any force? 00:05:55 Can you tell me what you're supposed to do?

00:05:58 Yeah?

00:06:00 Yes, you, yes. Student: [inaudible].

00:06:05 Prof: Right.

00:06:07 What r should I use for this force?

00:06:09 Student: [inaudible]. Prof: Pardon me?

00:06:14 x1?

00:06:16 Okay, why did you choose that? Student: [inaudible]. 00:06:22 Prof: Okay.

00:06:25 Her answer was, "That's the pivot so I chose that 00:06:27 distance."

00:06:28 The point is, when we say the body has no net torque, 00:06:32 which means it's not rotating around some axis. 00:06:36 If there is an axis around which a body is rotating, 00:06:39 then we know how to take the torques through that axis. 00:06:43 When a body is not rotating, it's not rotating around any axis 00:06:46 whatsoever.

00:06:48 So, you can say it's not rotating through an axis I draw through here. 00:06:54 So, I will take all the torque with respect to that point. 00:06:56 Someone else can say, "No, it's not rotating through that 00:06:58 either.

00:06:59 I will take all torques from that point."

00:07:02 If you change your mind on where you're taking the torque, 00:07:05 then the torques are all changing because in the force times distance 00:07:09 the distances are going to change.

00:07:10 So, now we have a predicament where for every possible point around 00:07:14 which I compute a torque, I will get a condition involving the 00:07:18 different forces and the distances from that point. 00:07:22 So, I can write down an infinite number of equations for each 00:07:26 different choice of pivot point.

00:07:28 But I only have two unknowns, F2 and N.

00:07:32 Two unknowns can only satisfy two equations, not 2,000 equations. 00:07:38 So, we've got to hope that the extra equations I get by varying the pivot 00:07:42 point all say the same thing.

00:07:45 And that's what I will show you.

00:07:46 I will show you, first, that if you have a bunch of 00:07:51 forces that add up to zero, then it doesn't matter around which 00:07:56 point you compute the torque.

00:07:59 If it was zero around one point, then it will be zero around any 00:08:02 other point.

00:08:04 Suppose I pick some point here and I take each force, 00:08:09 multiply the distance of that force from that pivot point and 00:08:13 they added up to zero.

00:08:15 Now you come along and say, "No, I want to take the torque 00:08:19 around that point."

00:08:21 What you will be computing will be the same force, 00:08:24 but to every xi you will add a distance a

00:08:28 where a is the distance by which you moved your axis. 00:08:31 Now, let's open this out.

00:08:36 You'll find it is Fixi +

00:08:39 a times sum of all the Fis.

00:08:47 This is zero because for me the torque around that point vanished 00:08:52 and this is zero because all the forces add up to zero. 00:08:56 In other words, if you find any one point and make 00:08:59 sure the torques around that point don't cause--that they 00:09:02 all add up to zero, then the torque around any other 00:09:06 point will also add up to zero, provided the sum of the forces adds 00:09:10 up to zero.

00:09:12 And of course, that's the problem we're looking

00:09:13 at.

00:09:14 So, you've got to realize that when you take the torque, 00:09:18 you may pick any point you like.

00:09:20 Now, that's where a strategic issue comes up.

00:09:23 If you were asked simply to find what F2 is,

00:09:27 not to completely solve the problem,

00:09:31 which would be also finding what the normal force is-- 00:09:34 if someone said, "Just find F2,"

00:09:36 I don't care what N is" then, there's a particular choice of pivot 00:09:41 point that is optimal.

00:09:43 That choice is the one in which this normal force doesn't get to enter 00:09:47 the torque equation and I think you all know what I mean. 00:09:52 Pick that point as the point around which you take the torques; 00:09:55 then N drops out.

00:09:57 So, this is the usual model when you take torques.

00:10:02 Always take the torque through a point where an unknown force is 00:10:06 acting because the unknown force doesn't contribute to 00:10:10 the torque equation.

00:10:12 So in principle, you can take the torque around any 00:10:15 point and you can satisfy yourself in your spare time; 00:10:17 it doesn't matter which point you take.

00:10:19 It will be the more complicated way to solve the problem because 00:10:22 N will come into the picture and no one asked you 00:10:25 what N is.

00:10:26 So, we take the point of support and then, of course here is 00:10:31 the point of support, and what did I have,

00:10:34 F1 here a distance x1,

00:10:37 I had F2 at a distance x2.

00:10:40 This is the point around which I'm doing the torque, so N 00:10:43 doesn't count.

00:10:44 Then, you get a condition F1x1 =

00:10:48 F2x2.

00:10:51 From that you can solve for F2.

00:10:53 For example, if this was one meter and this was six meters, 00:10:58 and this was say 10 Newtons here, then 10 times one has to be six 00:11:02 times F2.

00:11:05 So, 10 = 6F2 and F2 will

00:11:10 be 10/6 Newtons.

00:11:12 I don't want to use N because N is also standing 00:11:17 for the force here.

00:11:19 That's how we find F2.

00:11:21 Once you've found F2,

00:11:24 in case you want to know what is the support going through and how much 00:11:27 load is it carrying, you can come back now and add 00:11:30 F1, which was given to you,

00:11:34 and the F2 you just solved for.

00:11:36 Then if you want to, you can find N.

00:11:39 Even if you were asked to find N, it's best to first take 00:11:43 torque around a point N so N doesn't enter,

00:11:47 find F2, then it's very trivial to add

00:11:49 F1 and F2

00:11:51 to get N.

00:11:51 This is the easiest prototype.

00:11:53 So, you guys have to know how this is done.

00:11:55 Everything I do will be more and more bells and whistles on the 00:11:58 simple notion of how to take the torque.

00:12:01 All right.

00:12:03 So now, I'm going to take a slightly more complicated problem. 00:12:07 So, here is a wall and there is a rod here of length L

00:12:13 and mass M.

00:12:16 And it's supported by some kind of pivot on the wall. 00:12:20 And I want to hold it up by supporting it here with 00:12:24 a force F.

00:12:26 And the question is, "What force should I apply?" 00:12:33 Okay.

00:12:34 So, we know that if I draw the free body diagram of the rod, I 00:12:40 have my force F; I have gravity acting down

00:12:45 everywhere on the rod, then the pivot, namely the wall, 00:12:49 it can exert a certain vertical force.

00:12:53 It can also exert a horizontal force.

00:12:56 No one tells you the wall is frictionless or anything or the 00:12:59 pivot is frictionless.

00:13:00 So, the wall is exerting in general a force, which is at some oblique 00:13:04 angle, which I've broken up conveniently into horizontal 00:13:08 and vertical parts.

00:13:12 And torque due to gravity is what I want to discuss now. 00:13:19 First of all, I am only asking you to find the force I have to apply to 00:13:23 support the rod.

00:13:25 So, just like the other problem, I'm not asking you to find v 00:13:27 and h.

00:13:29 So the trick, once more, is to jump straight to the torque 00:13:33 equation.

00:13:35 Forget about the force equation.

00:13:37 Go to the torque equation and take the torque around this point, 00:13:40 then this guy is out.

00:13:41 No matter what's going on -- doesn't matter -- the torque due to 00:13:44 the pivot is out.

00:13:46 So, we have to equate the torque due to the force I apply to the torque 00:13:51 due to gravity pulling the rod down.

00:13:53 Torque due to force I apply; that torque, is equal to F 00:13:58 times the length of the rod because there's F times r times the 00:14:03 sine of the angle between them, and the sine of the angle is 9, 00:14:07 the sine of 90, which is one.

00:14:08 How about the torque due to the pull of gravity? 00:14:12 Anybody know what we're supposed to do here?

00:14:14 Yeah? Student: [inaudible].

00:14:20 Prof: Okay.

00:14:23 Now, why is that?

00:14:24 Why is it okay to replace the torque by the center of mass? 00:14:29 Yeah?

00:14:31 Either of you can answer, yes. Student: [inaudible]. 00:14:35 Prof: Right.

00:14:38 But why am I allowed to pretend?

00:14:40 See, we can pretend a lot but it occasionally has to correspond to 00:14:45 reality so that it works.

00:14:47 So, why is it okay to pretend?

00:14:49 That's a correct answer. Student: [inaudible].

00:14:52 Prof: Right.

00:14:55 But if I really wanted to prove it, somebody has a way to really nail it 00:14:58 down and show that it's okay?

00:14:59 Yes? Student: [inaudible].

00:15:03 Prof: Thank you.

00:15:05 Yes.

00:15:08 I know where you're going.

00:15:11 Look, what you want to do is you always have to go back 00:15:14 to first principles.

00:15:15 You cannot take anything on faith because you picked it up on the 00:15:18 street or someone whispered this to you.

00:15:20 That's not good enough.

00:15:22 One lesson I want to show you throughout the lectures is, 00:15:26 I cannot invoke new laws or new principles.

00:15:30 The only principle we know is F = ma.

00:15:32 You've got to go back to that all the time.

00:15:35 Of course, there is the other principle that gravity is an extra 00:15:37 force on the body.

00:15:39 But this is an external-- This is an extended object. 00:15:42 It's got a center of mass but gravity is not necessarily 00:15:44 acting there.

00:15:46 It's pulling each part of it down.

00:15:47 So, I want to show you that in the end of course the answer is as if 00:15:51 all the mass were concentrated here for the purposes of torques but I 00:15:54 want to show you why.

00:15:56 So, the trick proposed by this gentleman here--

00:15:59 Imagine dividing the rod into tiny pieces,

00:16:02 each of which is small enough to say it has a definite location 00:16:04 xi.

00:16:09 The torque, due to gravity acting on that little fellow-- 00:16:12 It's the mass of that little guy times xi

00:16:15 times g.

00:16:17 If mg is the force, then mg times

00:16:19 xi is the torque.

00:16:20 You want to sum it over all the little pieces.

00:16:23 Sum over i is summing over all the little pieces. 00:16:26 So, let me make my life easy by dividing by the total mass and

00:16:30 multiplying by the total mass.

00:16:32 If I started this combination mixi

00:16:34 summed over I divided by M, that is the

00:16:37 center of mass.

00:16:38 So, it does look like Mgx, where x is the center of mass 00:16:43 of the rod.

00:16:45 Now, the rod really doesn't have to be uniform for me to say this. 00:16:51 It doesn't have to be a uniform rod.

00:16:52 The mass can be distributed any way it likes. 00:16:54 Each mi need not be the same.

00:16:56 But it's very important that gravity be constant. 00:16:59 We demand that the force of gravity have a constant value over 00:17:03 the length of the rod.

00:17:05 And it's not obvious that's always true.

00:17:07 I mean, if I took a tiny rod, force of gravity is constant--took a 00:17:10 very big rod, comparable to the size of the Earth, 00:17:13 of course the pull of gravity is varying on the length of the rod. 00:17:17 But for all rods on the laboratory scale, that's not an issue. 00:17:22 And that's where this result comes from.

00:17:25 Okay.

00:17:26 So, now that we know where the result comes from, 00:17:28 we will now apply it in the torque equation. 00:17:30 This is the torque due to the force I apply Fx. 00:17:34 That's equal to Mg times x, where x is the

00:17:38 location of the center of mass.

00:17:41 That is going to be half the length of the rod. 00:17:44 I'm sorry-- this Fx is FL.

00:17:48 We cancel the L, then you find the force I

00:17:54 apply is Mg divided by two.

00:17:57 Yeah? Student: [inaudible].

00:18:01 Prof: If the pivot is exerting what force?

00:18:08 Student: [inaudible]. Prof: Yeah.

00:18:10 But it won't cause a torque. Student: [inaudible]. 00:18:19 Prof: I am finding--no.

00:18:23 Here, what I'm calculating is not the net force. 00:18:26 I'm finding the torque around this point due to the pull of gravity on

00:18:29 the different segments that make up the rod. 00:18:33 Student: [inaudible]. Prof: Oh.

00:18:41 But the pull of gravity is straight down and the separation 00:18:44 is horizontal.

00:18:46 Is that the question now? Student: I want to know about 00:18:50 the force you're applying, not the gravity. 00:18:52 Prof: Well, I'm--oh, the force I am applying. 00:18:54 Oh, I am applying the force here, straight up.

00:18:57 Oh, you want to know-- Student: [inaudible].

00:19:01 Prof: Ah.

00:19:04 Well, you're quite right.

00:19:08 You're saying we don't know a priori whether the force I'm 00:19:11 applying is vertical or horizontal.

00:19:12 Correct?

00:19:15 That's actually a valid question.

00:19:17 It's clear that I have to apply vertical force and possibly 00:19:21 horizontal force to compensate.

00:19:23 This one, if there is a horizontal force in the wall. 00:19:26 You're quite right.

00:19:27 Absolutely correct.

00:19:28 So, this F-- I should also divide the force that 00:19:31 I apply into a vertical part I call F and a horizontal part that 00:19:34 I call H prime.

00:19:36 Then, it'll turn that in this problem an H prime 00:19:40 is not needed.

00:19:42 You can sort of tell intuitively if you're holding up a plank, you know.

00:19:46 One end is anchored and you're trying to support it; 00:19:48 it's enough to apply a vertical force.

00:19:50 But I don't have to presume that.

00:19:52 So, if you want, you can think of applying horizontal 00:19:54 force.

00:19:55 But it's not necessary to hold this in equilibrium. 00:19:58 Maybe the main point is, a purely vertical force will suffice 00:20:01 to keep this in equilibrium, is the point that I'm making. 00:20:05 In fact, if that force is equal to Mg/2, it won't rotate. 00:20:09 Now, we can go back-- now that I find the force I have to 00:20:13 apply in this direction, what force is at the pivot? 00:20:18 Well, the vertical force on the pivot is pointing up. 00:20:21 My force that I calculated is pointing up.

00:20:24 And together they should add up to Mg.

00:20:28 Now, F = Mg/2 so the vertical force will be Mg/2. 00:20:37 Now, horizontal force-- In this problem, there is no need 00:20:40 for you to apply the horizontal force and there's no need for the 00:20:44 pivot to fight you back with the horizontal force. 00:20:47 It's not needed for equilibrium, but supposed you insisted. 00:20:50 Not only are we lifting it but we are leaning into the rod. 00:20:53 Yes, then, you will apply horizontal force.

00:20:55 It's not going to contribute to the torque equation, 00:20:58 but if you push it in, wall's got to push you out, 00:21:00 and H prime will be equal to H.

00:21:02 In fact, let's turn to a problem where the support,

00:21:07 in fact, exerts a horizontal and vertical force.

00:21:11 So, that's the next level of difficulty.

00:21:14 This is one of the standard problems in the section and 00:21:18 it goes like this.

00:21:19 Here is a rod and it's supported by a wire.

00:21:27 And the wire has a certain tension T.

00:21:31 It is anchored to the wall; the length of the rod is L, 00:21:34 the mass of the rod is M.

00:21:36 And at the--This end is some pivot, which is exerting some unknown force, 00:21:40 which can have a horizontal and vertical part.

00:21:44 We'll have a horizontal part H and a vertical

00:21:47 part V.

00:21:49 Now, I know for sure there's a horizontal part.

00:21:51 I think you can tell-- can you guys tell in your mind where 00:21:55 there's a horizontal part?

00:21:56 This force that's supporting it is pushing the rod into the wall and 00:22:00 also pulling it up.

00:22:01 The part that's pushing it to the left has to be compensated 00:22:04 by H.

00:22:05 But suppose the question they ask you is, "What is the tension 00:22:09 on that rope?"

00:22:11 And suppose that's the crucial issue because if the tension is too much 00:22:15 the rope is going to break.

00:22:17 If you want only the tension, once again the trick is, forget the 00:22:21 force equation and go to the torque equation because, 00:22:25 if you go to the torque equation, you banish both V and 00:22:27 H.

00:22:27 I hope that's clear to all of you.

00:22:29 A force going through the pivot point in any direction is incapable 00:22:33 of producing any torque.

00:22:35 That's so intuitively clear, we need not belabor that. 00:22:38 If you want the formula in F times r times sin θ,

00:22:41 r is zero.

00:22:42 So, we don't care what the angle is.

00:22:44 So, once again, we'll take torques around this point 00:22:48 and see what it will tell us.

00:22:49 So, the torque due to gravity, we all agreed, we can now replace by 00:22:56 MgL/2, turning it this way.

00:23:00 So, this force exerts a torque which is equal to the value of the tension, 00:23:05 the length of the rod and the sine of the angle between the two, 00:23:10 which is this angle here.

00:23:11 So, the nice thing is, you can solve for the tension 00:23:15 directly in one shot, which is just Mg/2 sin

00:23:19 θ.

Report this document

For any questions or suggestions please email
cust-service@docsford.com