Torque. Problem 1

[00:00:06]Now let's apply the formulas that we wrote for  torque to a problem. We will call it problem one.

[00:00:15]For this problem one, we will have again  a rod. It doesn't mean that all problems with torque about a rod, and this rod is  pivoted about the center of mass right here.

[00:00:33]it is pivoted about the center of mass.

[00:00:38]This rod has mass m, and length R.

[00:00:49]There is force of gravity applied to the center of  mass acting on this rod, mg. There is normal force with which this pushes on the rod. And  in addition to that, there are two given forces applied to this rod. One is on the  left. It is vertically down, and has magnitude, given magnitude of F1. It's given and another  one - I don't put vector sign here because I just showed the direction so this arrow refers  to this vector. And then there is another force, and this force is F2. It is applied  to this end. So this is r over 2, this is r over 2, and both forces F1 and F2 are  given. What do I want to find? I want to find total torque created by all of these forces  about the pivot point, about point o.

[00:02:05]So what is torque created by  gravity? Torque created by gravity will be zero. Why? I didn't even write the  equation that we will be using, I forgot. Tau is r cross F, F is force of gravity, r is from the  pivot to the point where the force is applied, so torque created by gravity is zero.  What about torque created by normal force?

[00:02:40]Normal force is up. This is the pivot, from  the pivot to the point, r is again zero, so torque created by gravity is equal to torque  created by normal force about this pivot point, and it is zero because r is zero. They are  applied to the pivot point. Now we will find torque created by F1 and torque created by  F2. What will be torque created by F1, tau F1?

[00:03:20]This is F1, and magnitude is given. r is from the  pivot to the point where the force is applied, this is r1. And direction is shown - it is to  the left. I have F1 and r1 perpendicular to each other - great! So torque is F1, magnitude  of r1 is capital R over 2. Let me do this F1, and then R two. This is magnitude  of torque - r times F - and they're perpendicular to each other. All I need to find  is the direction, and to find the direction, we always bring these two vectors together; we  start them from the same point. This is F1, r1 is to the left. Now you use your right hand,  and I use my left hand because you see my mirrored image. Before I start applying putting my hand  here or there, let's narrow down our choices.

[00:04:33]So what are our choices? Either into the board or  out of the board because torque is cross product, it is perpendicular to the plane of original  two vectors. You take your right hand, you put your fingers along r1, you  bend them towards F1, from r1 to F1, where does my thumb go - in or out?  From r1 to F1, it is coming out of the page, so the direction of torque created  by F1 about this point is out. How do I show this?

[00:05:19]This will be out of the page, and if I introduce  the coordinate system - which coordinate system would you like to use? Not important.  So I will use probably this standard coordinate system. For this problem it's not  important, but sometimes it really matters.

[00:05:38]so x y and z follow right hand rule. This is x,  this is y and z is out. And if I use unit vectors i j and k again, as in the previous video, then  I can write it in terms of unit vector k. F1 R over 2 unit vector k, which is in the direction  of positive z. We found tau F1. What is tau F2?

[00:06:13]Torque created by force F2  is equal to: force is F2, r is from the pivot to the point where the  force is applied. This is r2, so I have F2 times r2, magnitude of r2 is capital R over 2,  and these two F2 and r2 are perpendicular to each other. I found the magnitude of torque, now I need  to find the direction. And to find the direction, as always, we start these two vectors from the  same point. F2 is down, r2 is to the right, use your right hand now. Give me one second.

[00:07:18]So my choices are in or out. You  probably already guessed the answer, but I'll still do it. So my fingers are along r2,  I bend them towards F - I cannot switch the order, it is from r to F, and my thumb shows  direction in. So the direction of this torque is into the page. This is direction  of torque, and I will put it here. It is into the page. or if I want to write it in terms  of unit vectors, as I did here, it will be F2 R over 2 and minus k - it is into the board,  along negative direction of z. So now I need to find torque total, which means I need to find  the vector sum of all torques exerted on this rod, which will be tau created by F1 plus tau, as a  vector, created by F2 plus tau created by gravity and plus torque (tau) created by the normal  force. Well this one is zero, this one is zero, torque are created by F1 is F1 R over 2 along k,  torque created by F2 is F2 R over 2 along minus k.

[00:09:00]I can factor out minus here, and I will also  factor out R over 2. So I have F1 minus F2 times R over 2, and it is along k. Now let's look  at this once again. So for example, I have F1 = 5 newtons, and I have F2 = three newtons.  Five minus three is two, I plug in R, and I get torque in positive z direction. Now what happens  if this is 2, for example, and this is 5? 2 minus 5 is a negative number - it's minus 3, so  I'll have here minus 3 times R over 2 k, which means that I will have torque in  negative direction. So we derived this torque, we calculated the torque created by these  two forces, and your solution will take care.

[00:10:08]Finally at the very end you will plug in the  numbers, what is given, what are the magnitudes of F1 and F2, and you will figure out if it  is really in positive direction, which means positive z direction with my choice of coordinate  system, which means that F1 is greater than F2. Or if F2 is greater than F1, it will be in negative  z direction with my choice of coordinate system.