00:02
All right, so let's get right into this problem.
00:04
So how can we do giancoley's 9 .1 .19? and so let's just read through the problem real quick and write down some stuff we need and then continue.
00:17
So a traffic light hangs from a pole as shown in the figure below.
00:22
The uniform aluminum pole a .b is 7 .2 meters long and has a mass of 12 kilograms.
00:28
The mass of the traffic light is 21 .5 kilograms.
00:31
Determine a, the tension in the horizontal massless cable cd, and b, the vertical and horizontal components of the force exerted by the pivot a on the aluminum pole.
00:44
Okay, so a few things we should write down.
00:48
All right, let's take down the length of the aluminum pole.
00:52
I'm going to do this on the side.
00:53
Let's see how far can i go? so just on the side here, let's write l.
01:04
Is that enough? maybe a little bit more to the left.
01:08
Okay, so the first quantity we're given, the length of the uniform aluminum pole.
01:13
So let's write that down as l.
01:16
7 .2 meters.
01:19
And it has a mass.
01:21
Let's give the pole a mass of little m, because it's the smaller mass, 12 .0 kgs, kilograms.
01:30
The mass of the traffic light, let's go big m because it's bigger, 21 .5 kilograms.
01:38
And that's mostly what's given to us.
01:40
If you look at the picture, the angle between the massless cable and the aluminum pole is 37 degrees, and that will be handy later.
01:48
So let's write that down.
01:51
Or at least, yes, i'm going to write these down, but it would be important to remember them as well.
01:58
H, just to say this height, from the pivot to the top of the massless cable, is going to be 3 .8 meters.
02:06
And so i think that's all, that's all of our quantities that are going to be important.
02:16
So i guess next we would want to talk about how you approach this problem.
02:21
One thing that's very important when you're talking about forces is newton's second law.
02:27
It says that the force is equal to, well, the sum of the forces of a system is equal to the mass of that system multiplied by the acceleration of that system.
02:37
And so if you have some forces, maybe you want to find out force one or something.
02:43
Maybe you have two forces acting on some mass, and you want to find out the acceleration.
02:48
Then you could do something like this, or maybe you want to find out one of the forces you know the other one.
02:54
Then you could say that if the system wasn't moving, you could say that the one force is the opposite of the other and things like that.
03:02
So it's very important to use this kind of formalism to talk about forces.
03:08
At the same time, you can talk about rotational force, which we just call torque, is equal to i -alpha.
03:16
These are just rotational analogs, moment of inertia, and angular acceleration.
03:20
And so if both of these are equal to zero, which is what happens whenever a system is in equilibrium, then you can use these equations to help find certain quantities.
03:31
And torque, i didn't write, but it's also equal to, at least in the vector form, r -cross -f.
03:38
This means just distance crossed with force.
03:44
And so if you don't know what a cross product is, it would be good to look up and understand, but it's not essential to the problem.
03:55
We can just talk about the magnitude and it'll be fine.
03:58
So the magnitude of the torque is just going to be the magnitude of the r vector, the distance vector, times the magnitude of the force vector, times the sign of the angle between the two.
04:10
So if you have something like, you know, a fulcrum and a seesaw, then a force, let's say somebody's sitting on this side, some kid is sitting here, then there's going to be a force due to gravity based off that kid, right? and it's going to be acting at some distance, so the angle between them is this 90 degree angle here, but if he was, i don't know, for some reason hanging off the side differently, let's, if we draw the, if we draw like this, and maybe he's going at an angle now, then it would be that angle between them, right? and so the force would be a little bit different.
04:54
Anyway, this will all be very useful to solve this problem.
04:58
So maybe i'll keep everything up here, and then i'll just go to the next page.
05:05
So i'll have, oops, this will be our distance vector, force vector right there, and the angle between.
05:13
So with all these things in mind, how would we solve this problem? essentially, when you're dealing with forces, you want to use what we call it fbd, this is a free body diagram.
05:23
So if you have a mass, if you have a mass on a table or something and you want to find out some of the forces on it, then you draw it as a free body diagram would be taking this and drawing it as a dot, and then saying, okay, well, what forces are happening on this thing? so you have an upward force, of course, the normal force due to, this is why the box is not falling into the earth.
05:46
And then the earth is exerting some force on it, right? you have some gravity here, force due to gravity.
05:53
And since the box is not moving, we could say that the sum of the forces is zero.
05:57
So n minus m g, which is what fg would be equal to, is equal to zero.
06:04
And so we see that the normal force is just the positive of the force due to gravity.
06:14
So this can work for a lot of different situations, and we're going to use that here.
06:18
So the pole is in equilibrium so that the net torque is zero, but also it's, since it's also, since it's an equilibrium, the net force is also equal to zero.
06:28
So we can use both of these equations.
06:30
This is going to be very nice for us.
06:34
So from the free body diagram, which we'll draw later, we can calculate the net torque about the lower end of the pole with counterclockwise torques as positive.
06:45
What this means is you have to kind of choose a direction when you're talking about some of the forces is kind of easier than that you do have to set up a coordinate system, but it's not as important as, when you're talking about torque.
07:00
When you talk about torque, you want to choose either counterclockwise or clockwise as your direction.
07:06
When you talk about the forces, so that one of the forces, if you have two people, let's say now, so one of the forces will be positive and one of them will be negative, right? depending on where you place the pivot and all.
07:20
So let's go ahead and let's do that.
07:21
Let's draw a free body diagram on the next page here of our system.
07:28
And let's look at the forces.
07:34
Draw a pull here.
07:36
This is at a length l, right? then we know we have this angle at the top, so i'm just going to draw a lot of the situation here.
07:46
I'm just going to call this theta for now, but we do remember this is 37 degrees, right? let's see, the pivot at the bottom, a is going to act, you know, it's going to push in the same direction as the pole, right, fp.
08:06
But we want the vertical and horizontal components, so we only want this piece and this piece, right? fpi and fpx.
08:23
And geometry tells us that if you have a line like this, right, and another line, similar, two parallel lines, then the angles, the opposite angles should be the same there.
08:37
So these are both going to be theta.
08:39
Maybe i'll draw that a little bit closer in so it's not matching with the force.
08:44
Theta.
08:45
It looks like a dot.
08:49
Theta.
08:50
Okay, well, you get the idea.
08:52
Let's see if i can erase and make it look a little better.
09:12
Okay, let's get back in.
09:17
Just have this here.
09:21
Now it looks worse, but at least you can see the theta.
09:24
So we still have this force here in the diagonal direction, fp, and we're going to find, we're going to look for the x in the x and the left.
09:30
A while later.
09:32
So that'll be something we'll do for the second part of the problem.
09:36
Right.
09:37
The first part, we're still talking about this some of the torques.
09:41
So the question you want to ask, right, is what's going to cause a torque here? so like i said, when you're dealing with a fulcrum, it depends on where you place this.
09:52
When you have a seesaw on a fulcrum, it depends on where the people are sitting or where the forces are on the line.
10:01
And so i'll say towards the middle of this, well, no, at the exact middle, right, the center of mass of the pole itself is going to feel a force due to gravity.
10:12
So we're going to have mg here.
10:15
And the traffic light is going to feel something similar, right? so the traffic light is the very top here.
10:22
But now it has a different mass, right? so these are just the forces.
10:24
This is just fg for the big mass.
10:28
And this is just fg for the little mass, right? so traffic light will feel big mg and the aluminum pole itself will feel little mg.
10:42
And then there's one other force i'm forgetting, right? is that i have this upward force due to the, in between the center of the pole and the traffic light, you have the cable, right? you have this massless cable sitting here.
10:56
So basically for that, for that part, we want to have this force due to the cable right here.
11:08
Because that's pulling it horizontally, right? so, right? so we can do some of the torques now, basically.
11:23
So the sum of the torques is going to be zero.
11:25
I mean, like we said, the system is in equilibrium.
11:31
And then let's talk about which forces are causing torques.
11:35
Well, definitely the cable force is going to, if the pole was free to move around some forefront, the cable force is definitely going to cause a torque.
11:50
And since it's on the top side, and we're taking counterclockwise to be, we're taking counterclockwise to be our positive direction.
12:02
That would mean that clockwise is going to be negative.
12:07
Then that means that since the cable's arrow is above the line, it's going to be positive here.
12:13
So force from the cable, which is going to be the tension, basically, that we're searching for, is going to be multiplied by the distance that it's at.
12:23
And so if we remember from the problem, this distance, we write it in green, the distance from the cable to the bottom of the pole is just going to be h.
12:35
So that means the torque due to the cable is just going to be fc times h.
12:42
And then the other forces are going to be negative because they're on the underside of the pole.
12:53
So we're going to have minus.
12:56
So basically we're dealing with these three forces causing torques, the force of the cable, the force due to gravity of the traffic light, and the force due to gravity of the pole itself.
13:11
So those three forces are causing tors in the pole.
13:17
So the second, let's do the torque due to the pole itself.
13:23
And so we have to think, okay, well, what's the distance that this, what's the distance that this force acts at? and i forgot to do this earlier, but let's figure a general way to write the x distance here...