00:01
All right, so we've got this problem involving a beam and some wires and interestingly a coefficient of friction between the wall and the beam itself.
00:07
So instead of having a hinge at the wall and the beam, there's actually a, the thing is kept in place because the beam is pushing hard up against the wall.
00:14
This says right here, as shown in the figure below, but there is no figure given in the problem.
00:20
So let me just kind of recreate what i think it should be based on the information that we're given here.
00:25
We've got a wall here.
00:26
This is some brick wall or something like that.
00:29
And then we've got a beam which is sticking out.
00:30
I'm going to assume it's perpendicular to the wall, sticking out like this.
00:33
And it's not attached here.
00:35
So if i just let this bean go, it would plummet downwards.
00:39
And there's a cable which is attached from the wall over to this far edge of the bean right here.
00:46
And i'm going to assume that this 30 degree angle right here refers to this angle here.
00:51
Now, i'm not sure that that's the case because i can't actually see the problem.
00:54
But we'll assume that that's what's going on.
00:56
And if you continue reading here, determine the minimum distance x from point a, and we're going to assume that point a is right here.
01:02
We're going to assume the beam has got no thickness or anything like that, so we've got point a right here.
01:08
Distance from point a, which an additional weight to w, twice the weight of the rod, can be hung without causing the rock to slip at point a.
01:15
So we're going to plug something on top of here.
01:17
Let's say we've got a box of nails or something like that, that we're going to put on top of this object, on top of this beam.
01:23
And without the box, we had the force of friction kind of supporting this end, which is why the beam didn't slip out and fall down that by a, but we're going to try to figure out, given how sticky the surface of the beam is compared to the wall, where can we stick this box, which is twice as heavy as the beam, on the beam, so that everything still stays stationary, meaning this thing doesn't slip.
01:47
Now, we've got a couple of hints here.
01:51
Since this is an equilibrium problem, we expect to use the first and second conditions for translation on rotational equilibrium, respectively, to solve the problem.
01:58
So what we're going to have to do is set up some free body diagrams, about the forces that are acting on this object, and also thinking about what's happening with the torques acting on the object.
02:06
And it's actually combining those two big ideas that we're going to be able to figure out where this important distance is.
02:12
And let me just specify what i mean by this distance from here, how to, and we'll just go like this.
02:18
And i'm going to assume this box is not, in fact, an extended box that we're going to treat as a point particle at this particular location.
02:25
And what we'd like to know is what this distance is.
02:27
The other thing we're given is we're given the length here of the beam.
02:31
Given as five meters, i think, but we'll throw this in at the very end in order to figure out what's happening.
02:37
Now, this is kind of like a picture of what's going on.
02:41
And if i'm going to draw a free -bided diagram, let me draw the beam like this.
02:45
I'm not going to draw just a simple box or a dot because this is an extended object, and we're going to have forces exerted on it at different locations.
02:52
And what i'd like to find out, let's first of all think about the forces that are acting on it.
02:56
Well, i know that if this condition is satisfied, that the forces must be balanced.
03:02
Because i know that this object is not accelerating.
03:04
If it is stationary to begin with, then i look at it later and it's still stationary.
03:08
The velocity hasn't changed and therefore we've got no acceleration.
03:10
So all the forces on it must be balanced.
03:12
And if you look at this and think about, right, what are the forces that are acting on it in terms of field forces, things that are exerting a force without touching it? we've got the force of gravity which is acting on it.
03:21
And that acts on the center of mass of the object.
03:24
And this beam also has some other forces exerted on it.
03:26
There's the weight of the chunk of stuff which is sitting on it.
03:29
And i'm going to put that as a thing like this.
03:32
Here's where the force is exerted.
03:33
I'm going to make this arrow it twice as big because it's two times as heavy as the beam itself.
03:38
And then there's some other forces.
03:39
There's this force, which is acting this way from the cable.
03:43
And then there's the force between the wall and the beam itself.
03:48
And this i can think about as having two parts.
03:50
There's a part which is pushing away from the wall this way.
03:54
And then there's the force of friction between the wall and the beam that's acting up this way.
04:01
All right.
04:02
And because this object is in equilibrium, i know that all of these forces taken together must give me zero, which means all of the up and downs are balanced by, sorry, all of the ups are balanced by downs, and all of the lefts are balanced by rights.
04:16
Now, some of these forces are pretty clearly straight up and down, like the weight of the beam and the weight of the box on the beam.
04:23
And the force of friction, if we assume this wall is perpendicular to the ground, and this is also straight up and down force.
04:29
The force of friction always acts parallel to the surfaces in contact.
04:32
And this force is perpendicular to the wall.
04:34
This is also sometimes known as the normal force.
04:38
And then this one right here is at some weird angle.
04:40
It's at this 30 degree angle right here.
04:42
But this force i can think of as if it is broken down into two pieces.
04:46
So rather than draw it as a full -blown force with the black marker, i'm going to switch to a green pen here and say, i'm going to think about this force as if it's made up of two pieces.
04:56
There's a piece, which i'm going to draw this a little bit better, which goes straight up.
05:01
Let's try this again.
05:03
There's a force which is straight up.
05:05
There we go.
05:07
And there's a piece of this, a component of this force, which is in this direction, to the side, like that.
05:13
And how big these are, depends on how big this is and at what angle this is.
05:20
Now, if i take this off to the side, and we'll look at what's going on with the total torques in the up and down direction.
05:28
The total forces in the up and down direction are the force of gravity acting on the beam, and this thing.
05:35
That's everything pulling down on it.
05:37
And upwards forces, we've got this force and this force.
05:41
This force of friction between the wall and the beam and this component of the tension force right here.
05:46
I'm going to go through and label what we've got going on as far as the names of the forces here, just so we can see it.
05:52
This is a force of gravity, sometimes labeled fg.
05:55
They tell us that it's got a value of w.
05:57
This one, they say is twice as big as this one, so i'm just going to label this one to w right here.
06:02
We've got this one, which is the force of friction.
06:04
The force that occurs between two surfaces in contact, and the force of friction will oppose to act to oppose relative motion between the surfaces, and then we've got this one the normal force.
06:13
Now, i haven't put the normal force with the tail at the location where it is.
06:16
It would normally be like right down here, but things are getting a little crowded, so i kind of shifted it up a little bit.
06:21
And then we've got this one, the force of tension.
06:24
Now these little pieces are not additional forces also acting on the beam.
06:28
These are just another way of thinking about this force right here.
06:33
So this one, and i'll keep on going with the weird kind of greenish color for this.
06:39
If we think about what's happening with this one, i'm going to call this one f, t, but it's only the force of tension in the y direction, so ft y.
06:49
And in a similar way, this guy right here is ftx.
06:52
Kind of writing out space ft in the x direction like that.
06:57
All right.
06:59
Well, all right.
07:01
What do we know about an object that's at equilibrium? all of the downs, which i'll just switch back now and start doing some calculations, all of the downs, we can actually say this.
07:15
Let me write this one first.
07:16
The sum of the forces must be zero.
07:18
And that's both in the x direction and the y direction.
07:21
And if i take the y direction first, all of the up and down stuff, i know also the sum of the forces is zero.
07:26
All of these downwards forces, which is going to be 2w, the force of gravity acting on the box, and plus a plain old double.
07:36
With this guy, plus the other forces we've got in the up and down direction, like the force of friction, plus not quite the force of tension, but just this piece of the force of tension, ft in the y direction, ft in the y direction, has got to equal zero.
07:56
And if we've rearranged this a little bit, we end up with ff plus fty, those are the upwards ones, are going to equal w plus 2w on this side.
08:09
If i move them to this side, i've got to subtract them from both sides, negative 3w.
08:14
Well, there we've got, we've got something.
08:17
We've got an expression between how the weight of the boxes and the beam is related to these other particular forces, like the force of friction and the force of tension in the cable.
08:28
Well, let's expand these out a little bit.
08:29
Fty y.
08:31
Fty y is this piece of this force vector.
08:33
Let me quickly draw the force vector over here.
08:36
Here's the force vector.
08:37
I want to know about this piece of it.
08:39
In other words, same thing as this one if i shifted over that way.
08:46
This is not a very happy triangle.
08:50
This is my angle theta.
08:53
And if i look at this diagram, i'd like to know how big this is if i know that this piece is ft.
08:58
Well, this is the hypotenuse of a right triangle.
09:02
Here's my angle theta.
09:03
This is on the opposite side.
09:04
This one has got a magnitude of ft sine theta.
09:08
So another thing i could say, ftyy, is equal to ft, t sine theta.
09:20
Now we happen to know what angle theta is.
09:21
I'm going to just leave that alone for a little bit.
09:24
If in fact your problem in the textbook has got angle theta in a different spot then you might have to read, you know, instead of using angle theta, or 30 degrees, you might have to figure out what that is based on the new diagram.
09:33
But i'm going to leave it like this just for now, just get in the habit of using letters and then fill in my number stuff afterwards.
09:40
And then the force of friction, all right, the force of friction, i know some stuff about the force of friction.
09:44
The force of friction depends on two things.
09:46
It depends on how hard the surface are pushed together and it depends on how sticky the two surfaces are compared to each other.
09:52
So i'm going to write that as musas, this is the coefficient of static friction times fn.
09:57
The reason why i'm using coefficient of static friction is because these surfaces are not sliding.
10:01
That was one of the assumptions we were going to make about how we went about solving this problem.
10:05
And then fn is just a fancy way of saying how hard the two surfaces surpassed together.
10:09
This is the normal force.
10:12
And notice that normal force always acts perpendicular to the surface in contact.
10:16
Here are my two surfaces.
10:16
The direction perpendicular is that one.
10:19
And so this is the normal force right here.
10:22
And so here we've got an expression for f f f t y, and we can now fill those things in right here.
10:30
Ff was going to be mucous s times fn, plus ftyy, which was f t sine theta, equals negative 3w.
10:48
And this normal force right here, notice this one right here, has got to balance this one because the only force acting on this beam to the right is this one, and the only force on the beam acting to the left is this one, and therefore these two must have equal magnitudes, which means that any time i see an fn, i could throw in ft in the x direction.
11:10
Well, let's go back to this diagram right here.
11:12
This one was ft in the y direction, and this one is going to be ft in the x direction.
11:18
And from the same reasoning that i managed to figure out using trigg how big fty was compared to ft, i can figure out how big ftx must be compared to ft, to ft.
11:27
And in fact, this is ft sine theta.
11:30
This one is going to be ft cosine theta.
11:32
So i can actually say, mu's of s, ft cosine theta, plus ft sine theta equals negative 3w.
11:48
All right.
11:49
Now, i don't know how big ft is yet.
11:53
So i can't actually kind of crank stuff out quite yet.
11:59
And notice the other thing we've got is we have no, we can't we can't solve this for x because we've got no x in here yet.
12:07
So now we've got to shift now into kind of the second part and look at what's happening with the torques exerted on this.
12:13
And to do that, i'm going to start up with another, let's make a new place to work.
12:21
We've got the same kind of scenario we had before with a beam supported by the wall.
12:27
And that, in fact, now is going to look like this.
12:35
It's going to be the same diagram we had before.
12:37
But instead, this time what i would like to do is, i would like to instead, look at the torques acting on the beam.
12:49
And whenever you're doing a torque problem, the first thing you have to do is, what is the object that you're thinking about the torques acting on? and in this case, it's the beam.
12:56
I don't want to analyze the torques acting in the box.
12:58
I don't want to know about the torques acting on the wall.
13:00
I want to know about the torques acting on the beam.
13:02
And in order to think about this object and keeping in mind that the sum of the torques, again, equals zero, we've got to think about all the torques that are acting on it.
13:13
Well, we had some points of contact, and those things are potentially exerting a torque on it, and we have the force of gravity acting on some things.
13:20
So if we start to identify what's going on, let me quickly throw down, we had this little chunk of stuff that was right here.
13:26
We had a cable that's exerting a force out that way.
13:30
So some of the forces acting on it, let's first think about the force of gravity acting down on it.
13:34
Now the force of gravity acts on an object as if all the mass were concentrated at the center of mass.
13:39
And so we have this force of gravity acting down this way.
13:42
So we've got a torque due to gravity acting on the beam.
13:45
Torque of gravity acting on the beam.
13:48
But we've also got a torque due to gravity acting on this chunk of stuff, which is sitting on top of the beam.
13:53
So torque due to gravity acting on the chunk of stuff, i'm going to say chunk.
14:00
So those are two torques...