00:01
There are two conditions to have an object in equilibrium in terms of forces on it.
00:08
The first condition is that all the forces on the object have to add up to zero, and we can split that into two separate equations, that the forces in the x direction have to balance or add up to zero, and the same with the y or the vertical direction.
00:28
So there's kind of two separate, similar conditions within that first one.
00:36
The second one has to do with rotational equilibrium.
00:42
So if it's an extended body, then to be in rotational equilibrium, the torques have to add up to zero.
00:51
And a reminder that is a torque is a force applied off the pivot point of the object.
01:00
Times the radius out from the pivot, times the sign of whatever angle is between the force and the radius.
01:13
And we'll look at examples with that theta equals 90 degrees.
01:19
So sign of theta is one when you have an angle of 90 degrees.
01:29
And those are easier to see what's going.
01:32
On.
01:34
So we have four different examples of a wheel with a weight of 50 newtons.
01:42
Yeah, that's the weight is a force downwards.
01:46
And we will call positive y up and positive x to the right.
01:55
And you usually call torques counterclockwise positive.
02:02
And torques that go clockwise are considered to be negative in terms of rotations.
02:09
But let's take a look at the first case.
02:12
What i notice is that in terms of the y forces, there is 20 newton's upwards.
02:24
So if we look at the y forces all added together, there's 20 newtons positive upwards and 20 neutens downwards with a negative.
02:36
So yes, that adds up to zero.
02:41
Now let's take a look at the x direction.
02:45
Both of those 10 neutrons are applied to the positive direction.
02:52
So they both come in with a positive value.
02:58
And so the sum of the x forces does not work out.
03:04
So we'll put an x by that and we'll put a checkmark by the conditions that work.
03:12
In terms of torque, those 2 10 -dutons forces are creating a torque because they are perpendicular to the radius of the wheel.
03:25
So if we look at the summation of the torques, let's give the radius some value.
03:36
R equals the radius of the wheel, and it will be used to determine rotational motion.
03:49
But we have two torques due to the 10 newtons.
03:52
The top torque is applying a force 10 newtons times r in the clockwise direction, so that is negative.
04:06
And the bottom is producing 10 newtons times the radius.
04:12
I don't think they gave us a radius in the counterclockwise.
04:17
So that is positive.
04:20
And we do wind up with the torques equal to zero, the net torque, adding them up.
04:27
So only one condition is violated in that first situation.
04:33
And that is that the forces in the x direction do not cancel.
04:41
Okay, so moving right along.
04:43
The next one, case b, we can see that the up and the down forces are balancing.
04:50
There's a 20 dutons up and a 20 dutons down, and that produces zero.
05:03
If we look at the x direction, we have a 10 dutons to the right, which is positive, and a 10 dutons to the left, which is negative, and that is also equal to zero.
05:24
So good.
05:26
Two conditions met so far.
05:29
But if we look at the torques, and i should probably show little arrows, which way the forces would tend to spin the wheel if they were actually pulling on the wheel...