00:01
So in this problem, we have a box that is 10 meters by 10 meters.
00:08
And so the first thing we want to look at this and see where is this point o.
00:12
Now it looks like it's in the center and it is.
00:15
It's a square, so all sides are equal.
00:17
We have this 45 degrees that's going through a corner.
00:21
And you could work this out with a right triangle of an angle of 45 and two sides that were the same.
00:30
But we can just know directly that the origin is in the.
00:34
Center and so each distance from the side to the center is five meters we're also towed that a clockwise rotation is negative and that is the notation to denote that which means that a counterclockwise notation rotation is positive so we want to find the torque on the square and a torque needs a distance and a force.
01:14
The distance is the perpendicular distance between the line of action of the force and the origin that we're calculating moments about.
01:23
So for this bottom 10 newtons, if we were to extend this force through, we can find that it's 5 meters perpendicular distance.
01:37
So for the 10 newton at a diagonal and the 10 noon off to the right here, we see that their lines of action as extended in the problem go through the origin, which means that the torque for those two forces is zero.
01:53
For this distance, for this 10 newtons at 30 degrees, there are two ways to resolve this.
02:01
One way is to extend it and measure the perpendicular distance by the definition.
02:07
Now you could do this.
02:07
It's probably a little bit longer.
02:10
So what we do instead is break down the force into its component parts in the axis system, where we have some f of x and s.
02:22
F of y.
02:26
Now in here we don't need to specifically make a coordinate system but we can say right here x and y if we want to or probably more conveniently at the origin...