00:01
So in this problem we should be thinking of our free body diagram.
00:04
So we do that and this time we do need the fact that it's a cube to work out our problem.
00:11
I'm going to take the axis to be the bottom corner here, x and y.
00:16
So we want the situation where the box just tips.
00:19
If we imagine the situation where the box is tipped, let's draw it like that with our force applied somewhere near the top corner there, we see it rests solely on this back corner.
00:33
Now the box hasn't tipped yet, but right at the limit of where it would be tipping we're going to present our forces as acting at this point.
00:43
There's a normal force upwards from the contact to the ground and then some friction force equal to mus n to keep it stationary.
00:55
Then we have our applied force f, we know it's 15 degrees.
01:04
Now the center of mass, if this is a uniform cube, is going to be right in the center and that's where the weight is going to act, equal to mass times the acceleration of the gravity.
01:16
So now we need the dimensions.
01:18
The whole cube is 1 .2 meters and then so of course mg is in the center, 0 .6 meters.
01:30
So to tip the box we need to get it rotating, to rotate it over, and so we want to look at the moments.
01:41
Newton's second law, the sum of moments is equal to i alpha.
01:46
Now the limiting case to tip it over is when the acceleration is equal to zero, because any more force we have an acceleration and some of the force goes into the acceleration, but right at the limit they barely tip it over we can model it as no acceleration.
02:00
So the sum of moments is equal to zero and we are going to take counterclockwise as positive by convention.
02:09
Now we're going to take moments about the origin and we're doing that seeing as a moment is a force times the distance.
02:19
The normal force and friction force act through this point, so they make no moment because the distance is zero...