0:00
This would be two perspectives.
00:02
Here we have a cross section where here each side of the trough exerts a normal force on the crate.
00:08
And that's what you see in the first diagram.
00:10
And the second diagram, that's simply the free body diagram.
00:13
And so we can say that the force sub nr would be equaling two times the force normal times cosine of 45 degrees.
00:30
And this is equaling the square, rather radical two multiplied bottom.
00:34
The force normal and so we know that here the gravitational forces of course m g here and we can apply newton's second law in the x and y directions so for the second diagram we have mg sine of theta minus f equaling m a and so that the resultant normal force minus m g cosine of theta should equal 0.
01:11
And so at this point, because the crate is moving, each side of the trough exerts a force of kinetic friction.
01:18
And so the frictional force f has a magnitude 2 times the coefficient of kinetic friction multiplied by the normal force...