00:01
So here we're going to draw the free by the diagram for the system at an angle 32 degrees below the horizontal.
00:08
We have force f going straight down would be mg and going straight up would be force normal.
00:15
This would be m and then always opposite to the direction of motion would be force of friction.
00:23
We can say that the sum of forces in the y direction, this is equaling zero because this system has translational equilibrium in the x and y direction.
00:31
So this would be equal to force normal minus m g minus f sine of theta.
00:40
So we can then say that force normal simply equals m g plus f sign of theta.
00:46
We can then say that here applying the sum of forces in the x direction this also equals zero because again it has translational equilibrium in both the x and y directions.
00:56
This would be equal to f cosine of theta minus the force of friction kinetic.
01:06
We know that then force of f cosine of theta would be equal to the force of friction kinetic.
01:13
This would simply be the coefficient of kinetic friction times force normal and then substituting for force normal and then we're solving for f.
01:24
So f is going to be equal to the coefficient of kinetic friction times mg divided by cosine...