00:01
For this problem on the topic of force in motion, we are shown two blocks of masses 16 kg and 88 kg, the figure that are not attached to each other.
00:08
However, the coefficient of static friction between the blocks mu -s is 0 .38, but the surface beneath the larger block is frictionless.
00:16
We want to find the minimum magnitude of the horizontal force f that will keep the smaller block from slipping down the larger block.
00:25
Now, the free body diagrams for the two blocks are drawn, and we treat them individually.
00:29
The force f prime is the contact force between the two blocks and the static friction force is f s is at its maximum value.
00:38
So this frictional force fs is equal to this maximum frictional force fs max, which is the coefficient of static friction mu s times f prime.
00:54
Now treating the two blocks together as a single system, we can apply newton second law with positive rightward, positive x right and we find an expression for the acceleration.
01:05
So the force f is equal to the total mass times the acceleration a, which means that the acceleration is the force f divided by little m plus big m...