00:02
So problem 22 says a force of magnitude 7 .5 newtons pushes on three boxes with masses, 1 .3 kilograms, 3 .2 kilograms, and 4 .9 kilograms as shown in the picture 5 .29.
00:17
Find the magnitude of the contact force between boxes 1 and 2 and between boxes 2 and 3.
00:24
So picture your problem, list out your known variables, and then you can apply the equation.
00:34
Force equals mass times acceleration.
00:41
Now, the total force equals the sum of all the masses, so mass 1 plus mass 2 plus mass 3 times the acceleration on the whole system, since we're sliding all the boxes along together.
01:03
We can rearrange this, then acceleration equals the force divided by the sum of the masses.
01:19
So then we can plug in what we know.
01:21
We know that the force is 7 .5 newtons, and we know that the masses are 1 .3 kilograms plus 3 .2 kilograms plus 3 .2 kilograms plus 4 .9 kilograms.
01:53
And when you do it the math, you'll find that the acceleration on the whole system is 0 .78 meters per second squared.
02:14
Wants to know what the contact force between the boxes at these different seams are.
02:20
So to solve for the individual forces at the contact points, we can apply our force total equals the difference in the forces...