AP Physics 1 Sample 3-mass pulley problem

All right, so this type of problem involves a pulley system and you're asked to find the acceleration of the system when it's released and the tensions in the strings. And you're given the masses of the three different blacks. So again, the first thing that we're going to do is draw a free body diagram for each of the three blocks. On the first block you'll have the force of gravity. Down. And the force of tension up. On the second block you will have the force of gravity pulling down on it as well. And the force of tension from the first block. Notice that I call this M two G. And this one M1 G. And then I'll also have a force of tension pulling up on the second block. That same force of tension will pull up on block three because they're connected to the same string. And block three will also have a force of gravity pulling down on it. So these are my free body diagrams. And now I'm ready to do the sum of the forces on each of the blocks. So the sum of the forces on block one. Will be and I need to define a positive direction. Since mass three is more than the masses of M1 and M two add it up. I'm going to say that the acceleration goes in the direction of M three. So that way my acceleration will be positive. If you choose to make the acceleration go, the opposite direction, that's no big deal, but you're going to end up with a negative acceleration in the end. So my sum of the forces equation for F one for block one ends up being the tension force up minus the force of gravity down equals the mass of block one times the acceleration. For block two, I have the force of tension up. This T one points down so it's minus T one and the force of gravity points down minus M two G equals the mass times the acceleration of that block. And then since I ran out of room here, I'm going to stick some of the forces on block three up here. Block three is on the opposite side of the pulley. And so down as the positive direction here. So this is going to need to be M three G minus T two. Equals M three a. So I'm going to notice a couple of things in order for me to do some magic. Algebra. I could notice that I have negative T two here and positive T two here. So if I add these two equations, T two will go away. So I'm just adding everything on this side of the equal sign. I have minus T two and T two, so that goes away. And now I want to notice that I have positive T one here and negative T one there. So I could add these two equations and get T one to go away. So then I can factor out my G on this side of the equation. And factor out the a on the right hand side of the equation. And I'm ready to plug in some numbers here. M three is 7.8. So 7.8 minus two -5 gives me .8. And 7.8 plus. Two plus 5 for 7.8. Gives me 14.8. .8 times 9.8 is 7.84. So. The acceleration ends up being 7.84 divided by 14.8. Which gives me .53. So then to find T one, I would use this equation right here. T one equals M1 a plus M1 G. And that gives me 20.66 newtons. And to find T two, I'll be using this equation here. I'm going to add the T two to the other side. And then subtract M three a. So I get 72.3 newtons for T two. All right, don't hesitate to ask questions if you did not understand something. Or watch this video again.