 So now we're going to take a look at a two pulley system. We've got three objects we're looking at. We've got a mass here in the middle, the cart, that is 100 grams. And we're going to assume there's no friction between the cart and the table. We've got a lighter mass over here of 100 grams. And we have a heavier mass over here of 200 grams. They're attached to the cart by two frictionless pulleys. Let's think about a free body diagram for this situation. We have three forces of gravity. We have a force of gravity on the cart. We've got a force of gravity from the lighter mass. And we've got a force of gravity from the heavier mass. There's also a normal force acting up on the cart. There is a net force. We'll have to see which way that points here in the second. And there's a force of tension in the ropes, but we'll look at that one a little bit later. Let's see what happens if we let go of the system. It accelerates. So when that accelerated, we saw that there was a net force acting to the left. Now, all that a pulley is going to do when we're looking at the forces is it's going to change the direction of the force from going downwards, in the case of our mass, to going sideways. All that pulley does is it straightens out or changes the direction of the force. So what we're going to look at now is straightening out that pulley in our free body diagram so we can solve for the net force and then we can eventually calculate the acceleration. Now we're going to solve this problem. Here I've got the free body diagram for the question we were just looking at. And we're going to take a look at what these pulleys are going to do to start off with. So the pulleys, remember, are just there really to change the direction of the forces which are acting over the pulleys. So in this case, the force of gravity kind of gets transmitted all the way up through that pulley to the cart. Same with the force of gravity too. Now what we want to do is take our forces of gravity and we're just going to redraw them so that they're all nice and co-linear. We're going to straighten out those pulleys is sort of what I would call this in class. So there's the force of gravity one. Instead of having it point down, I'm going to have it point to the left. And I'm going to make a smaller, about half the size, since this is half the mass and it should be half the force, force of gravity two. Pointing to the right instead of down to straighten out that pulley. Now I can start to think about doing a net force statement. I've got a good free body diagram. Now I'm going to do a net force statement. And I've got to think about what forces am I going to put into my net force statement. Well, I got four choices I suppose. I've got the force of gravity one and two, the normal force and the force of gravity of the cart. Here's the thing about the force of gravity of the cart and the normal force. Those forces would not cause this cart to go left or right. They're in the wrong dimension, the vertical forces. So those forces aren't going to appear in this net force statement. I'm going to put in here force of gravity one and force of gravity two. Which are the forces which decide which way the cart is going to move left or right. So there's my net force statement. Now I'm going to put in some algebra for that. In net force statements, we always put in MA for the net force. The equation to calculate a force of gravity is MG. And what I'm going to do here is I'm just going to put M1, because I'm going to put in the mass of the first object there. And then M2 to help me remember to put in the mass of the second object. Now I can start to fill in some numbers here. I've got to convert from grams to kilograms for my masses. And I'm going to put in the acceleration due to gravity. Now notice what I've done. I've put in the acceleration due to gravity negative 9.81 meters per second squared is a negative. Watch what I'm going to do with the other mass and acceleration due to gravity though. I'm going to put that in as a positive. That might seem sort of random. Like why did I choose to make one positive and one negative? Well the acceleration due to gravity here for the force of gravity one was describing a vector that was going to the left. It's a leftward negative vector. So I decided to put in that acceleration due to gravity as a negative so that when I went and multiply these numbers together I get negative 1.962 Newtons. I would get a negative force out. You can put the negative sign in later on if you wanted to in this step but I like making the acceleration due to gravity a positive or negative number. Kind of a nice way to keep track of your vectors as you work through the problem. That now the force for force of gravity two is a positive number and it's going to the right. Wrong units though. I'm going to mix those up. That's a force in Newtons. So now I have my forces worked out on the right hand side but I've got to deal with the net force. Well in place of acceleration, I'm just going to put acceleration. That's what I'm solving for. And in place of mass I'm going to total up all of the masses because all of those masses have to accelerate here. So I'm going to put in not 200 grams, not 100 grams but a total of all of the objects, even the mass of the cart 400 grams or 0.400 kilograms. Alright, let's see what this works out to. So into my calculator I'm typing in negative 1.962 plus 0.981 by 0.4 to get an acceleration of negative 2.45 meters per second squared. Now before I'm done there's one extra thing that I sort of do into this problem. I might record this as plus or minus, positive and negative 2.45 meters per second squared because the acceleration of the system, which is what we're solving for in this question means the acceleration of the cart or either the masses. So some of the masses go up, which is positive. One goes down, which is negative. The cart goes to the left, which is negative. So when I write positive or negative in my answer I'm showing that I could be describing the acceleration of any of those objects going left, right, up, down, no matter what the case may be. So now we've got to think about the force of tension. The easiest way to do that is consider just one of the masses. It doesn't matter which one. I've got the 200 gram mass, so I have the heavier one. And we're going to think about the forces that were acting on that mass just before or just as it got moving. And we're going to think about that mass just hanging on the string. The tension is the same in either piece of string. So you can pick either mass to do this. I'm going to pick the 200 gram mass. And I'm going to make a free body diagram of the forces acting on it. So one of the forces we have acting is that force of gravity. There's no need to call it one anymore since I'm only dealing with one object right now. And there's a force of tension as well. And I always ask students, do you think I should make the force of tension bigger or smaller or the same size as the force of gravity? And here it has to be smaller because I know overall there has to be more force pulling down than up. The net force on this object has to be downwards. So there's my free body diagram. Those are the only forces that are acting just on that 200 gram mass. Tension going up, the net force going down, and the force of gravity going down. Now I can do another net force statement. It looks a lot like the last one. In fact, it is a lot like the last one. I've got force of gravity and force of tension this time equaling the net force. Now in place of net force, I'm going to put MA just like before. Force of gravity, MG just like before. And there's no formula for the force of tension. We're just going to solve for that. The mass I'm using now is just the mass that I chose for the question. I chose the hanging 200 gram mass. So that's the only mass I'm putting in. I'm not putting in all three masses now because I'm only considering the one 200 gram mass. I also already know what the acceleration is. This heavier mass will accelerate downwards at negative 2.45 meters per second squared. So I got that from the acceleration from the previous problem. Now I'm going to put in the mass again, multiplied by the acceleration due to gravity. Again, I made it a negative because the force of gravity is going to be going downwards. And plus force of tension, which I'm going to solve for. So into my calculator I go, I'm going to type in 0.2 times negative 2.45. I'm going to add to that 0.2 times 9.81. So what we get works out to a force of tension of 1.47 Newtons. And one of the things you can kind of do to make sure you're in the right track is notice that the sign your calculator gave you on that was a positive, meaning that it matches perfectly with your free body diagram. That's a great check to make sure you're doing these free body diagram problems correctly. For more examples like this one, check out my website, LDindustries.ca.