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>> Hello class Professor Anderson here, let's take a look at a word problem dealing with Newton's Second Law and let's take this one with rubber bands okay. So we have 2 rubber bands that are going to be tied to an object and they're going to accelerate that object at a. How many rubber bands would it take to accelerate m over 2 at 5a? Anybody have a thought on this one? How I might approach it? Megan what do you think, how should I approach this? >> (student speaking) I feel like you should set it up like m1 over m2 equals a1 over a2. >> Okay, m1 over m2 equals a1 over a2. >> (student speaking) Yeah. >> Possibly. But if I did that what would I really be saying? Ben what do you think? >> (student speaking) That it changes. >> That something changes right? What changes? >> (student speaking) That the mass of the acceleration [inaudible] change. >> Okay certainly the mass has changed, the acceleration has changed. If both of those things have changed, then what else in our picture has changed? >> (student speaking) The force. >> The force right? So we have to be a little bit careful about it. Now, when you're facing this sort of problem, right you have a bunch of words sitting in front of you right and you're like ah now what do I do? I think you should probably draw a picture. Okay. Let's do a picture of this object. So in the first case we have an object, there's our object. And we have 2 rubber bands that are tied to it. And they're going to pull on it. Okay. And they're pulling on it each with a force f. All right this our mass m. So this could be your block on ice, maybe we're looking at the side view of this thing, we're not going to worry about any friction or anything like that. What can we say? Well, sum of the forces is equal to the mass times the acceleration. In this case I have 2 forces. Right? 2 rubber bands tied to it. So I get 2f equals ma. What about the other picture? Well the other picture looks like this. It's half the mass, m over 2 and we're going to have a bunch of rubber bands tied to this thing, we don't know how many. But let's say there are n rubber bands. And each one is applying a force f. All right. Sum of the forces is equal to the mass times the acceleration, but we know that the mass got cut in half and the acceleration went to 5a. And now what is the left side of this equation? Well when I had 2, it was 2f. If I have n, it is of course just nf, nf is equal to m over 2 times 5a. And now we're trying to solve this thing for n. Hmm, how can I do that? Well, looks like this equation we can just solve for f, f is ma over 2. And now this equation I can solve it for n, n is 5ma over 2, all of that over f and now look what happens, I can stick that f right in there and I get n is equal to 5, the ma over 2 from there cancels with the ma over 2 from there and I just get n equals 5. Okay, how many rubber bands are needed? 5 rubber bands are needed. The other way you can see this is the following. If I had 2 rubber bands accelerating m, then 1 rubber band could accelerate m over 2. And if it only took 1 to go at a, then it would take 5 to go at 5n.

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