Hey, everyone. So we've gotten some practice with using \( f = ma \). And in this video, we're going to take a look at how we solve for forces using Newton's second law. So we're going to be using the same list of steps and equations. Let's check it out. So we've got this 10 kilogram box, and it's accelerating to the right. It's being pushed by 2 forces. So I've got this box; here is 10. Here, I actually know what the acceleration is, \( a = 9 \). And I got 2 forces. One of them pushes to the left with 30 Newtons. So this is 30 here. And so what I want to do is I want to find the other force. So what does that mean? Well, I've got one force pushing it to the left, but it accelerates to the right. So I know I have to have another force that's pushing it to the right as well. So this is my mystery force here. This is \( f_a \), and I'll call this \( f_b \) just like we have before, and I'm trying to figure out what this \( f_a \) is. So I know I'm going to have to use \( f = ma \), but first, what I want to do is I want to choose the direction of positive. And just like we have before, we're going to usually choose the right direction. So the direction of positive is going to be to the right like this. And now we get into our \( f = ma \) here. So we have \( f = ma \). And remember that when we are expanding the sum of all forces, we have two rules. Forces along our direction of positive are written with a plus sign, and against our direction of positive, are written with a minus sign. So, for example, here, we've got our \( f_a \) which is positive even though we don't know what it is, and then we got our negative \( f_b \) here because it points to the left, and this equals \( ma \). Now we just replace the values that we know, right? So we have \( f_a + (-30) = 10 \times 9 \). So when we move the 30 to the other side, what you solve for, you're going to get \( 90 + 30 = 120 \). And so that's your answer. You've got 120 Newtons here. So now let's take a look at our answer choices. Well, because all of our answer choices are positive, then we don't have to worry about any signs or anything like that. And we could just go ahead and choose answer b. That's going to be our correct answer. Let's move on to the second one. Here, we have a very similar scenario here. The 10-kilogram box accelerates to the left this time. So we've got this 10 kilogram box. Now we know the acceleration is to the left. Even though it points to the left, we're still going to write it in our diagram as 6, but we're going to indicate it with the correct direction. And we have 2 forces. We've got one that's to the left, oh, sorry. One that's to the right. This is \( f = 70 \). We want to calculate the other force. So we have another force just like we had before that accelerates to the left. There has to be a force that's pushing it to the left. So this is our mystery force here, and I'm going to call this \( f_b \) and this is \( f_a \). So now we want to figure out \( f_b \) in this scenario. So we got to choose our direction of positive. We actually don't need to because we're going to assume the direction of positive is to the right; that's given to us in the problem here. So now we write our \( f = ma \). So now we just expand the forces that we know; remember, along the direction of positive is written with a plus sign, and then we got our negative \( f_b \) here, and this equals mass times acceleration. Now we just replace the values that we know. We know this is \( 70 + (-f_b) = 10 \times (-6) \). Well, remember, assuming the direction of positive is to the right, our acceleration actually points left. So this actually brings us to an important point here. Whenever we write \( f = ma \), we're always writing the letter \( a \) as positive, meaning you would never write \( m \times (-a) \), for example, if you knew that it pointed to the left. However, when you actually know the value of the acceleration and the direction, you're going to plug in the correct sign. So for example, here, we've got 6, but it's actually going to be negative 6 because our acceleration points to the left. Alright? So now what we've got here is we're going to rearrange and solve for this force. So we've got we're going to move this over to the other side and this is going to be \( 70 + 60 = f_b \) which is 130. So if you take a look here, we've got \( f_b \) is 130 Newtons. So let's take a look at our answer choices. We actually have two. We have 130 and negative 130. So which one is right? Well, one of the things you might have noticed here is that in previous videos, whereas when we solve for \( a \), we get a positive or negative, which is basically just the direction. Our final answer gives us the direction. When you're solving for forces, however, you actually always get a positive number. Notice here how in these examples, when I solve for a right-pointing force, I got a positive, and a left-pointing force, I still got a positive number. I always get positives because, basically, you're solving for the magnitude of these forces. So what you do here is there's actually a couple of ways to solve this. You could indicate the direction by actually writing it into your answers so your professor knows that you know the direction of the force. Another thing we can do here is look at the problem itself. We're going to assume the direction of positive is to the right, which means if we get a left-pointing force between these two answer choices, it's actually going to be negative 130 Newtons. That's it for this one, guys. Let me know if you have any questions.
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6. Intro to Forces (Dynamics)
Newton's First & Second Laws
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Newton's First & Second Laws practice set
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