Intro to Springs (Hooke's Law) - Video Tutorials & Practice Problems

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Intro to Springs

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Hey, guys. So in this video, I want to start talking about Springs and Spring Forces. Let's check it out. So when you push or pull against the spring with a force, let's call this force FAA and Applied Force. Um, the spring pushes back. This is just Newton's third law action reaction. You push against something pushes back against you. So the force of the spring will be the negative off the force that was applied on it. And again, Newton's action reaction. That's why, um, to say magnitude, but opposite direction. Okay, so this negative here means just opposite direction, so there's nothing new there. What is new is that this also equals to negative K X. And I'll talk about what K X is in. Just a second notice that there's another negative here and again. That negative just means opposite direction. That negative is you see this more later is almost sort of a conceptual negative to remind you that it's not the direction, but in a lot of calculations, we'll just get rid of that negative to make things a little bit simpler. All right, So what is K and X? I'm gonna start with X X is the deformation that the spring will experience. So the idea is that if you push or pull against the spring, this spring will either be compressed, which means it's now shorter or it will be stretched, which means, obviously, that it is now longer. But X is the deformation. It's not the actual length, but it's the changing length. So in springs, what matters is not how long it is, but how much longer or how much shorter it gets. Okay, so Expo you can think of it is the absolute value of the of the final length, minus the initial length. The reason why the absolute values. Because if you get a negative, we just think of it as a positive because the direction doesn't matter. Okay, so, for example, here I have this first strong here have a spring attached to a wall. Um, this green thing is sort of like a base that you can sort of drag on the spring. And if the spring here is just sort of left hanging there, um, it will have no deformation. It's the original length of the spring X equals zero. We call this the relaxed position of the spring, right? The spring is just chilling, basically without having been compressed. Um, in this situation over here, this red dotted line here, uh, indicates the original length of this spring, and now I'm to the right. And that's because we're to the right of of this spring here of the original length. And that's because we pushed to the right with a force f A. If I push to the right action, reaction to spring pushes back this way, Force of Spring is this way. The gap between its original length and its new length. We're gonna call. It's the deformation. We're gonna call that X. Okay, Now, since I have two different situations, I'm gonna call this X one. We're gonna use X two for the bottom here, here at the bottom. I was originally the spring was originally here, and then I pulled it this way. So my applied force, um, is f A like that, and the spring will push back with F s this way. And the gap between its original length and its new length is Deformation X, which I'll call X two. Okay, so that's how F and X kind of work. Um, que is this spring? That's the second variable here. K is the Springs force Constance. Okay, now there's actually multiple names for this. I've seen force Constant. I've seen spring coefficient. I've seen stiffness, constant or stiffness coefficient. But basically, this is the coefficient of how hard it is to compress the spring. Que has to do with how stiff or how hard to compress or the form in general. Ah, spring is the higher the K, the harder it is to the form of spring. So, for example, if you've ever played, I'm sure you have with one of those pens that you push on it, the tip comes out. Um, if you've seen inside of it, there's a small spring that allows us to happen, is very simple. And it's a thin spring. And if you've ever played with it, uh, it's really easy to squeeze and sort of stretch it out and moves very easily. So that spring has a very low K. Now, if you had springs, for example, at the bottom of an elevator shaft, they're designed as a safety. If the elevator comes down falling the spring, maybe tries to hold the elevators in place Those springs. We need them to be very stiff. Otherwise, the elevator would just crush them and they wouldn't do anything right. So the K has to do with how hard it is to reform something real quick. The units for K R. Newtons per meter and you can see this from the equation. Force is Newton equals K X X is meters, so K has to be Newton's per meter so that this equation is dimensionally consistent. Okay, So if you ever forget, um, if you ever forget the units, you could just get it from the equation by isolating by leaving the X solving the X, the K rather leaving by itself and seeing that the units video unit analysis, you get Newtons per meter or you can just remember that. Okay, So, force one more point to talk about the Force force of a spring is a restoring yeah, force. And it's really important. Uh, it's a restoring force. What does that mean? It's always going to the spring is always trying to get back to its original length. Um, if you pull the spring the spring to the right, the spring will pull back left. And if you post the spring to the right, I'm sorry to the left. It pulls back to the right, and if you pull it to the right, pulls back to the left. You can see here that we went to the right of equilibrium, and the spring is trying to pull back to the left. Here we are to the left, and the spring is pulling to the right. So if you if you're the right, you're gonna be pulling to the left and vice versa. Okay, so the spring is always pulling back. The spring forces always pulling this spring back to its original length. X equals zero. It starts relaxed. It likes being relaxed, and it's going to try to get back to its original length. Um, the force is always opposite to the to the deformation X. Okay, let's do a quick example here and see how this works out. I have a 1 m long spring that is laid horizontally so very similar to these diagrams here with one of its ends fixed. So there's sort of a wall and you have a horizontal spring. There's usually a little bar here. Uh, sort of a platform that you can drag the spring or sort of put put, push a box against the spring against this. Okay, so it has an initial length of 1 m. As you do this, I want you to remember that's what matters in Springs is not the actual length, but the changing length s It's not the length initial length final, but the X, the changing length. Okay, Now you're gonna come here and pull with the force of 50 which will cause the spring to stretch to a final length of 1.2 m. And remember, what matters is not the length initial final, but the changing length. In this case, if I go from 1 to 1.2, that changes. Obviously. Just 0.2 m. Okay, This is my applied force over here. And I want to know what is the Springs force? Constant Force Constants K. Your book. A professor might use a different word. A different, uh, terminology instead of force constant. But it's the same thing, right? So what is K Well, right now, we only have one equation that deals with springs. We're gonna have more later, but for now it's just force of the spring. Is the negative of the force applied or the negative of K X. The force applied in the force of the spring of the same. I'm just going around 50 here and we're gonna ignore the negatives. Que is always a positive, so I can just ignore the negative. You can think of it in terms off this negative. Cancel this negative, but I think it's easier to just think in terms of the negative. It's just a conceptual negative. If I'm calculating your number, I'll drop the negatives and I'll get just positives. Okay? And that works. Vetter X is 0.2 right here and now we can solve this. K is 500 divided by 5000.2 and you get to Newtons. Newton Premier rather okay now values typical values of K that you would see in physics. Problems are usually gonna be in the hundreds maybe in the thousands, but it's usually in the hundreds. Okay, It's usually not gonna be a two or a five or 10,000. It's gonna be a few 100. Um, eso That's generally if you remember that it might give you a little bit more confidence when you're solving these problems. But again, they could really give you whatever number they want. Right? Professors? So how much force is needed to compress it? 2.7. So I wanna know how much force I need. So what is the FAA needed to compress it? 2.7 now? Really important. Here is this word here compressed 2.7 Because I said compressed 2.7. Um, compress 2.7 means that my length final is 0.7, Which means my ex is 0.3. I started at one. I'm going to 10.7. My next 0.3 had I said compress. Bye 0.7. It would have mean that my ex is 0.7. Or if I set a compression off 0.7, then it would also have meant that the excess 0.7. So be very careful here. Am I telling you how much it's changing by or am I telling you what the final length is? Okay, in this case, we're dealing with this one here, but you have to be very careful. X is 10.3 and I wanna know f the equation again is fs minus F a equals negative K X I wanna know what the forces. So I'm just gonna say f a equals K X dropped the negatives. Um, okay, we're talking about the same springs. Okay, Is 2 50 in the access 0.3. So this is 75 Newton's. Okay? 75 Newtons is the force needed to get a compression of, uh, 0.3. They should make sense. A force of 50 got me a x of point to, ah, force of 75 gets me a compression of 750.3. A little bit more force. Ah, little bit more compression or deformation. Technically right. Um, in fact, this is 75 is 50% more, and this is 50% more. And that should make sense because this is a linear relationship. If one doubles, the other one will double as well. Okay. Eso That's it for this one. I want to quickly talk about vertical springs. So let's look at how this works. If you attach a mass toe a vertical spring and let the mass come down slowly, Um, I'm gonna talk about this little detail here towards the end. Um, but basically, if you put a mass What's gonna happen is the weight of the mass will stretch the spring. Ah, spring, if left alone won't self stretch. Right? Um so you need a force to make it stretch. In a situation like this, this object has a mass. It's being pulled down by the earth, has a weight of M G. And that M G is what's going to be responsible for pulling the spring down, and it will stretch the spring until they reach equilibrium. Now, remember, equilibrium means acceleration equals zero, and that also means that the forces will cancel. Okay, so here's the idea before you apply before you put a mass here. This thing is that an X equals zero. It's relaxed because it's X equals zero. The force of the spring is also 00 and zero. Okay, now let's let's go through this real quick. If you have a spring here, you put an object and you slowly let it fall. Once it falls a little bit, it's gonna have a little bit of a compression or a deformation is going to stretch a little bit. If it stretches down, the spring is going to pull back up trying to restore itself to its original length. Okay. And the idea is, now you get a little bit of enough a little bit of X and you get a little bit of a force of a spring. Let's play with some numbers here. Um, let's say M G is 100. Okay, you start with the force of zero for this spring, but then you let it fall a little bit. Now the force of the spring. Let's say it's 10 and you have 100 going down and, ah, force of spring of 10 going up right now, you let it fall a little bit more. This 10 becomes a 20. It's still weaker than mg. So if you release, it falls a little bit more and so on, so forth until this number is eventually 100 right? Why? Because this ex kept growing. So this f keeps growing right and the X will keep growing. This thing is going to keep getting pulled down until these two numbers are the same once they both are 100. Once the force of the spring rich is 100 because it's been deformed so much that it's pulling back with a full 100. Now these forces are the same. They cancel. And if you let go, it doesn't actually stretch anymore because the forces cancel. Okay, so you can imagine that the if you were to sort of graph this force, it would look like this. And eventually you reach 100 as your deformation gets bigger and bigger. This is sort of a force deformation here, right? So once you get 100 they will exactly cancel each other and because they will exactly cancel each other, I can say that K X, which is the spring force, equals K X and M G, which is a gravity will equal each other. And that's the equation that you need to remember. So they will reach equilibrium. And I will have that k X equals M G. Okay, this is what I like to think of as the vertical spring equilibrium equation. Okay, now, that's not the official name of the equation, but I give it a name because I really want you to remember that every time you have a spring at vertical equilibrium with a little mass, uh, it's just hanging there. Stopped this equation will be true. Now, this is only true if you let the mass come down slowly. So the idea is that you put a mass in here and then you hold on to the bottom of the mass and you don't just release it. You sort of let it come down slowly until it's reached equilibrium once. Once it's critical. Librium, if you keep moving on, hand down, this thing will stay there right now. If you just release what's gonna happen is instead of getting to equilibrium, it's going to pass equilibrium and sort of go up. And now it's going toe oscillate around equilibrium point. That's not what we want. That's a future chapter. Okay, so for this equation toe work, you have to let the mask come down slowly. My last point is that this set up also applies if you have a mass on top of a spring, so imagine have a spring like this, and then you have a mass on top of it. If I release the mass is going to do this right point. But if I release slowly and just kind of let it slowly compress eventually I'm gonna release, and it's not gonna move anymore. It works exactly the same. K X equals M. G still applies in that situation as well, because the forces air canceling just the same. Cool. Alright, so that's it for that. Um I want you guys to do this practice problem. Remember the stuff we talked about? The X is the difference in size or length, not the length themselves. Remember that this equation applies if you have vertical equilibrium and all the other stuff we talked about, let's try to do this and see what you get.

2

Problem

Problem

A vertical spring is originally 60 cm long. When you attach a 5 kg object to it, the spring stretches to 70 cm.

(a) Find the force constant on the spring. (b) You now attach an additional 10 kg to the spring. Find its new length.

(For the multiple choice selection, answer only part (b). Use g=10 m/s^{2}.)

A

(b) L_{new}=80 cm

B

(b) L_{new}=90 cm

C

(b) L_{new}=100 cm

D

(b) L_{new}=110 cm

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