Spring Force (Hooke's Law)

by Patrick Ford
521 views
1
8
Was this helpful ?
1
Hey, guys. So in this video, we're gonna talk about Hook's Law in the equation for Forces of Springs. We're gonna talk about all the variables Hook's law represents and how to use it to solve problems. So let's jump right in. So whenever you're pushing or pulling against the spring with some applied force, that spring pushes back on you in the opposite direction, and it's because it forms an action reaction pair. So, for instance, in this example, I'm gonna take this spring, I'm gonna push against it with some applied force. That's the action. And then the spring pushes back on me with an equal but opposite direction force. That's the reaction force. And so we say that the Spring Force is equal to the negative of your applied force, and that's equal to negative K. Times X, where X represents the objects deformation here and then K is just a constant. It's just a force constant. So the only thing you need to know about this negative sign here that negative sign just means that that force points in the opposite direction. So it means, and because that that negative sign often just goes away and we're only just ask for the magnitude of these forces. We're gonna use this more powerful equation, the absolute valuable. These forces are equal. And that's just equal to K Times X. The negative sign just goes away. So to see how all this stuff works, let's just go ahead and take a look at an example. So I got in this first example here. I'm pushing this spring to the left, so that means that my applied force is equal to 1 20. And so that means that the Spring Force is in the opposite direction. Also equal toe 1 20. Now I'm told that the K constant, which is that force constant, is equal to 20. And what that K constant really just means is it's a measure of how stiff the spring is, how hard it is to push or pull it. And so I have asked for is how much it compresses by. So I've got these two forces and I'm pushing on this spring and have compressed it by some distance X, and that is the objects deformation. That deformation is always measured relative to this dotted line here, which is basically the relaxed position that's right here. So that is the relaxed position. By the way, the other word that you might hear for that often is the equilibrium, positions, equilibrium. And that's where X is equal to zero. Okay, so let's go ahead and just set up my equation here. So I've got that. The magnitude of these forces are equal to each other, and that's just equal to K times X. I'm looking for X, and I've got K right, and I also got the force. So my applied force, which is the Springs force, was just equal to 1 20. My K is equal to 20 and then I've got X is that is my variable here. So you go ahead and solve for that and you get a deformation of 6 m. That's how much I compressed it by. So let's take a look at a different example here, because here I was pushing against the spring. Now what I'm gonna do is I'm gonna pull against it. So now what happens is I'm taking this spring. I'm pulling it out. So my applied forces here, which means that the spring force is equal and opposite to that. But now I'm actually looking for how much force I need to pull it out to some distance here. So let's take a look. I've also got that The K constant is equal to 40 in this equation. So let's just write out my formula f s equals f A equals K Times X. Now I'm looking for the actual forces here. I need I know what the K is now. I just need to know what my deformation is. So let's take a look at these numbers. I've got 10 m 16 m. Which one of those things represents my distance, my deformation. Well, before we actually had a compression distance that was given to us. But here all we're told is that the spring was originally 10 m and then once you've pulled on it, the spring has now become 16 m long. So really, this distance in between here the amount that you've pulled on it or performed it is that X right? So that's equal to 6 m. So what that means is that the X is not about the springs length. The X is actually the difference in the springs length from the final to the initial So now I've got my ex. That's just equal to 6 m. Now I just go ahead and solve the problem. So I've got f s is equal to eso. Yeah, so I've got 40 and then I've got six here. So I've got the spring Force is equal to 240 Newton's, so I've got that. So let's take a look at what happened here, right? And this first example, I had an ex or deformation of six. I had a cake Constant of 20. And then the force was equal to 120. Well, in this example, I've got the same exact information. Six. I've got the cake. Constant was forties double then and the restoring force or sorry, the spring Force was equal to 240. So what happens is I doubled the spring constant, the force constant, And then I doubled the force so you can actually see that just just using the equation itself. F s equals K X. So if the X is the same, and then this thing gets doubled than than the force has to be doubled as well. So what that means is that this k represents how hard it is to to form a spring. You have to use more forced thio reform it the same exact amount. And so the units for that the units for K are gonna be in Newtons per meter. How much Newton's you have to do probably meter of displacement. Now, you should remember that. But if you ever forget, it's just f s equals K X. You could get to that. What are the units for F s? That's just in Newton's. And then you've got K and then the units for X R in meters. So you just move it over to the other side and figure that out. Okay, so in both of these examples here, right, Whether I've pushed up against the spring or I've pulled against it, the force that acts in the opposite direction always wants to pull the system back to the center. And so what that means is that F S is a restoring force. We call that restoring force, and it always opposes the deformation. It always opposes your push or your pull. So that's everything we need to know about Hook's law. Let's go ahead and you do use some examples