https://youtu.be/g0npcBEyvl0
PHYS 1101: Lecture Twenty-Three, Part Three
Now let’s move on in the lecture and study those aspects. The first thing we need to look at is the size of this force and the direction. How do we mathematically represent that?
I’ve got a sketch here of my wall and my spring and my object in red. The important features you need to define are first the equilibrium position. This is, if the stretch, the spring, is at its natural unstretched length, where does that object sit? We’re going to call that our origin. Call that x=0. Now at any time, at any instant, if this object is at a position, coordinate x, we’re going to call positive x to the right, we need an expression that captures the force from the spring on the block. It’s to the left, so it’s in the opposite direction of x, and it does get bigger the more the spring is stretched.
The right equation that does this for us is right here. It’s called Hooke’s Law, named after somebody by the name of Hooke. Here’s the force. This is the vector information in one dimension. If this is positive or negative, it tells us if the force points to the left or the right. With my origin defined at equilibrium, if I set it a positive x coordinate, then I would put in a positive number here for x. The negative sign here would flip then the direction. It would make this whole thing point in the opposite direction, so that’s then in the negative. That’s what gets my force always to be in the restoring direction, trying to move the object back to equilibrium.
You can appreciate, I think, if we’re at equilibrium and I had instead compressed the spring, if I was sitting here at a negative x value, the -x times this negative ends up with a positive force, and my force direction would flip. I would then have a force to the right, which is what I need for compression. The spring is trying to push me to the right to equilibrium.
Okay. The other thing I want to emphasize is the third variable here of all these that you see is called the spring constant. This is just a number that’s a characteristic of the stiffness of the spring. I’m sure you have the sensation, or appreciate, if this were a very lightweight spring it wouldn’t be able to supply a very large restoring force. If this were a shock absorber on a car, on the other hand, it’s incredibly stiff. Boy, for a very small stretching of this shock absorber, there’s a large restoring force back to equilibrium. k captures that difference, so it’s just a property of the spring. It’s a number. It’s either given to you in the problem, or you may need to solve for it. They may ask you given a certain force and an x characteristic, how stiff is this spring? What’s the spring constant?
Okay. That’s first important equation number 1 in this lecture. A force caused by a spring is described by, it’s equal to, -k times x. x has to be the position coordinate from equilibrium. Okay. Question number 11 for your quiz: what must the units be for k? What units do I have to have here so that when I multiply by meters, a distance, I end up with units of Newtons?
Okay. Let’s see how this equation works through this example. I have a hand exerciser that utilizes…it’s going to be a really stiff coiled spring. It says here that a force of 89 Newtons is required to compress the spring by this amount. Determine the force needed to compress the spring by a different amount. Okay. I would just draw my generic spring picture to represent this. Here’s the end of my spring, and the equilibrium position I’m going to put here, that’s my origin. That’s what equals 0, so I have compressed this spring to this point here. This is my x coordinate with the spring compressed. If it’s compressed by 0.0191 meters this will tell me then that it’s sitting at an x coordinate from the equilibrium that’s going to be minus 0.0191 meters. Okay.
What I actually want in this problem, I need the value of f for an x coordinate of 0.0508 meters. So this problem is fundamentally asking me about the connection between the force caused by a spring given how much it’s compressed, so my fundamental equation that relates those quantity is Hooke’s Law, the force caused by a spring is equal to -k times x, where I’ve defined these variables carefully for you above.
Ultimately I want to solve for f, the value of f, when x has this value of -0.0508 meters, but I don’t know what k is. I do have this extra information, though, that when the x coordinate is -.0191 meters I do know what f is. It’s 89 Newtons. So, in fact, I’m going to go up here beside that value of x and write that this leads to a force of 89 Newtons as part of my knowns.
So the spring is the same. It’s the same stiff spring. I’m not changing its physical properties, so the k value is just a property of the spring. It has the same k value when I compressed it this amount compared to when I’m going to compress it about five times as much. So I can use the initial compression coordinate of -0.0191 meters to determine k. So 89 in Newtons is what I end up with when I have an x coordinate of -0.0191 meters.
The negative signs here would indicate the direction of the force. It is positive. That’s what this would be if I compress this spring. The force on the end of here of the spring would be to the right. The spring would be pushing to try to go back to equilibrium. That’s why this ends up being positive, and I have this negative canceling with that negative. Do a little algebra here, so we know that k then has to be 89 Newtons divided by .0191 meters. That gives us a spring constant of 4,660 Newtons per meter.
Now I do have k, so I can go back to my general equation for a spring and solve for f. So I want f where x is now larger. f is going to be minus this k times that new x. It’s going to be minus 4,660 times my new compression coordinate, my new coordinate, 0.0508 meters. Again the negatives cancel, telling me that my force that the spring applies is positive. When I compress it even more, the x coordinate moves further to the left, so it’s a larger negative value, and the magnitude of the force gets bigger, but it’s still to the right, it’s still positive. When you multiply that out, I end up with, the force that I want is 235 Newtons, 235 apples. That’s a lot.
The next quiz question has you thinking through the nature of these forces caused by springs. You’re going to compare two scenarios. I’ve got the same spring in both problems, so it’s going to have the same k value. Let’s just call it k. Put it over both boxes here. I’m sorry. I should put it over the springs. ks are the same. What we’re doing in this problem is displacing the boxes to the right and then releasing them, meaning here’s my box now moved over to the right. My springs are going to be stretched if they’re on the left. They’re going to be compressed if the spring is on the right. So this left spring is stretched. This spring is compressed, and at the instant they’re released, which scenario represents a greater net force due to the springs? Is it a, b, or are they the same? So you’re looking at the combination of the forces from both springs.