https://youtu.be/TkB8WzjwaOI
PHYS 1101: Lecture Twelve, Part Three
So, let’s move right into our lecture and start with the first example that I want to work for you. I’ll do these examples; a lot of it, just writing out for you and discussing how I’m solving it, the approach to take. But there will be some quiz questions sprinkled throughout.
The first problem is similar to a homework problem that you have already gone over. Let’s look at the scenario where I have two forces that are acting on this four and a half-kilogram block. It’s resting on a frictionless surface. Let me draw in the surface, I think I cropped it out on my picture accidentally.
We’re asked “What’s the magnitude of the horizontal acceleration of the block?” I’m going to go through it with the problem solving steps in mind to emphasize how to apply these steps. Fall back on these steps whenever you get stuck as you work through these problems.
You always have to apply Newton’s Laws to a single object. Often, it’s obvious what that object is. Our case, it’s the block. More advanced problems, it’s going to be less obvious. But be sure you’re focusing on only a single object.
Often, people can intermix it when they go to identify the forces. They’ll picture some forces on the object. But then, actually, forces perhaps that this object is applying to a surface or to something else, can incorrectly get assigned to a force that that object is experiencing. You don’t want to mix that up.
What’s your object? What’s the acceleration for that object? And what are the forces acting on that object? Not what forces it’s responsible for, but what is it experiencing?
Okay, am I after acceleration? Do I have acceleration or not? All I want to do at this point is have a visual picture of equilibrium or not. What’s the magnitude of the horizontal acceleration? Well, let me look here.
If I have acceleration, it’s because I have a net force. A net horizontal acceleration means horizontal components. What do I have horizontally? Well, this whole force is horizontal. This force is up at an angle. But it will have a horizontal and a vertical component.
So, what do I have looking just at the horizontal part? Well, here’s something to the right, let’s call that positive. And here is another force component to the right. Nothing opposing it, it said it was frictionless. So, this thing will accelerate.
I have a motion diagram that looks something like this. Where it’s going faster as time goes on. Meaning my acceleration has to be in the same direction, so it’s to the right. If I go with a standard sine convention positive to the right, that means A will be positive, my V’s are always positive. Fnet will be positive.
So, next step, identify all the forces on the object. Because we’re only asked about the horizontal acceleration, I’m only inclined to focus on horizontal forces and I’ve really already identified them.
Let me put a dot here for the object. And I’ll put in parentheses, we want to emphasize the f(x) part. So, I’m going to draw on my free-body diagram here, that I have a horizontal force that’s 3.7 newtons. But then I also have a force up at this angle of 5.9 newtons that then has a horizontal component that I’m going to need to calculate.
Okay, break a’s and f’s into components. I know my acceleration only has an x component, so let’s go break up and pull out our horizontal components to our forces. Well, the only trick I have to do to get all the horizontal components is trig on this vector. My hypotenuse is 5.9 newtons. That triangle looks something like this, 43 degrees. The resultant hypotenuse is up into the right.
So, my x component, horizontal component is to the right. And the length of this side of the triangle is going to be given by the cosine. I can use the cosine to get that. So, I would write F2x is equal to the hypotenuse times the cosine of 43 degrees.
The magnitude of F2 is 5.9. And then, I have to multiply again by my cosine of 43. And I end up with 4.315. And it is positive because that component is pointing to the right. F1 is entirely in the horizontal direction, so I would just write down here that my other horizontal component is equivalent to that full vector, that resultant. And it has a magnitude of 3.7.
Next step is to make the list of knowns in the variable that you want. I wrote them up there, let me rewrite them again. F1x is 3.7 newtons. F2x was 4.315. And I know I want the acceleration. I also know that the mass of this block was 4.5 kilograms. Okay? Work with the starting equations to solve.
I care about a(x), so the a(x), the horizontal version of Newton’s Second Law, is what will constrain me and define what a(x) has to be. So, a(x), sorry; there’s quiz question six for you. “Which starting equation will you use?” Let’s make that worth four points versus one.
Let’s see 100 percent on that one. Okay, which equation are we going to use? Then quiz question seven is “What is the horizontal acceleration?” So, when you go into work with this equation, to determine what a(x) is, you have to add up all of the horizontal components. It’s a vector equation, so the sine has to be right.
So, let’s say now with that same problem and scenario that we were asked about the normal force and how large it is on the block. Well, the steps above that I did were the same. Other than the free-body diagram I focused on above, I only concentrated on the horizontal components.
When I started asking about the normal force, now I have to be sure I have all of the vertical forces represented. My surface is frictionless. I do know though that I’ve got the force due to gravity down. And then, I’m touching a surface here, so I have to have a normal surface, normal force pushing up from the surface.
So, here is a rough visual, then, of this new free-body diagram where I’ve got F1 and F2 shown. But to that now, I’ve completed my full free-body diagram and drawn in all of the rest of the forces; a normal force and the weight.
I know that the effect of all of these, when added up, has to lead to an acceleration to the right. It equivalently has to lead to Fnet to the right.
Let me just point out and remind you that sometimes, it’s useful to get this visualization of forces balancing up/down, yet adding right/left to leave me with something headed to the right. That I can go in and take forces that are off at an angle. And instead of leaving that vector at the angle, replace it with its two vector components instead.
So, I’ve copied that free-body diagram here to the right. I’m blowing it up so I can show you what I mean by that. Let me go in and note that with a slightly different blue color, this force, F2, has a vertical component that’s approximately that high. And a horizontal component that’s approximately that large to the right.
That means that with those two components, I can replace the effect of the resultant. If you do it this way, now, I can more quickly visualize if the up/down and the right/left forces add up to my acceleration picture. Do they balance and give me the zero? Or do they add up to give me the net a that I need?
It looks like if I imagine adding everything up, then I do end up with something big and long. And that does look like the length of the force due to gravity. Likewise, I only have forces to the right, nothing to the left. So, I do end up with an Fnet to the right and acceleration to the right. Up/down balances, horizontal leaves me with an Fnet to the right.
So, what do I want in this problem? What’s the variable I have to go for? I’m asked for the magnitude of the normal force? That means the variable f(n). I need to determine the magnitude of this vector here.
So, the quiz questions I have now for you just have you finishing out this problem. Question eight is “Given the knowns and given the variable that we need to solve for, which is the magnitude of the normal force represented by the variable f(n), what’s the right starting equation to use?” It has to be the horizontal component equation for Newton’s Second Law or the vertical component of Newton’s Second Law.
Question nine. Just consider, for a moment, the vertical component equation. What should you substitute in for this numerator? The Sigma f(y) for that problem. For this same equation, what should you substitute in for a(y)? And then, question 11, follow it through. What is the value of f(n) that makes this equation true?