https://youtu.be/BEgIfZC2SXs
PHYS 1101: Lecture Eleven, Part Three
To begin the new material, I’m going to explicitly define for you, a tool we’ve already been using that is so important and useful to you for defining and identifying these forces and starting to picture how you’re going to apply Newton’s Second Law to solve the problem. In this tool I’m talking about is called a free-body diagram. And it’s going to help you be sure you have all your forces identified and drawn correctly. And here are my steps for it.
Just jot down a dot for this object, so picture this as one snapshot out of a motion diagram. Go through your list of identifying forces and be sure you’ve got each one identified. For every force, you need to picture roughly how long to draw the arrow. And very importantly, get the direction right. Identify the direction for this free-body diagram.
After you’ve identified direction, think about the size. Go back and forth and be sure that the vectors that you’ve drawn for these forces match, at least roughly, the acceleration scenario that you have. Really just be sure if the force is balance, which is what gives you no acceleration. So, that would say that the resultant force, just by i, balances to 0. I’m sorry, that goes over here. That’s acceleration of 0.
Force is balancing, no net force, no resultant. So, no acceleration. Or that you have a net force, which then means you have some net acceleration. The direction of this acceleration has to be the same and it will be the same as the direction of that Fnet. Just be sure your picture matches it. So, here’s an example illustrating it.
First, what’s the object? Let’s say we’re focused on this book sitting on a table. Identify my forces and draw forces. Well, here’s my dot for the book in this free-body diagram. I know I have a force due to gravity down. Now, let me picture going around the surface of this book and seeing what’s in contact.
The only contact I have is the table. Table, we’re going to learn any surface has a normal force. There’s a real contact pushing force from any surface. That’s the only thing in contact, so that’s the only other force that I draw.
Quick visual, the size of this force roughly balances the size of that. These two cancel, there’s no other forces. So, my visual says that the acceleration is 0. Does that match the real motion of my book? It does; the book is just sitting there.
So, here’s a couple of what should be fairly easy questions to solidify that idea. Our object is always the book for these questions and I’m showing you different scenarios. And I just want you to tell me how many forces are on the book. I’ve already told you that a surface is responsible for one force pushing up against the book.
So, this first question is the same scenario I just walked you through. How many forces are on this book? Question seven, the only addition now, is that someone’s hand is touching or pushing down on this book. As you follow this imaginary circle around the book, now, I have another contact force. What would be the direction of that? You have to represent the net effect of this hand pushing down with one force. Now how many forces do you have?
Question eight, now two people are pushing down. Again, circle around the book how many places are making contact? There’s one, there’s one, here’s the table. I know I’ve got the force due to gravity. Now how many forces on the book?
Question nine. For each of those scenarios above, what’s the value of the acceleration for the book sitting on the table? Question ten. I know the value of acceleration and you’ve answered question nine. Do this mental check. Is the net force picture consistent with the acceleration? So, what is the net force on the book for each of the three scenarios above?
For the next two quiz questions, I want to emphasize a point about this equation to prepare you for these questions, help you out. So, this is our acceleration, this vector results the vector’s sum of forces divided by m.
So, as I’ve pointed out, that numerator could be replaced by a tail to tip addition or just showing the vector sum, just writing it out for the numerator. All of this gets substituted for this symbol, Sigma and f.
So, this equation, you can think of in two ways. There’s this vector picture or resultant vector picture of this equation, which is drawing the real tail to tip addition of all of these forces. Figure out what the resultant would mean, Fnet. And then, is Fnet pointing in the direction in a? Or conclude, then, that A has to point in the direction of Fnet. That’s what I mean by this first bullet.
But we learned when we started adding vectors together, that we could also use a component approach to addition. We could focus on the horizontal part to this sum and then the vertical part to this sum to represent the resultant or Fnet.
We could apply that same reasoning to this equation, meaning that equivalent to real resultant vector picture of this equation; equivalently, we know this equation has to be given by and will also be true that the sum of all the horizontal parts to all the forces have to add up and then divided by m to give you a(x). And all of the y components to all those forces have to add up. And then, when divided by m, have to give you a(y).
Remember when we write them like this though, that in these one-dimensional pictures or these one-dimensional equations for horizontal and vertical, that these are still vectors and that we have to use the scalar component for these which means there’s going to be a sign associated with each to indicate right/left, resulting in a positive/negative right/left acceleration. And then, similarly, the positive and negative y components represent the up/down direction of each. And when that’s all added up, I’m going to end up with the sign for a(y), that’s going to tell me is a(y), up or down.
I just move this down here a little bit to write out more explicitly for you, what this would be. The Sigma f(x) means you literally replace the whole numerator by this term in parentheses; it’s going to be the sum of all the x components for every force. So, go around your free-body diagram. For every force, determine what its horizontal component is.
If it’s a vertical force, it’ll be 0. But if it’s off at any angle, you may have to some trigonometry. Get the value for each of these correct and the sign. Likewise for the vertical components to determine the vertical acceleration, I have to substitute for the numerator, all of the vertical scalar components for every force.
Once you’ve done that sum, let’s take a(x) of all the positive/negative values. You’ve added those up, you end up with something that’s either positive or negative or maybe it’s 0. Whatever it is, you then have to divide that by the mass, so many kilograms. That then gives you the units for acceleration, gives you the value for acceleration. The sign will tell you if a(x) is up or down.
So, test out that picture, that way of doing it. On the left, I show you two forces. And I’ve got them drawn in such a way where I actually tell you, am I giving you the legend on a map? I tell you that a force that is this long represents one newton.
So, I can look at this force here and conclude that it is 1, 2, 3 newtons is the magnitude of this force. And then, based on how I’ve carefully drawn the second force, you can read off what the x and the y components are and how big those are. So, question 11 and 12 just asks you to carry out the operation of calculating these components for these forces that I show you.