https://youtu.be/ZXrBw1hQwWM
PHYS 1101: Lecture Eleven, Part Nine
I’m going to finish the lecture out by just doing an example for you, trying to put these pieces together. Here’s our problem, a football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes. The friction force on the sled is 1000 newtons, and the angle between the two ropes is 20 degrees. How hard must each player pull to drag the coach at a steady 2 meters per second?
There’s a typo here, steady. Okay, first of all, I know I have to start with my F equals MA problem solving steps. These are the basic equations I have to start with and work with. I know that because there’s a mention of forces.
Here it is, it talks about friction force, ropes, it talks about, I need the force that they’re pulling on these ropes with, how hard must each player pull on it. That’s going to be the tension in those ropes. I have to start with my F equals MA problem starting steps. So, here, I’ve pulled out these steps, and let’s go through them one at a time.
What’s the object? Should you be focusing on the coach? One of the players? The rope? The sled? Look to the final question, how hard must each player pull to drag the coach at a steady speed of this. Okay, it’s the motion of the coach sitting on the sled. I want you to focus on the coach and the sled as really being one object. And, I emphasize including the sled, because I’m given information about forces on the sled.
I’m not really told about forces directly on the coach. So, I’m going to consider the sled with the coach sitting in it as really being just one object. I’m just going to make it a red box as my object because I’m told about forces on this object and I need to know how big some of the forces must be on this object in order to make this true, that it’s moving along at a steady pace.
Okay, step two, what’s the acceleration? Is it an equilibrium or isn’t it? Reading this last line, I know that it is and equilibrium. I have to picture this object moving along at a constant velocity; it’s moving at a steady pace.
Here’s my quick motion diagram, velocity is not changing, acceleration is not 0, I’m worried about equilibrium here. Equilibrium means Ax is 0 and Ay is zero. So, when I look at these equations, then the left side is 0, in both scenarios, that’s what gives me then, that the forces have to balance. All the horizontal parts have to add up to 0, and all the vertical parts have to add up to 0.
When the left side is 0, the mass just drops out, doesn’t matter. You algebraically can picture multiplying both sides by M to get rid of it. On the right 0 times M is still 0 on the left. So, I’m going to jot down, right here, and looking ahead a bit, but because I’m in equilibrium, my Ax equation becomes 0 equals the sum of all the horizontal x components. My mass cancels out. Similarly my Ay equation becomes 0 equals the balance of all the y, the vertical components. The plus and minus quantities of the forces vertically have to add up to 0. Same thing in x.
Okay, the next step, I have to identify all the forces on the object if I’m going to be able to implement the right side of these equations properly. What are all the forces? Here’s my other free body diagram, so I’m going to use this dot to represent the sled and the coach.
What are my forces? I know that I have a frictional force, the direction which would oppose the motion. And I know the magnitude of that force. That’s a force I’m going to draw pointing straight to the left in the direction opposing the motion that I’ve sketched. It’s kinetic friction because there is real sliding between the bottom of the sled and the ground.
What other forces? What else is touching this object? If this is my sled, aside from the rough grass underneath the in contact, I know that there are two players pulling on this sled and the angle between the ropes that they’re pulling with is 20 degrees. So, I could sketch onto my object here. Let me go ahead and do that. Here’s my sled, I know I’ve got a player pulling on the ropes, the angle between both of these ropes is 20 degrees and they’re pulling it across this rough surface, lawn, whatever it is.
Okay, for each of these ropes, I have a tension force. I’m going to call these two tension forces T1 and T2, for the two players. And now just looking at that, I’m going to do a quick visual vector balance here, and see if it matches my A. I know that the net force has to be 0. Everything has to cancel out.
What do I have horizontally? Horizontally, I have a large force to the left. What’s the horizontal parts of these two? Those are what have to add up to balance this negative horizontal part. I have, for the top force, I have a horizontal part about that long. For the bottom force, I also have a horizontal part about that long. So, that length plus this, the two positive vectors, are they roughly the length of this one vector? Yeah, it looks pretty good.
That looks good for Ax equals 0. How about Ay? Ay would be the vertical parts of these forces. This tension force has a vertical part, roughly that long. This tension force has the vertical part that’s down, and roughly that long. This has no vertical component, so my only vertical components are minus this amount, plus this amount, and that’s a good drawing. Those look like they balance. So, I’m going to check these off. Ay looks like it’s 0 for my drawing.
Okay, the implications of that, just from this drawing then, is that the hypotenuse of these two vectors have to be the same. If the angle is split equally, 10 degrees, to make up my 20 degrees, then the hypotenuse has to be the same. But, we’ll see that in the math here in just a minute, the verification of that. I just moved my free-body diagram down and made it a little bigger for us to look at.
Okay, break A’s and F’s into components, that’s step four. Often, these problems, you’ll end up having vertical and horizontal vectors. Which is nice, and you don’t have any trig to do. For this problem, I do have some trig. I’ve got these ropes pulling on the sled off at an angle of 10 degrees. I’ve got to do that trigonometry.
Let me re-draw that triangle. Here’s the triangle for T1. I know the magnitude of the hypotenuse, I’m going to call T1, and I know that this is 10 degrees. So, T1y, and T1x, I need to get for my trigonometry. T1x is the adjacent side, so that’s equal to the hypotenuse T1 times the cosine of 10 degrees. T1y is equal to positive T1 times the sine of 10 degrees.
Okay, here’s a safety tip for you. This is an example of a problem where we’re supposed to solve for the tension in this line, the hypotenuse, the magnitude of this result in force. When I get down to doing trig, I can’t calculate a final number for that, because I don’t know what the hypotenuse is. But by writing it out this way, I have represented that hypotenuse with a variable, T1, because when I go to plug it into my equations, I’m going to have to plug it in the horizontal parts for my Ax, and the vertical parts for my Ay. And, this is what they represent, the hypotenuse times that angle.
By doing that, I have a variable that represents what I want, and I have it in my equation now. You’ll see more, maybe in a minute.
The other tension is down and at an angle, T2. So, similarly, T2y and T2x, if this is 10 degrees, this is 10 degrees, so T2x is positive, the hypotenuse times the cosine of 10. And, T2y, it’s now going to be negative hypotenuse cosine of 10.
So, I’ve really done step four now. I’ve thought about every vector in my problem, and I’ve thought through any trigonometry if any of those vectors are at an angle. Acceleration vector is 0, my friction force is all in the horizontal direction. The only ones that have any possible trig to do or to represent are T1 and T2. And I’ve done that. I’ve represented my components in terms of the hypotenuse, its resultant, and the geometry, the angle.
So, let’s make a list of our knows and the variable that we want. I’m going to do that, x and y. X components, I have a minus. Well I know that I, kinetic friction is going to be, the magnitude of it is 1000 newtons, and when I go to put that in my equation, I need to be sure to make this negative. I could write a negative here if I wanted to. At this point, that’d be okay.
Also, in the x direction, I have positive, I have a T1x, which is represented by T1 cosine of 10 degrees, and I have a T2x represented by T2 cosine of 10 degrees. On the y equations I only have a T1y, is T1 sine of 10 degrees. Oops, I just realized a type here, T2y is the sine.
Okay, what I’m working these kinds of problems, my rule of thumb is, I often write down, as my list of columns here, just the values and the magnitudes of these vectors, and then when I go to put them in the equations, that’s when I explicitly show the proper sign to represent the direction.
Now, what variable do we need to focus on? We need T1 and T2. We need to know how hard each player must pull to drag the coach. How hard they’re pulling is represented by the tension that they put in these ropes, and I need to know for each of the players, what that tension has to be. It’s going to be so many newtons. It’s going to be some point of holding up so many apples to represent how hard that pull is, how strong that pull is.
So, now, we start with our equations. I’m going to do x, the starting x equation on the left. And then one at a time, I’m going to go through the x and the y. Let’s start over here on the y. No special reason to, I’m just going to. On the y equation, maybe I’m drawn to that because I know I only have two forces here to include.
So, what is my y equation? It’s Ay is equal to the sum of all the y forces divided by N. I already walked you through the situation that if we’re in equilibrium, the component for acceleration is 0. If I multiply both sides by M, I’ve removed M from the right, M times 0 is 0, so this whole left side continues to be 0. For equilibrium, I end up with simpler equation to work with, which is just that the components have to balance.
So, what does that say for my case? 0 has to be equal to, and here, now, is where I have to be careful about being sure that whatever I write in for each force, I have the right sign to represent direction. T1y and T2y, well, T1 pointed up and to the right, positive Y direction to find up, that means I have to make T1y positive, and T2y negative.
So, I’m going to write plus T1y minus T2y and then I’m going to write in, I’m going to substitute in, from my list of knowns what we know about those. I know that T1y is equivalent to this trigonometric expression of the hypotenuse times the sine of 10 degrees. And then for T2y, it’s the hypotenuse, the tension in that resultant, time the sine of 10 degrees as well.
Okay, doing my algebra, I see I can add this whole term to the other side, I end up with T2 sine of 10 degrees is equal to T1 sine of 10 degrees. The sine I can divide through on both sides, that cancels out, and it ends up just telling me what my diagram already suggested had to be true. These two players have to pull with the same amount of tension, with the same force. T1 and T2 have to be equal.
So, you know what I might do at this point is just call a simple variable T equal to the tension that’s in each of those lines. They’re all the same, but I’m just going to call it T so I don’t clutter up the subscripts.
Okay, but that still doesn’t give me a number for T, it just tells me that they have to pull the same. Now, when I’ve run out of steam in one direction, I then am going to go over and start exercising my other equation and see what it tells me. I know I’m in a equilibrium horizontally as well, any equilibrium scenario, that equation ends up being the balance of the force components in that direction.
What did I have horizontally? You can scroll back or look back in your notes. I know I had my frictional force, which was 1000 newtons, that it was pointing to the left. So, I’m putting the sign in here, when I plug in here explicitly. I have minus 1000 newtons plus T1 times the cosine of 10 degrees plus T2 time the cosine of 10 degrees.
Now, I had just learned that T1 and T2 are the same thing, I’m just going to substitute in then, and call that T. These two terms are exactly the same, so that’s equivalent to writing that I have two of those, combining them together. I’m trying to solve for the tension in the ropes, that’s variable T. Notice it’s the hypotenuse of that vector that did end up in my equation to give me a variable to focus on the answer the problem.
I’m going to add that to the other side. I end up with 1000 newtons is equal to 2 times T times the cosine of 10 degrees. Dividing through by 2 times the cosine of 10 degrees and I end up with that T is equal to 1000 divided by 2 times the cosine of 10 degrees. And, that turns out to be about 508 newtons. It’s about 508 apples. That’s big, that’s your answer. That is how hard each player must be pulling on the rope. It’s a force that has to have units of newtons. Checking through my units, I see that the answer was in newtons, the units are good. 508 apples, that’s probably reasonable to pull a heavy coach and a heavy sled across a grassy lawn, maybe.
A lot of friction I had here, I had the equivalent of 1000 apples worth of friction. It’s probably reasonable. So, for here I would check it. That’s all I can do. I think it’s probably right, and I’d go with that answer. And that brings me to the end of Lecture 11.