https://youtu.be/WCkV72FYIM8
PHYS 1101: Lecture Fifteen, Part Three
Let me directly show you how to do that in the context of an example. Let’s say we’ve got a problem where a 5-kilogram box starts from rest, and it’s pulled across a frictionless floor with a rope that’s at an angle of 30 degrees to the horizontal. The tension in the rope is 50 Newtons. How fast is it going after it’s moved 2 meters?
First point, notice the words here, the connection between position. It’s moved 2 meters. I have information about speeds, the initial speed, I’m being asked about a final speed, and I see the position change by 2 meters. So, the speed and position focus to this problem is my first indication that conservation of energy is going to be the best tool to apply to solve the problem. That means that my starting equation is going to be this net work on the left, is equal to, goes into, changing the kinetic energy. It’s equal to the final kinetic energy minus the initial.
The key, though, once we have multiple forces, is that work net must include all contributions to the speed change. So, we’re going to have to go through and think through carefully for this object, identify all the forces on it, and then consider them one at a time and say to what extent does that force speed up or slow down this object. So, here are the steps to do that.
Draw yourself the free body diagram first. So, play back the movie. Quiz question number 3 is about that. What is the right free body diagram for this box? The next step to doing that is identifying the displacement vector, because we’re after only the part of the force that’s in the direction of motion. Only that part is going to contribute to speeding up or slowing down this object. So, you need to, very carefully and specifically, identify the direction that this object moves.
So, as you read that problem again about the box starting from rest, it’s pulled. Let’s assume it’s pulled to the right. What’s the right displacement vector? Remember, a displacement vector is a vector that starts where the box is initially. It’s a straight line that then ends at the end of the arrow, being the location where the box ends up. For this box being pushed to the right, what’s the right displacement vector? Got choices A through D. We were told that the box is going to be moved 2 meters. That’s telling us, then, the magnitude of this displacement vector.
To plug into the left side of our equation, we have to add up the work done by all of these forces. So, follow these bullets, and you won’t go wrong. So, have your free body diagram drawn on your page for this object. Be sure you’ve got every force represented. Go around that free body diagram, and every force that you run into, you’re going to add a term to the net work. The net work, in other words, has to be the sum of the work that each force does. Notice, for convenience or just to help me remember, I jot down a little subscript here that reminds me that this is the work that the normal force, this force, would do.
I’m going to add to that, the work that the gravitational force, the weight, may do. I’m going to add to that, the work that any other force, like the tension, the rope, may do, and etc. This should include every force on your object. Now, some of these are going to be 0, but write them down like that, just so you explicitly are just careful not to leave anything out.
Then, what you have to do is to calculate each of these. You have to go in and find out and ask yourself what part of that force is in the direction of the displacement. That’s the part of the force that is the force times distance that will then represent how much work that force does on the object. That’s this last bullet. You’re doing that because it’s only that part of the force, in the direction of s, that’s causing the speed to change. It’s slowing it down or speeding it up.
So, here’s what you’ll find. This is a reminder of the meaning of the sign. In other words, some of these forces will contribute 0 work. Some could contribute positive work. Some terms may end up being negative. What’s the meaning of those signs? Well, a force will do positive work if it contributes to increasing the speed. Effectively, this means that that force is adding energy to the object. You’ll see that we’ll calculate that some forces do negative work. What that physically represents is that that force is working to decrease the speed. That translates to removing energy from the object.
So, let’s go through these forces one at a time for this specific problem, and think this through. We’re slowly building up the intuition, the ability to properly use the left side of the equation or customize the left side of the equation to our problem. If you read the problem again, here’s a good representation of the free body diagram. The box is sliding to the right along a frictionless surface. I have the force of gravity down. The rope is pulling up at an angle. Then, I know it’s touching the ground. There’s no friction. The ground always has a normal force. A surface always pushes against the object, up and away from that surface, normal or perpendicular to the surface.
Okay, so here I’ve got drawn what I recommend above that you do. I’ve got my free body diagram, and I’ve got my vector that represents the displacement. So, let’s go through these forces one at a time and ask yourself these questions 5 through 6. Start with the normal force. Does the normal force have a component that’s in the direction of s? This is asking you to consider an axis, like an x-axis for example, that’s aligned to s. Is there an x component, then, to this force? That’s what it means to say that does this force have a component in the direction of this. Question 5 has to do with FN.
Question 6, then, asks about the tension. Does the tension have a component in the s direction?
Then, continuing around, we get to the force of gravity. Does the weight have a component in the direction of s? If you answered “no” to any of those, either of those 3, any of those 3 questions, then that means that the work then that that force does has to be 0. It can’t be doing anything to speed up or slow down that object. It’ll have 0 work.
The next quiz question words it slightly differently. Of that list of forces, there were three of them: the normal force, the tension, the force due to gravity. Which force or forces directly contributes to the change in velocity or the acceleration? What is the motion diagram for this box? What direction is the acceleration vector? What force or parts of the forces is in the direction of a?
What you’ll find, or you should find, is that your answer to question 8 should be consistent with the only force that has a component in the direction of s. Okay. Well, let’s take one of those forces. Let’s take the tension that’s caused by the rope that’s pulling on this box. That force is up and to the right, and we’re going to use T to represent that hypotenuse, if you will. In order to calculate the work that this force does on this object as it moves from here to here, is we need to find the part of the force that’s parallel to s, which means finding this component.
So, we’ve done a lot of trigonometry in this class so far. You should be able to do this. I want to picture my axes being one parallel to s and one perpendicular. You need to find this component of that vector. Well, that’s the horizontal component. That’s the cosine. It’s the side that’s adjacent to this angle. So, the x component is the hypotenuse times the cosine of 30 degrees. It’s only this part of that force that has any impact on speeding it up or slowing it down. That’s the part that we need.
So, here’s the steps that you can follow to always get the right answer, to always be able to get the right part of the force that’s doing work. Step 1: slide the force of interest, such as this tension, slide it over so that the tail of the force matches or is lined up with the tail of s. So, in other words, this case, my displacement was to the right. That’s my displacement vector. Draw the tension with the proper angle, the proper relative length compared to the other forces. Slide it around so that the tail is next to s, like hands of a clock. Then go in and start at s and swing up until you get to that force. That angle that you’ve just swept through, call it