Anderson Video - Work due to Multiple Forces

Professor Anderson
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>> Hello, class. Professor Anderson here. Let's take a look at one of your homework problems. This is a work problem. And in this problem it says you have an object that is going to be pulled a distance, d. And it's being pulled by three forces: F1 at an angle theta 1 with respect to the horizontal, F2 at an angle theta 2 with respect to the horizontal, and then finally F3 directly back along the direction of travel. OK, so it starts there, it goes all the way to d. And we need to calculate the work done by these three forces. So how do we do that? Well, what we remember from class is that work is equal to the integral of F.ds, how far you move the object. Now if it is a constant force, then that F can come right out of the integral. And so this becomes F times the angle between them, cosine theta, times the integral of ds from 0 to d. Remember, the dot product there tells us that we have to include the cosine of the angle between those two. Integral of ds just becomes s evaluated from 0 to d. And so this whole thing becomes what we already know, work is equal to Fd cosine theta. Now again, this is only in the case of a constant force, constant angle that that force is applied, but in this problem that is exactly what we have. All right. So what's the work due to force 1? It is F1, it moves a distance d, and we are given that angle. That angle is between the force and the distance, the displacement vector. And that angle in this case is theta 1. What about the work due to force 2? Well, that is F2d times cosine of theta 2. All right, no problem there. What about the work due to F3? Work due to F3 is F3d cosine of something. Cosine of the angle between the displacement and the force. The displacement is to the right, the force is to the left, and therefore that angle is 180 degrees. Cosine of 180 degrees is -1, and so we just end up with -F3d. All right, you should have all those numbers in your homework and you can plug them in, try it out. If you have any questions, come see me in office hours. Cheers.
>> Hello, class. Professor Anderson here. Let's take a look at one of your homework problems. This is a work problem. And in this problem it says you have an object that is going to be pulled a distance, d. And it's being pulled by three forces: F1 at an angle theta 1 with respect to the horizontal, F2 at an angle theta 2 with respect to the horizontal, and then finally F3 directly back along the direction of travel. OK, so it starts there, it goes all the way to d. And we need to calculate the work done by these three forces. So how do we do that? Well, what we remember from class is that work is equal to the integral of F.ds, how far you move the object. Now if it is a constant force, then that F can come right out of the integral. And so this becomes F times the angle between them, cosine theta, times the integral of ds from 0 to d. Remember, the dot product there tells us that we have to include the cosine of the angle between those two. Integral of ds just becomes s evaluated from 0 to d. And so this whole thing becomes what we already know, work is equal to Fd cosine theta. Now again, this is only in the case of a constant force, constant angle that that force is applied, but in this problem that is exactly what we have. All right. So what's the work due to force 1? It is F1, it moves a distance d, and we are given that angle. That angle is between the force and the distance, the displacement vector. And that angle in this case is theta 1. What about the work due to force 2? Well, that is F2d times cosine of theta 2. All right, no problem there. What about the work due to F3? Work due to F3 is F3d cosine of something. Cosine of the angle between the displacement and the force. The displacement is to the right, the force is to the left, and therefore that angle is 180 degrees. Cosine of 180 degrees is -1, and so we just end up with -F3d. All right, you should have all those numbers in your homework and you can plug them in, try it out. If you have any questions, come see me in office hours. Cheers.