 So I want to take a look at an example problem. And the example problem is basically the design or selection of a belt given this scenario. So we have a motor driving a belt and pulley system. It's rotating at a speed of 1750 rpm. The wrap angle of wrap on here is 165 degrees. And we're talking about the pulley that's on the motor. And we've mentioned already that it's the smaller pulley that we usually want to consider. So we're using V belts with a beta angle of 18 degrees, a size standard given as 5V, the unit weight, the power that we want in, and the maximum tension that we can support with our belt. And a friction coefficient of 0.2. The book uses f here. I'm going to replace that with mu because that makes more sense to me. And then the question is how many belts do we need in this system in order to be able to handle this supplied mode? So one of the first things we want to do is start filling in our equations. So the first thing I'm going to determine are centrifugal tension, which is m prime omega squared r squared. And we can start populating that we have 0.012 pounds per inch. Note that this is given in pounds. So it's a weight rather than a mass. And mass is what we need. So we'll divide that by 386 inches per second squared, which is the acceleration of gravity in inches per second squared. I'm going to multiply that by 1750 rpm times 2 pi over 60 to do the unit conversion to radians per second. And all of that is squared. And times r, we're given the diameter of 3.7 inches. So we'll divide that by 2 and square it. All right. So doing that calculation, then we come up with 3.57 pounds of tension due to just the rotational velocity. All right. Our next equation, p1 minus pc over p2 minus pc equals e to the mu phi over sine beta. Okay. And we've specified that p1, our maximum tension, is 150 pounds. So we can substitute that in for p1. Now we don't have p2. So in this case, that's actually what we're solving for is this p2, which gives us an understanding of what's going on. So we have 150 pounds minus 3.57 divided by p2 minus 3.57. And that equals e to the friction, which is 0.2 times the angle. Angle was 165 degrees. That is 2.88 radians divided by sine of 18 degrees. Okay. So just in case the question comes up in your head in terms of why did I have to convert my 165 degree angle to radians, but I didn't convert the 18 degrees for beta. And that is because the 165 degree angle is just being multiplied within that system. It's mu phi over sine beta. So I have to have it in radians because multiplying by degrees doesn't really work mathematically. I didn't have to convert it when I carry out the sine, assuming that my calculator is in degree mode and therefore the answer will come out when I take sine of 18 degrees. So just in case that was a point of confusion, that's why I had to convert one of the angles but not convert the other one. Okay. So rearranging and solving this equation for p2, basically multiplying this denominator p2 minus 3.57 across to the right hand side, dividing this whole right hand side back over, and then adding the 3.57 over once I've carried all that out, I find that p2 is equal to 26.3 pounds. So this is useful because now I can take and plug that into my torque equation, p1 minus p2 times r, and I get 150 minus 26.3 times 3.7 over 2. And this comes out to be 229 inch pounds. And next step is to calculate how much power I can therefore carry because this is the torque I can expect to transmit with a single belt. And the power that I can get from that then is t times n over 52, 52. So this is a standard power equation with some unit conversions built in. And it's equal to 229 inch pounds divided by 12 inches per foot to put that in foot pounds times 1750 rpm. So this equation with this 52, 52 works when I'm using foot pounds for torque and rpm for velocity, rotational velocity. And this comes out to be 6.36 horsepower, again built into the unit conversion there. And that is per belt. So I've calculated what my power capacity is per belt. And if we go back up here, the problem said that we need to input 25 horsepower into the system. That's what we're intending to drive with the system. So we simply take the 25 horsepower divided by the 6.36, which is I guess horsepower per belt. And we get 3.93, which of course we can't have a fractional part of a belt. So that means we need four belts. Of this particular design or of this particular, these particular dimensions, we need four belts in order to be able to drive 25 horsepower through this system. So of course, this is all based on the fact that the specific belt that we would use was specified in the problem. And we specified what its maximum tension carrying capacity is. You know, if we had options, we could potentially choose a different belt with a higher tension rating, which would allow us to maybe use fewer belts. But given the limitations of the problem, this is the requirement for belts to be able to carry 25 horsepower. Alright, thanks.