Hi. This is Module 15 of An Introduction to Engineering Mechanics. We're going to have to have one learning outcome today. We're going to solve another problem finding an equivalent system. We did this last time but I'd like to do one more problem because I think that this is a skill that's very important. So again, if you'll recall the conditions for having equivalents systems, s 1 and s 2, is if the sum of the forces are the same on both system, and the sum of the moments about a common point are the same on both systems. So this is another system. where I have a number of forces acting on it and I want to found an equivalent force and couple at point P which together form a, an equivalent system. And so we'll call this system one And I want to find an equivalent system two. Let's draw a picture here again. Just rough sketch. So I would like to find an equivalent force and couple system at point p here. That makes these two systems equivalent. And so let's start off with a condition. The first condition that some of the. Forces on system 1, equals some of the forces one system 2. ,, . And so on system 1. I have a 200 Newton force in the i direction. So 200 i. I've got a 1,000 Newton force down, in the j direction, so minus 1,000. J. I've got 700 Newton force, which i'm going to break into its components. Its x-component is going to be 700 cosine 30 degrees, and its y component is going to be 700 sine 30 degrees. So we're going to have minus 700. Cosin 30 degrees in the i direction, and minus 700 sin 30 degrees in the j direction. And all that's going to equal the sum of the forces on system 2. And, so I'm going to have, I collect my i terms. I've got 200 and then 700 co-sign 30 is minus 606, all in the i direction. And then I've got plus, minus 1000. And then 700 times sin of 30 is minus 350 in the j direction. Sum of the forces in system 2. So the sum of the forces on system 2 equals minus 406. i and minus 1350 j and the units are Newtons. So, I can draw that up on my system over here, I have got minus. 406 or 406 to the left Newtons. And minus 1350, or 1350 Newtons down. So I've taken care of the first condition. Some of the forces on both systems equal the same thing. I've redrawn that up here under system 2 406 Newtons the left 1350 Newtons down. now I have the e equate the these some of the moments about common point p. So I've got, some of the moments. About p on system 1, equals some of the moments. About point p on system 2. Last time, Last module, I did it using, the vector definition of moments. since we're working in 2D. For variety here, let's go ahead, and, use the scalar method this time. Time. And so, when I use the scalar method, I'm going to have to define a direction as positive. I'm going to call, just for assembling my equation I'm going to call clockwise positive. And so, we have, let's see, we've got. About point P. First of all we have this 200 Newton force, but you can see that its line of action goes through the point p and so it's not going to cause a moment about point p. The next force we have is the 1000 Newton force. It's pulling down, so if point p is up here it's going to cause a. Counter-clockwise rotation. So, that's going to be negative in accordance with the sign convention that I'm using for assembling the equation. So, I'm going to have minus 1000, and I've gotta multiply by it's mominar, which is the perpendicular distance between the line of action of that force, the line of action of the 1000, newton force, and the point p. The perpendicular distance is 4. Okay? Then we have 1 more force here. But the 700 Newton force. Its, its y component will be along this line of action. And that line of action also goes through point p. So it will not cause a moment. The only force component of the Sumner-Newton force that's going to cause a moment is the, the y component and you can see this y component's line of action is in this direction so it's going to cause a clockwise rotation about point P so it's going to be positiv e in, in terms of my sign convention, so I'm going to have plus. It's magnitude is 700 Times the cosine of 30 degrees and its moment arm if this is the line of action the perpendicular distance from the point P to its line of action is going to be 3 meters and that together is going to be equivalent to the sum of the moments needs to be equivalent to the sum of the moments about point P on system 2. If you multiply 700 times the cosign of 30 times 3, you're going to get 1820 and so the sum of the moments about P On system 2 is going to equal minus 21 80. So, the minus sign says that it's opposite the direction I assumed is positive assembling the equation. So, it's going to be counter clockwise. Counter clockwise is going to be in the. Minus k direction. So it's going to be minus 2180 k. And the units are Newton-meters. Or I can express it as 2180 Newton-meters counterclockwise. Both of those are correct. So now I've Satisfied the other condition, and I'm going to have a counterclockwise moment at point P, which is 2180 newton meters. Okay, so if I go on through the next slide, you can see that I've put all of those conditions on there. the last thing I want to do again is I'm going to. resolve the 406 Newton and the 1350 Newton force into one, one force. And so we will note that we have 406. And 1350 and that's going to be equal to a single force down into the left, its magnitude will be square root of 406 square plus 1350 square which ends up being. 1410. And it's at an angle. I could put it on the slope. Or I can also express this as an angle. Let's express it as an angle. The angle, theta, will be the inverse tangent of the opposite side, which would be 406. Over the adjacent side which would be 1350 and if you run that in the calculator you will find out that 16.7 degree with 3 significant figures. And so I can re-express my system S2 visually at point P Is having my 2180 Newton meter copper and then a single force with a magnitude of 1410 Newton acting at an angle of 16.7 degree s. From the vertical. And that completes the problem, so now you should be really strong at finding the force and couple, equivalent force and couple for two different systems. And we'll see you at the next module.
