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Hi. This is Module 15 of An Introduction to Engineering Mechanics. We're going to

Â have to have one learning outcome today. We're going to solve another problem

Â finding an equivalent system. We did this last time but I'd like to do one more

Â problem because I think that this is a skill that's very important. So again, if

Â you'll recall the conditions for having equivalents systems, s 1 and s 2, is if

Â the sum of the forces are the same on both system, and the sum of the moments about a

Â common point are the same on both systems. So this is another system. where I have a

Â number of forces acting on it and I want to found an equivalent force and couple at

Â point P which together form a, an equivalent system. And so we'll call this

Â system one And I want to find an equivalent system two. Let's draw a

Â picture here again. Just rough sketch. So I would like to find an equivalent force

Â and couple system at point p here. That makes these two systems equivalent. And so

Â let's start off with a condition. The first condition that some of the. Forces

Â on system 1, equals some of the forces one system 2. ,, . And so on system 1. I have

Â a 200 Newton force in the i direction. So 200 i. I've got a 1,000 Newton force down,

Â in the j direction, so minus 1,000. J. I've got 700 Newton force, which i'm going

Â to break into its components. Its x-component is going to be 700 cosine 30

Â degrees, and its y component is going to be 700 sine 30 degrees. So we're going to

Â have minus 700. Cosin 30 degrees in the i direction, and minus 700 sin 30 degrees in

Â the j direction. And all that's going to equal the sum of the forces on system 2.

Â And, so I'm going to have, I collect my i terms. I've got 200 and then 700 co-sign

Â 30 is minus 606, all in the i direction. And then I've got plus, minus 1000. And

Â then 700 times sin of 30 is minus 350 in the j direction. Sum of the forces in

Â system 2. So the sum of the forces on system 2 equals minus 406. i and minus

Â 1350 j and the units are Newtons. So, I can draw that up on my system over here, I

Â have got minus. 406 or 406 to the left Newtons. And minus 1350, or 1350 Newtons

Â down. So I've taken care of the first condition. Some of the forces on both

Â systems equal the same thing. I've redrawn that up here under system 2 406 Newtons

Â the left 1350 Newtons down. now I have the e equate the these some of the moments

Â about common point p. So I've got, some of the moments. About p on system 1, equals

Â some of the moments. About point p on system 2. Last time, Last module, I did it

Â using, the vector definition of moments. since we're working in 2D. For variety

Â here, let's go ahead, and, use the scalar method this time. Time.

Â And so, when I use the scalar method, I'm going to have to define a direction as

Â positive. I'm going to call, just for assembling my equation I'm going to call

Â clockwise positive. And so, we have, let's see, we've got. About point P. First of

Â all we have this 200 Newton force, but you can see that its line of action goes

Â through the point p and so it's not going to cause a moment about point p. The next

Â force we have is the 1000 Newton force. It's pulling down, so if point p is up

Â here it's going to cause a. Counter-clockwise rotation.

Â So, that's going to be negative in accordance with the sign convention that

Â I'm using for assembling the equation. So, I'm going to have minus 1000, and I've

Â gotta multiply by it's mominar, which is the perpendicular distance between the

Â line of action of that force, the line of action of the 1000, newton force, and the

Â point p. The perpendicular distance is 4. Okay? Then we have 1 more force here. But

Â the 700 Newton force. Its, its y component will be along this line of action. And

Â that line of action also goes through point p. So it will not cause a moment.

Â The only force component of the Sumner-Newton force that's going to cause

Â a moment is the, the y component and you can see this y component's line of action

Â is in this direction so it's going to cause a clockwise rotation about point P

Â so it's going to be positiv e in, in terms of my sign convention, so I'm going to

Â have plus. It's magnitude is 700 Times the cosine of 30 degrees and its moment arm if

Â this is the line of action the perpendicular distance from the point P to

Â its line of action is going to be 3 meters and that together is going to be

Â equivalent to the sum of the moments needs to be equivalent to the sum of the moments

Â about point P on system 2. If you multiply 700 times the cosign of 30 times 3, you're

Â going to get 1820 and so the sum of the moments about P On system 2 is going to

Â equal minus 21 80. So, the minus sign says that it's opposite the direction I assumed

Â is positive assembling the equation. So, it's going to be counter clockwise.

Â Counter clockwise is going to be in the. Minus k direction. So it's going to be

Â minus 2180 k. And the units are Newton-meters. Or I can express it as 2180

Â Newton-meters counterclockwise. Both of those are correct. So now I've Satisfied

Â the other condition, and I'm going to have a counterclockwise moment at point P,

Â which is 2180 newton meters. Okay, so if I go on through the next slide, you can see

Â that I've put all of those conditions on there. the last thing I want to do again

Â is I'm going to. resolve the 406 Newton and the 1350 Newton force into one, one

Â force. And so we will note that we have 406. And 1350 and that's going to be equal

Â to a single force down into the left, its magnitude will be square root of 406

Â square plus 1350 square which ends up being. 1410.

Â And it's at an angle. I could put it on the slope. Or I can also express this as

Â an angle. Let's express it as an angle. The angle, theta, will be the inverse

Â tangent of the opposite side, which would be 406. Over the adjacent side which would

Â be 1350 and if you run that in the calculator you will find out that 16.7

Â degree with 3 significant figures. And so I can re-express my system S2 visually at

Â point P Is having my 2180 Newton meter copper and then a single force with a

Â magnitude of 1410 Newton acting at an angle of 16.7 degree s. From the vertical.

Â And that completes the problem, so now you should be really strong at finding the

Â force and couple, equivalent force and couple for two different systems. And

Â we'll see you at the next module.

Â