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Welcome back to module 12 of An Introduction to Engineering Mechanics.

Â Today, we're going to define the moment or torque due to a couple. And we're going to

Â find out how to calculate the, the moment, due to a couple using both a scalar method

Â and a vector method. And then we'll solve a problem determining the moment due to a

Â couple. So a moment of a couple is a tendency of a

Â pair of forces to cause a rotation of a body and those forces are equal in, in

Â magnitude, opposite in direction, and they act along parallel lines of actions. And

Â so a good example is a lug wrench here, which I talked about in the introduction

Â to the course. We've already seen how to calculate the moment about a point for a

Â wrench, and so now we get to do the lug wrench problem. And so for this lug

Â wrench, a couple is a pair of forces, equal in magnitude, okay? Opposite in

Â direction, and they act along parallel lines of action. And they're going to have

Â the same effect, as long as they're the same distance apart, they're going to have

Â the same effect no matter where I, I, have them act on the body. And so the couples

Â are what we called a free vector, because they have the same rotation effect

Â anywhere on the body, and to find their, , the magnitude of a couple's, the moment or

Â torque due to a couple, using the scalar approach, it's just the magnitude of those

Â forces times the perpendicular distance between them. And our vector approach is

Â just like before. You take For instance, the moment about point b, as I show up

Â here for a, a, a couple system of force in this direction, and a force in this

Â direction. I can go from b to a, r from b to a, and cross it with f. I could also do

Â r from b to d, or c to a, or c to d, and I'd get the same answer. So we can use

Â both the scalar and the vector approach. So let's use the scalar approach and this

Â is the equation I just showed on the previous slide. here I have a pair of

Â forces F, each with a magnitude of 50 pounds. In part a which we're going to do

Â together, I want to deter mine the total moment in about point A. In parts b, I

Â would like on your own for you to do the movement about point O and, and then

Â answer part C as well, and I've got a PDF of this solution in the module handouts

Â for parts B and C. And so for part a, now I've got my, my lug wrench here, and let's

Â say I'm in a tight situation, so that I'm pulling on the force and on a, on a 4 on 3

Â slope back on this direction, and I'm pushing with a four on three slope in this

Â direction. So as a top view, you see the figure that I've shown on the slide here.

Â And so I want to find the magnitude of the moment about point a, and that's going to

Â be equal to the magnitude of the force times the perpendicular distance between

Â them, and I'll choose a sine convention of, let's choose positive clockwise.

Â Well, to find the perpendicular distance between these two forces geometrically is

Â difficult, so again, I'm going to make take advantage of Varignon's theorem, and

Â look at the components of the force. And so we've got the moment about point a is

Â equal to, well, let's say f, x, d, perpendicular x, plus f, y, d,

Â perpendicular y. And those are these components. This will be, I'm, I'm taking

Â the moment about a, so I'm looking at this force from the right. So this will be my

Â f, x component, and this will be my f, y component. And so, at a magnitude of the

Â moment about a is equal to the f-x component is the 3/5 component of f. So,

Â that's equal to 3/5 times 50 times the perpendicular distance between the line of

Â action of the force f-x, the component force f-x, to the point about which we're

Â rotating. If you take this line of action now, you see that the line of action

Â itself, of that component, goes through point A, so that has 0 perpendicular

Â distance. And then we're going to take the fy component, and you'll see that the fy

Â component of the force is acting down, so it's going to cause a clockwise rotation

Â about point a. And so that's positive in terms of my sine convention, so I'm going

Â to have plus f y is the 4/5 compo nent of the 50 pound force, and now its

Â perpendicular distance will be the distance, between the line of action of

Â the f y component and point a, and so in this case it's 8 inches. That's d

Â perpendicular y. So we've got 4 fifths times 50 times 8, positive because it

Â causes a clockwise rotation and so that's going to be equal to 320. And so the

Â moment about point A, due to this couple, is equal to 320, the units are distances

Â in inches, forces, and pounds. So, it's going to be inch, pounds, and it's causing

Â a clockwise rotation, because it came out as a positive value. Positive by our sine

Â convention was clockwise, so that's, also could be expressed in the k component is

Â minus 320 inch pounds k, because negative k is clockwise or back into the board. And

Â so that's our solution of the moment due to that couple, on this lug wrench, using

Â the scalar approach. Now, let's look at a vector approach. In this case, I'm going

Â to have r from a to b, I'm taking in the moment about point a, so the position

Â vector from a to b, which is just going to be 8i, crossed with the force f. And,

Â we're going to look at the force f on the right here. So, for the force f, we need

Â to express it as a vector. And, so, what we need is a, unit vector in the direction

Â of force f, so that's going to be e of the force f, and that's equal to the position

Â vector along the line of action of that force which is going to be minus 3i. This

Â should be old hat to you by now. Minus 4j over its magnitude, which is 5, square

Â root of 3 squared plus 4 squared. And so the force F as a vector is equal to 50

Â times minus 0.6i, 3 divided by 5 minus 0, 8j.

Â 4 divided by 5 which is minus 30i minus 40j pounds. So, now we have the force, F,

Â expressed in vector form. Minus 30i minus 40j pounds. We know that r from a to b is

Â going to be 8i, and so we can just substitute it in and we get. The moment

Â about a is equal to r a b is 8i crossed with F minus 30i minus 40j. And so, the

Â moment about a, if you do that cross product it is minus 320. Units are force

Â is in pounds, distance is in inches. So this is inch, pounds in the k direction.

Â And so we get the same answers we did with the scalar approach. So now you know how

Â to find the moment due to a couple or application of a lug wrench type problem.

Â