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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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