This course is an introduction to learning and applying the principles required to solve engineering mechanics problems. Concepts will be applied in this course from previous courses you have taken in basic math and physics. The course addresses the modeling and analysis of static equilibrium problems with an emphasis on real world engineering applications and problem solving. The copyright of all content and materials in this course are owned by either the Georgia Tech Research Corporation or Dr. Wayne Whiteman. By participating in the course or using the content or materials, whether in whole or in part, you agree that you may download and use any content and/or material in this course for your own personal, non-commercial use only in a manner consistent with a student of any academic course. Any other use of the content and materials, including use by other academic universities or entities, is prohibited without express written permission of the Georgia Tech Research Corporation. Interested parties may contact Dr. Wayne Whiteman directly for information regarding the procedure to obtain a non-exclusive license.
Module 12: Moment Due to a Couple
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En provenance du cours de Georgia Institute of Technology
Introduction to Engineering Mechanics
1290 notes
À partir de la leçon
Define and Calculate Moments
In this section, students will learn the definition of a moment. Students will calculate the moment of a force about a point, line or axis, and moment due to a couple.
Rencontrer les enseignants
Dr. Wayne Whiteman, PESenior Academic Professional
Woodruff School of Mechanical Engineering
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.
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