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 6: Systems of Particles Equilibrium
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From the course by Georgia Institute of Technology
Introduction to Engineering Mechanics
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From the lesson
Forces and Particle Equilibrium
In this section, students will learn the definition of a force and how to represent a force as a vector in two (2D) and three (3D) dimensions. Students will learn the concept of particle equilibrium and equilibrium of systems of particles. Concepts will be reinforced with example problems.
Meet the Instructors
Dr. Wayne Whiteman, PESenior Academic Professional
Woodruff School of Mechanical Engineering
Hi and welcome to Module 6 of An Introduction to Engineering Mechanics.
Today's learning objective is to calculate the forces acting on systems of particles
or multibody. Problems. This is a typical multibody problem. this
is an example from a, a textbook by Doctors Dave McGill and Wilton King.
Although I said in this Coursera course you didn't, there was no required textbook
I am going to use several figures and examples From, a textbook called
engineering mechanics, statics by, doctors McGill and King, and, I will give, I, I'll
put a little note on, below those figures, to sh, to show when I'm using those, but,
I know doctors, McGill and King, and they were kind enough. To allow me permission
to use material from their textbooks in this course. both of those gentlemen are
professors emeritus at Georgia Tech and so it's great that they're supporting this
Corsera course. Okay so here's the situation. We have this, this, this person
who is exerting a force in this rope. And holding up these systems of pulleys where
there's a weight, which would be known, weight 1 here on the right and a weight 2
here which is greater than weight 1. So if we go over here to a model. Got a model of
this situation so I'd be the, the giant person if you will, over here on the, on
the left hand side. That's holding this cable, as it's attached, and holding this
system up with its 2 weights. So, let's go back over here, and look at the
assumptions that we would make in solving this problem. Typically, for these
problems, we, we, we make several simplifying assumptions, these, pulley
problems, to help us to. Getting a, a, a, a simple but an, a fairly accurate
solution. If we find that these assumptions are, are not good, and we need
more accurate an, a, more accurate answers, we may need to go back, and, and
relook and include some of this some of these forces and things that we're
neglecting. So, one of them is that we assume that there's frictionless bearing,
bearings in these problems. We assume that there's a, a constant tension throughout
the cable. Or the belt or the rope, and we neglect the pulley weights. And so let's
look at these assumptions again over here on, on, on my model. And so again, if I'm
the, the, the man here the giant person over here on the left. I'm holding this
cable and we're saying that the, or this string if you will and we're saying that
the tension in this string is, is the same throughout, over to here, and the tension
in this String or cable is the same throughout. We're assuming that these
bearings are frictionless, okay. We're also neglecting the pulley weights. We're
saying that the pulley weights are, are much, insignificant when compared to the
weights themselves. And so those, those are some, some good assumptions to
simplify the problem and still give us a fairly accurate answer. So in solving this
problem, we take our system which has to be completely in static equilibrium and we
apply the equations of equilibrium, and we talked about these last module. we must
have a balance of forces in the x direction and a balance of forces in the y
direction and so let me draw my x and y directions here. Here.
Lets call the left x, and to the y up. And so this whole system has to be an
equilibrium, and each piece of the system must be each, an, an equilibrium, to
satisfy static equilibrium. And so, what I'm going to do, is I'm going to separate
this thing out, and I'm going to look at just 1 little piece here. I'm going to
look at this Pulley P1, and I'm going to draw a free body diagram of that. So this
is a free body diagram of pulley 1, or pulley P1 if you will. And so I separate
that pulley from the rest of the world here, so I'm drawing it Unconstrained.
Here is the center. And now I apply the external forces that are acting on that
pulley. And so I see that on the r-, on the right hand side here I have this, this
chain if you will and so if I cut that chain there's a tension in that chain
which is equal to The weight one, and that's going to be pulling down on the
right-hand side of this pulley. So, I've got, w1 here. One of our simplifying
assumptions was that the tension in that cable is the same throughout, so if I cut
it over on this left-hand side and apply a force reaction there, I get w1 down. I
have this a connection of the set of the pulley to the ceiling and so I have to cut
that and there'd be tension in that, I called that tension 3 and then I have a
tension in this other cable which is coming down from. the center of the pulley
and if I cut that, that tension, I'll call t2. And I want you to note on t2 that if I
follow the tension in t2 all the way around here, that it's actually the
tension or the force that the person must exert to hold the system in equilibrium,
so let's make a note of that. We'll note that T2 is the force the person exerts. So
that's going to be our answer. , So now we apply our equations of equilibrium, as I
mentioned. here they are on the right. We've done 'em in the last module. We
don't have any forces in the x direction in this problem, so we don't have to worry
about this first equation of equilibrium, so we'll just sum forces in the y
direction. And, we'll make sure they're balanced, so we'll set them equal to 0. I
have to choose a sign convention for assembling my equations. This is
arbitrary, and I'll show you that it doesn't matter when we do this. So, what
I'll do to start here, is I'll just choose up, as being positive. And so when I
assemble my equation, I have T3 was, is up, so that's going to be positive. And
then I have weight one is down, so that's going to be negative and in accordance
with my sign convention. T2 is down so that's negative, and the weight one on the
other side is down, so it's negative. And all that has to equal zero. And so I end
up with T3 = 2 W1. Plus T2. And so that's my equation that results
from that force balance. Just to show you that it wouldn't have mattered If I had
chosen my sign convention as being down arbitrarily, let's just redo that. So I
have some of the forces in the y = 0, I'll choose down is being positive. And I would
get, well in this case T3 is up, so it would be negative. -T3.
W sub 1 is down, so that's positive in accordance with this sign convention.
That's plus W1. T2 is down, so that's plus T2. W1 is down so that's plus W1 = 0. And
you'll note, that if I carry T3 to the other side, add the T1s, W1s together that
these two equations are exactly the same. And so you get the same result regardless
of the sign conven-, convention that you've arbitrarily chose to assemble the
equation. Okay, so, the equation that results though, we have a problem here
because we have, how many, equations? Well we have one, equation. And let's count the
number of unknowns. Well, we would know W1. We'd know that weight. Given in the
problem, but we don't, don't know, tension T2, which is what we want to find. That's
the force the person exerts, and we don't know tension T3. So we have 1 equations, 2
unknowns, and so we have unhappy face. and so what do we do next? And you may want to
think about this for a second and then come back. Try to think about what we
would do next. Okay, what you thought about that what you need to do in this
case is. Okay, every piece of this system has to be in eq. Let's take another piece
and draw another free body diagram, so we're going to try. Another free body
diagram, and the one I'll choose next is, is pulley p2. So, at the top here I've
written my analysis from pulley p1. I, I arrived at the equation t3 = t. 2W1 + T2,
so 2 equations, or 1 equation, 2 unknowns. Now lets do a free body diagram of Pulley
P2. ,, , And in fact I'd like you to take a few minutes on your own and draw that
free body diagram and then come back we'll do it together. Okay, now that you have
completed that here is my body pulley P2, here is the centre, I apply my external
forces I have, this was t2, and on Pulley 1 in tension, so it was pulling down on
Pulley 1, it's going to be pulling up on Pulley 2. This tension in that cable is
the same throu ghout, so it's also, if I cut it on this left-hand side it's going
to pull up on the left-hand side t2. In the middle here, if I cut this chain
remember the tension in the, in this chain was the same throughout and it was equal
to W1. So that's up W1. and then finally I have tension is this rope which is hooked
to weight 2 and so that's going to be down. And that's weight too, so that's a
good body diagram, and that's what you should have arrived at. Once we've got
that, we can assemble our, our equation again. We'll sum, there's no forces in the
x, so we'll only sum forces in the y direction. Set it = to 0.
I'll choose up as being positive. And so, I've got T2, positive, because it's up,
plus W sub 1, positive because it's up, plus T2 - W2 = 0. And if I solve for T2
now, I get T2 = W2 - W1 / 2. And so that's my answer. So if I'm given the values for
the weight 1 and weight 2, I, I subtract the 2, and divide by 2. That's going to
give me the force that the person must exert. And you'll notice that I get a
bonus in this, that I can actually, go up here, and put the result for T2, into this
equation, and find out what T3 is.
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