Module 14: Conditions for Equivalent Equations

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Do curso por Georgia Institute of Technology

Introdução à engenharia mecânica

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Georgia Institute of Technology

Introdução à engenharia mecânica

1120 ratings

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.

Na lição

Equilibrium and Equivalence of Force Systems

In this section, students will learn the equilibrium equations in two (2D) and three (3D) dimensions. Students will solve equivalent system problems. System force results will be defined and calculated.

Module 13: 2D and 3D Equilibrium Equations4:34

Module 14: Conditions for Equivalent Equations9:21

Module 15: Solve an Equivalent Systems Problem10:34

Module 16: Single Force Resultants-Coplanar System5:08

Module 17: Single Force Resultants-Parallel Force Systems7:10

Module 18: Distributed Force Systems9:22

Conheça os instrutores

Dr. Wayne Whiteman, PE
Senior Academic Professional
Woodruff School of Mechanical Engineering

Hi. This is an introduction to engineering

mechanics, and this is module 14. Today's learning outcomes are to, first of all,

describe the conditions for 2 systems to be con-, con-, considered, equivalent

systems, and then to solve a problem finding an unequivalent system. So first

of all, equivalent systems are defined as, as being such if the sum of the forces

acting on, on one system is equal to the sum of the forces acting on the other

system. And, if the sum of the moments about a common point p on the one, the

system one and the system two are the same. And, if both of those conditions are

met, we have an equivalent system. So let's go ahead and do an example. We're

going to find a force and a couple at point C here, which is together form a

system which is equivalent to this system, where we have a hundred foot pound couple

here and a two hundred pound force in the I direction and a three hundred and fifty

pound force in the J direction up here. So we'll call this system 1, and we want to

find an equivalent system too at point c. Let's go ahead and draw. I can't leave her

being there again, and this will be, be point C, and it's 3 feet from the left

edge. And so our condition is that the sum of the forces. First condition is that the

sum of the forces on system one equals the sum of the forces on system two. And so,

for the sum of the forces on system one, I only have two forces. I have a 200 pound

force in the i direction and a 350 pound force in the j direction. So we've got

200i plus 350j pounds equals the sum of the forces, on system 2 and so I can, go

ahead and draw that. On my system 2 now, so I have a 200 pound force here, and a

350 pound force here. So that's the first condition, now we have to apply the second

condition for the sum of the moments about the common point c between both systems.

And so I'll start off here. I've got my, my force condition on my system, s2. Now,

let's, moments about point c on system one. And make that equal to the sum of the

moments on sys tem 2 about point C. So, first of all I have on system 1, this 100

pound, foot pound couple acting, clockwise and so if it's clockwise by the right hand

rule. It's into the page, where that would be in the negative K direction. So I have

minus 100K, and then I've got two forces that are causing moments about point C.

And so I'll use the definition of the moment R cross F. So I've got plus R from

C to A for the 350 pound force. So that's crossed with 350 pounds in the j

direction. And then I also have my 200 pound force so plus r from C to A That's

the position vector from the point about which I'm rotating to a point on the line

of action of the force, crossed with 200i. By the way, just as a side note. Un, in

using R from C to A. For the 350 pound force I could have also used r from c to b

because point b also lies on the line of action for the 350 pound force, and I'd

get the same answer. And you should try that out on your own to make sure that you

do get the same answer. So r from c to a is just 7 units or 7 feet in the x

direction and 2 units or 2 feet in the y direction. So this is 7i plus 2j, R from C

to A over here is also 7i plus 2j, and if I do those cross products, I'll get minus

100. I forgot my vector symbol here so I'm going to put it in K so this is K. And

then if I cross this, I'm going to get plus, two hun-, 2,450 K. Because I crossed

j as k. 7 times 350 is 2,450. J cross j is 0. And then we're going to have, over

here, I cross I is 0. J cross I is minus k. So we're going to have a minus 400. K

equals, and up here I should have put, I also forgot equals the sum of the moment

about point c on system 2 right? So we had some of the moments about C on system 1 is

on the left hand side and then sum of the moments about system C on system 2 is on

the right hand side. So this is some of the moments about c on system 2. And so I

get some of the moments on system 2 about point c. Is equal to, 2450 minus 500 or

1950, in the k direction. And the units are force times dista nce, or foot,

pounds. Okay, so now we've satisfied the second condition. We have, not only, that

some of the force is equal to 0, or excuse me, that some of the force is equal to

same on both systems. I also have the, some of the Moments about point C the same

for both systems. And so, here it, it shows this, on, on my system 2, I've got

my 200 pound force, my 350 pound force, my 1950 foot-pound Moment at point C. we can

take one more step and, and, and do a resolution of, of the two forces into a

single direction force. So I'm going to say, note, I can take my. This is 200

pounds. This is 350 pounds, and I can make that a single vector. It's going to be on

a slope of 350 on, 2. 350 on 200, or 3.5 on 2. And its magnitude is going to be the

square root of the sum of the squares for these two orthogonal directions. So the

square root are the sum of the squares for 350 squared plus 200. Squared, which comes

out to be 403. So this magnitude is 403 pounds. And so I can draw a final picture

of my system 2, with a single force and couple is this. This is point C. This is

three feet. And we have our single force, which is up and to the right, 403 pounds.

It's on a 3.5 on 2 slope, and we also have our counterclockwise moment, 1950

foot-pounds. And that's our final answer. That's a force and couple at point C which

together form a system which is equivalent to the original system, system 1. And

that's it for today's module.

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