[ Music ]
Yes?
So are the homework problems in the 11th edition
the same as the late editions or--
No, it's not actually. That is exactly what
I said last time. The homework that are assigned
I put on the board is from 11 Edition. Which
is not the same as the 10th or 9th and 8th.
Okay.
So if it is on the-- if you don't have the
book. You should copy those homework from
the 11th edition.
Okay.
Did you do that?
No, I actually did the homework--
That mean you haven't started your homework
yet?
Yeah.
That's not good. That's what I said. Already
there were people in my office-- anyhow. I
have a-- I think I printed a few copy. No,
that was for the other-- the other class.
I do not have any extra copies. Yes?
Sorry you posted in the homework that in quiz
2-- that you would post problems from quiz
2. And it's not in that first handout that
you put on Blackboard.
It was on page 9. One of the homework is one
of the question or the second question on
page 9. There is 2 cylinder there, do you
see that? Against the wall, that's your homework
there. I'll put it up on the board--
It's due Thursday right?
What?
The homework is due Thursday?
Every Thursday except this week which was
exceptional. So I gave you extra homework--
this week is exceptional. From next week,
every Thursday your homework will be due.
And every Thursday we are going to have a
quiz. Everybody on that and all this routine
will be continued. That means we'll have that
question again and again. Yes?
We do both questions right?
No, question number 2.
Okay.
I put it on the board. I don't know. Do you
remember that?
I wrote it.
I think if you wrote it down. Question number
2 on page 9. That was one of your homeworks.
Question number 1 actually going to do it
today in class, okay? Still I am waiting for
the real question. These are not real ques--
real question means about the concept that
I explained last time. Which was not much
of course. Most of them you already knew.
But the most important part at the end of
the class. That's what we did. We said that,
okay if the r is the resultant of several
forces. And those forces represented by f
1, f 2, f 3, etcetera. You can put more forces
if you want. And since the resultant generally
is-- this is summary of what we said last
time. The resultant is equal to r x i plus
r y j. As a unit of either pound or newton
or kilo-newton, etcetera, etcetera. And then
since r must be equal to f 1, f 2, f 3 represented
with a polygon if you want. But we don't have
to do it again. We did it last time. Or at
the end result and I add it together. r x
end up to be equal to what? R x ends up to
be equal to? S
Summation.
Summation of all the--
Components.
X component of all the forces. Everybody on--
that's the statement. We don't want to write
it like that. So we simplify it in this format.
Summation of the forces, all the forces in
x-- that's it. You don't have to put i and
i here because these are all along the x axis.
Therefore all of them are parallel to each
other. But it is the scaler equation. It is
no longer a vector equation. It's that correct
or not? Yes?
Yes.
Okay. And what was r y equal to?
Summation.
R y was equal to summation of all the forces
in the--
Y.
Y, of course these forces doesn't have to
be f 1, f, 2. It could be weight, it could
be contact forces or any other forces. Doesn't
matter. This statement, this formula that
you see there. That you should get familiar
with. Means summation of the forces in the
x direction. As many forces as you have, is
that correct or not? Generally that's the
case because when you end up with a polygon
of forces like this 1, 2, 3, 4. And then you
end up with r, this is you r going from the
first to the last. To first head to the last--
I mean, head at 2, 2 to head this way. So
this is your r. Notice this r has r x and
r 1. There we came to this conclusion that
some of you already discuss it. During the
office hours with me. What would happen if
r suddenly become--
Vertical.
Vertical. So obviously what happens there?
Summation.
Summation of all the--
X axis must be equal to 0. That's conclusion
that you have. Or if r become-- horizontal.
Now to practice this I have given you several
examples. There is one example on this class
exercise. Go to your handout. There is one
example there which is similar to one of the
quizzes that I have asked in the past. Let's
study that together. There are a few points
there that would help you out with that. These
are the-- let's see. Class exercise which
is-- no, before that. Yeah, page-- let's go
to page 6. Number 1 I already did. Let's look
at number-- 3, problem number 3 on the handout.
Which is I am going to put that problem on
the board for you to see. So here is the x
axis given to you. There is the-- and of course
the y axis. There is force here f 1. Even
with the angle of alpha it's been defined
like that. Then, there is a force on the y
axis. Okay, we call it f 2. I'll put it in
the vector quarter. I'll give you the magnitude.
The magnitude is in your handout. Then, there
is a force here, f 3. But slope of f 3 is
512 and 13, they put 13. At this angle-- okay.
And then they said the following. They gave
you an x prime axis that passes through central
coordinate. And they have given you that angle
equal to 60 degrees. The question is the following
as you can see it in your handout. The magnitude
of f 2 is given equal to 200 newtons. The
magnitude of f 3 with a slope given is equal
to 180 newtons. Also they said that the resultant
force, look at the resultant force. It's very
important. The resultant force has the magnitude
of 800 newtons. That's the resultant force,
okay? It is along the x prime axis. So it's
going back that magnitude. So in other words,
this is if I show that with color. So the
resultant force is this force, r. Question?
This is the data, remember I said that. Always
when you read you want to do your homework.
You should always read the question carefully.
That's the first part that some people miss
a little bit, miss the point. They do not
get all the information. Put all the data,
all the information given in the problem in
front of you. Then define your goal. Now,
what is our goal? Read the question. Our goal
is to find the-- magnitude of--
Force.
Force f 1. So okay so here the question. So
the question we want to answer is f 1 magnitude
at its direction. Which is calculating the
alpha. So you want to answer these 2 questions.
Here is the concept, yes or no? Correct, how
do we go about it?
Take the parallel of x and y.
Yes. I can-- see sometimes this is I want
to tell you. Many instructor ask you to write
f 1, f 2 and f 3 in component force. I am
not going to ask you to do it because it is
simple. I am assuming you already know that.
Yes, but after we do that. Which we can do
it by inspection. It's very simple. The concept
that I am using here which we already explained.
F 1 plus f 2 plus f 3 must be equal to r.
So that's the concept we are discussing. R
equal to f 1 plus f 2 plus f 3. For r, x must
be summation of. X component of all the forces
and r y. But is r given or not? R is practically
given. Is that or not? What is r? So let's
write the r. This is reverse of some of your
homework as we say. So in your homework many
times they give you f 1, f 2, f 3. They calculate--
ask your r here. I have given you r. I am
asking you to find the force f 1 and its direction.
Anyhow, r equal to what? 800 times what? It
says 800 magnitude is given. Times cosine
of this angle I can do that angle. This angle
is 60 degrees. So this angle from here to
here is--
30 degrees.
30 degrees. So it becomes cosine of--
30 degrees.
30 degree i. This is the method we discussed.
Writing the forces in component form, correct?
And then it's positive because it's positive
and positive. Then plus 800 sine of-- cosine
of 60 degree. Or sine of 30 degree j and we
have a unit of newton. So don't forget that.
You can put it in front of each number or
we can put it in front of equation. I put
it in front of equation to apply to all the
numbers. Is that correct or not? Can we simplify
that? Of course we can. So let's go one step
further so 8-- that becomes equal to r equal
to cosine 30 degree. You put it in your calculator.
Come to 693 i plus 400 j newton. That is your
r. Next is writing down the other 2 equations.
Sigma f x and sigma. In this case, the resultant
is neither horizontal nor vertical. Therefore
it has both components, yes or no? But it
can give you 2 equation formulas. If it is
2 equations and I have 2 unknown. I should
be able to solve it, yes or no?
Yes.
Okay what are those 2 equations? Sigma f x.
Okay sigma f x which is actually equal to
r x. Should be equal-- now here is the key
we want to learn. Now look at this one. This
is f 1 times what? Times cosine of-- alpha.
Is the alpha is given? No, that's the question
you want to answer. F 1 is given? No. Both
of them is unknown. So let's put it that way.
Doesn't matter. So f 1-- before I do that.
Since this is the first time. Let me write
all the forces in the vector form. Because
this is the first time I wanted to show you
a few points there. Which is very important.
First f 1 before. I am going to now stop here.
Go back, write out the 3 forces in vector
form. We already wrote the r in vector form.
Why not writing f 1, f 2 and f 3 in vector
form? I show you a little-- not really. You
should mention algebra here. You see most
students when they come to this force, they
use the angle. You should-- I would rather
use the slope. I'll show you. It is very obvious
for most of you. But some of you for some
reason are not using it. And that's the--
that takes a few minutes extra work. So let's
put it this way. So f 1 equal to what? F 1
equal to f 1 magnitude cosine alpha of i.
Plus f 1 sine alpha j newton. Both unknown,
is that correct or not? F 2 is equal to--
it is on the y axis with magnitude of 200.
So it would be 0 i plus 200 j newton. Notice
on the j and i you can do that, newton. Is
that correct? F 3, this is the 1 r literal
concept. How do you calculate f 3? Yes?
1 a b times the cosine is-- adjacent with
their hypotenuse so--
Correct. That's like what I did for f 1. But
are you calculating this angle or not? That's
my question. Yes or no? Do you need to calculate
that angle or not? That' s the whole thing.
You know many people will do that. Many people
said, "Okay. This is tangent of this angle
equal to 5 over 12." So I saw a couple of
people come into my office. That's why I am
saying that. And then they calculate that
angle equal to let's say 18.3 degree or whatever
it is. Is that correct? Then you say it f
180 times cosine of 18 degree which is some
extra work. All I am saying that use this
slope. Is this slope 5, 12 and 13? Yes? Yes
or no?
Yeah.
Okay. So then this is the scenario. That is
5, 12, 13 this way. So since our force going
in that direction. What is cosine of this
angle? Cosine of that angle is 12 over--
13.
13 sine is-- 5 over 13. So why do I need that
angle? That's what I said.
Because--
Oh, that's what you were saying. So you were
saying the same thing.
Yeah.
Okay. Very good. So in this scenario, any
time there are many example in the book. That
they have given you slope. And you use it
as such. You don't need the angle. So exactly
what you said. I'm sorry. I didn't hear you.
But however, that is exactly like that. 180
times 5 over-- oops, sorry. X component is
12 over--
13.
13 i. If that is what you said, that's exactly
what I want everybody to do. Is that correct?
Are we all calculating alpha? 2 people came
to my office or 3 people. All of them were
calculating the alpha or that angle. Whatever
that is, this is not necessary. You do it.
There is nothing wrong with it. But it's not
necessary. You better use the slope. Many
times in the book. Many examples they have
slope. Or even if it is not round. You can
still calculate the length and use it the
same way, correct? X, y and either direction.
Anyhow, and the other one would be 180, 5
over 13 of J newton which can be simplified
into. As followed-- oh by the way. X in this
one, x is negative. Y is positive. So you
need to put a minus there, minus plus. And
therefore f 3 becomes equal to-- I'll write
it here easier. Become minus 166 i and plus
69.2 j. I'm using 3 digit, newton. So I have
all my forces and I have my resultant. And
all the x components together must be equal
to how much? 600--
93.
93. All the y component must be equal to--
400. That gives me 2 equations for 2 unknown.
Let's write 2 those equations down. The first
equation is that's what I was doing here.
So let's re-do that again. So start all over.
Maybe write it somewhere like I don't need
this part. So let's erase that part. So summation
of forces in the x direction which represent
the r x must be equal to. From first equation
I get f 1 cosine of alpha. From the second
equation I get nothing, 0. I rather you to
put the 0 there not to miss anything because
you know. You have 3 forces from the third
one. You get minus 166. And those altogether
must be equal to 693 newton. Can I solve for
f 1 and alpha? No I can't. It's that way.
Because one equation, 2 unknown. Then you
write sigma f y. so sigma y equals to r y.
This time we have f 1 times sine of alpha.
Because this is what we have there plus 200.
Plus 200 from second equation. Plus 69.2 from
third equation. And all of them together must
be equal to 400 newton. However, this is 2
equations for 2 unknowns. Now, when I gave
this problem as a home-- as a quiz. Many people
got to this point easily. But they had a little
trouble solving these 2 equations for 2 unknowns.
Because there is sine and cosine, f 1 involved.
Is that correct or not? Of course that should
not be any difficulty for you guys. So therefore
all you have to do. Put all the numbers on
one side. So from equation 1. So that's what
we get. From the first equation we get. F
1 cosine alpha equal to-- f 1 cosine alpha
become equal to 859 newton. Sum it up this
plus 166. 693 plus 166. From second equation,
you get f 1 sine alpha equal to 400 minus
that or whatever that is. Which becomes 130.8
newton or 131. Now, solving that is very easy.
You can square this everybody. And then add
it together because sine square plus cosine
square become equal to--
1.
1, or you can divide. Is that correct, you
can divide to get the tangent. Because f 1
will be eliminated. Let's divide equation
2 by 1. Which ends up to be f 1 sine alpha
over f 1 cosine alpha equal to 130.8. Divided
get 859 newton over newton. And as you see,
f 1 and f 1 drops out. That times alpha and
you calculate alpha to be equal to 8.7 degrees.
So alpha is 8.7 degrees. So in the book or
in the handout it says calculate that angle.
So obviously that angle if I call that angle
theta. That angle theta is 30 minus 8.7. So
that is one of the questions they ask there
too. And then finally I have to find the resultant
of f-- I mean. The magnitude of f 1. The magnitude
of-- I plot alpha to one of these equations.
The other that is obvious to you how to solve
it. Is that correct or not? So this is-- if
you look at your-- all the quizzes. You see
one of the quizzes is exactly the same one.
Look at all the sample quizzes that I have
given you. There is one quiz exactly like
that. So you can check it. The first 3 if
we go there. This one, number-- page. Go to
page 8 actually problem number, the second
problem on that page. The first problem is
the resultant is horizontal like the book.
But instead of 2 force, we have 3 force. This
should be no problem for you. The second problem
is exactly like that. But I gave you a different
location for r. Have in some of your homework
there are sometimes horizontal. Sometimes
it's-- vertical and sometimes not. So that's
all your first set of homework. We are done
with this part. If there is no more questions
there, let's move on to the next part. Which
is the equilibrium of particle, yes? So everybody
okay so far? Because this was simple enough.
So let's move on. Now, the equilibrium of
particle is that-- if you understood what's
going on here. Actually we already know that.
Okay what's the difference? Is-- we'll explain
that in a minute. Now, first of all particle
versus rigid body. In this chapter, next chapter
or this chapter or next chaper we are going
to talk about particles. And later on in the
book, we are going to talk about rigid body.
What's the difference between the 2? Any input
from you guys? What is the particle versus
the body? What is-- if you had in physics,
you had always this before. When we're talking
about the car. The car has a length. It has
a width. It has the engine somewhere, passenger
somewhere, load somewhere, engine load somewhere.
And you appreciate that the weight of it or
the forces on the wheel will be different
is that on an axis. Yes or no? However in
physics you assume the hold mass as a point,
at the center of the point. That is when we
are talking about the particle. In other words,
in particle we don't care about the dimension.
We are going-- we are assuming all the weight
is at the center of gravity of that object.
No doubt. When you go to chapter 4 or 5. Then
we are going to talk about equilibrium of--
body. That means we have to consider the length,
the width. How the distribution of load looks
like. So in this chapter we are talking about
particle, not rigid body. For the time being
that's out. Is that correct or not? See the
difference between the 2. That is when we
get to the result you will see that there
is a big difference between the 2. Now, here
is the body now okay? That body will be under
different load. One of the loads obviously
will be the weight. Is that correct or no?
This object may be connected to-- you want
to make a note of that. This is the next subject.
May be connected to a sort of a cable. And
that cable is pulling it with a force f 1.
Or another rope here pulling it with tension,
t. It could be the object may be in contact
with some other object. So you may have here
a contact force n. And also there could be
a spring somewhere and you are pulling it.
Or you are pushing it, you may have a force
of a spring like that f s. And therefore what
I am saying that there is the object under
several forces. All those forces you would
recognize. You have studied them in physics
or in the math classes. So you know where
they are coming from. Is that correct or not?
But the point is this. We are assuming that
the system is in equilibrium. When it is in
equilibrium there's newton's first law of
course. That means the sum of the forces must
be equal to--
0.
0. Therefore if that is the scenario, and
some of the forces must be equal to 0 in the
vector form. Means the polygon of all these
forces together where you start from point
a. You would end up to point--
A.
A. No resultant left. So therefore in effect
if I am saying that resultant of these forces
equal to f 1. Plus t plus f s plus w-- because
this is the vector polygon of the-- plus n.
That is I need to make this all as a vector
plus n. And all of them equal to 0. But I
should write it like that. 0 i plus 0 j. Because
all the forces have x component and-- y component.
Correct or not? But if r equal to 0, notice
in this problem I purposely did it this way.
In this problem r has both x component and--
Y components.
Yeah. There x component is. 0, y component
is 0. Therefore what? Therefore both these
2equations must be equal to--
0.
0. So that is what we call an equilibrium
equation. So the result of this analysis is
this. And this is the one you are going to
use from now on. So from now on you are going
to use sigma f x equal to 0. And sigma f y
equal to 0. That's all you need. These we
call it equations of equilibrium for a particle
if it is a 2D problem of course. If it is
the 3D problem, we have to change it, yes?
Okay. 2 equations. So obviously like this
problem I cannot have if I have 2 equations
I cannot have more than 2 unknown. Now, what
are those 2 unknown? This is the problem.
When they give us a problem like that and
we are investigating the equilibrium. Some
of these forces that right here are known
right down in your notes. Some of them are
known. Some of them are--
Unknown.
Unknown. The goal is to find the unknown forces,
yes or no?
Yes.
How? That's the question. To find out this
is something you haven't heard before. We
are going to draw a free body--
Diagram.
Diagram. Good, so you already know all of
that, yes? Have you seen free body diagrams
before? In physics? Very good. I'm impressed.
Free body diagram. So what's free body diagram?
How can you define that for me?
Just means the sum of all the forces I guess
[inaudible] draw it or picture it.
More or less that's correct. Yes?
Putting all the forces acting on a particle
on the [inaudible].
Yes partially correct. But before we go that.
It's very important that with those. I would
like all of you to write this statement down.
This is not in the book. You want to write
it and this is the meaning of free body diagram
I go-- I am going to make a reference to it
again and again. Free body diagram is part
of a structure or the entire structure. Both
part of a structure or entire structure. Disconnected--
from all its connection to other bodies. Disconnected
from all its connection to other bodies. And
those connections being replaced with appropriate
forces-- with appropriate forces. In parenthesis
put-- in parenthesis put (movement and movement).
Now here in particle we have no movement.
We don't know yet what movement is. So we
are not knowing. This is for chapter 4 or
5 when we get to the equilibrium of body which
movement come into the picture. For time being,
all you have to say is that. The free body
diagram is part of a structure or entire structure.
Disconnected from all its connection. And
those connections being replaced with appropriate
forces. Movement is out for time being. When
we get to chapter 5, I will talk about that
movement later on. Everybody understood that,
yes? Okay saying that is very easy. You already
know about free body diagram. I got a partial
definition all of them, correct? But probably
not complete. Everybody made it complete for
you. Now, let's go to that problem in the
next page. You see that the spring system.
And I'm going to draw it on the board. And
let's understand. This is so simple. But everybody
knows the answer. However, I want to show
you what the free body diagram is. And how
you should be applying that free body. Which
is the essential part of your work from now
on. Free body diagram is one of the most important
part of any analysis using-- not in this class
in future classes. In a string of material
in other classes what were design forces you
use? The first order of the work is to draw
a free body diagram. To find all the forces
applied to the structure. Obviously the first
goal is to find the unknown forces, yes or
no? Because some are known. Some are-- unknown.
Okay let's go to that problem. Erase the board.
The concept is here. Maybe we should keep
it there. So see go through the free body
diagram of this structure. And-- okay. Let
me be quick here. Draw it exactly like we
have it set up for you. Okay it's here. Here
is the ceiling. Okay and there I have spring
coming here. And here is another spring coming
here. And the slope of those. This is equal
to 30 degree. This angle is equal to 60 degree.
There is a spring attached to here like this.
There is a spring attached here like that.
Then there is another spring attached here.
Then suddenly there is a box here. I call
it box e. And then there is a spring here.
And then I have here box f. And this point
is called a, b, c, d for example. As you have
it in the picture in front of you. Okay, the
weight of the box e is given equal to I believe
200 pounds. And the weight of box f is given
equal to 100 pounds. These are the data. The
picture is given. The system is in equilibrium
means these boxes staying in this position
as it is shown, correct? And then, the weight
of each box is given. The question now, the
goal. The question is finding the force initial
spring. So let's number them in order not
to make a mistake. Let's say this is spring
1 which is between d to e or f to e. And then
spring 2 which is from here to there, from
e to a. And let's call this one spring 3.
Which is from e to c. And spring 4 which is
between a to b, etcetera, etcetera. How many
spring do we have? 4. So how many unknown
do we have? 4. Notice if I am doing free body
diagram, I cannot have more than--
2.
2 unknown. And already I have-- that means
that in many occasion I have to do more than
1 free body--
Diagram.
Diagram. This is as I said so simple problem.
I just want to see-- for you to see the relevance
between the-- What you already know and this
free body diagram business. What is the force
in spring 1? Care to tell me?
100.
Yes? How much?
100.
100, correct? Everybody agreed? Because if
I'm holding it from here. How much weight
is down there? It is equal to-- 100. When
I ask this question in class. Actually many
student will answer that immediately 100.
Is that correct or not? Did we use a free
body diagram? No. But we did it by inspecting.
But this is what we did actually. This is
the free body diagram of just what you just
told me. The weight of this box. I am separating--
this is-- that's what I'm-- separating this
box from all its connections. That's what
we wrote down. Is that correct or not? Separate
the box from all its connection. It's only
1 connection spring. Is that correct or not?
So here is the box. This is the spring. It's
being cut. And I now have to replace that
cut with the appropriate forces. Everybody
knows this springs has a tension or compression.
Depends whether it is being pulled or pushed.
In this case, it's being--
Pulled.
Pulled therefore there is a tension, yes or
no? So that is t 1. So this connection that
I cut here. Represents a tension in spring
1. Let's put it this way in a vector form
which is vertical, yes or no? And then since
there is the weight here going down. Or you
can put it at the center if you wish. Weight
and that weight is only-- both of them are
in y direction. So we don't have to write
the equation this and that. Is that correct
or not? Yes? Obviously, if w is equal to 100,
t must be-- t 1 must be equal to--
100.
You did it by inspection. That's perfect.
Is that correct? But this is the free body--
Diagram.
Diagram of box-- f good. So there on top of
that write it down. Free body diagram of box
f. Okay then we come to box e. Now, what should
I do? Box e has 1 connection or 2 connections?
2.
2 connection. Now I can have 2 choices here.
One according to what I just said. I use different
color. I can separate this from all its--
Connection.
Connection. But there is a spring on the bottom.
There is a spring on--
Top.
Top. Okay, what's the free body diagram? This
one I used, okay. So here I put it in red.
T 1 magnitude equal 100 pound. Now we go to
the free body diagram of-- free body diagram
of box e. Okay, now box e has-- instead has
a weight of 200-- pounds. There was a connection
here. There is a connection. You are getting
the point. I don't have to tell you what.
What is this one represent? What is this cut?
This cut representation in the spring 2. Okay
I have to put here. Tension, t 2 let's put
it in magnitude this time. Because they are
all vertical so I don't have to use the vector.
Is that correct or not? Yes? But what is the
pull on the other side? T 1. Should I put
it up more or down more? Notice this t 1 is
up 1. This t 1 should be--
Down.
Down. Very good. So no problem there. Oops,
got to put magnitude vertical. Now what's
the value of t 2? I don't have to tell you.
T 2 must be equal to t 1 plus--
200.
200. So t 2 must be equal to 300. Nevertheless,
you could do both of them with inspection.
Without going through free body diagram. But
I just showed you what we wrote down is applicable
to this simple problem as well, yes? Now that
we go to the upper part. Now what should I
do?
Separate them.
Separate what? No, I'm asking you the question.
How do I calculate? Now I found the tension
in the spring 1. I found the tension by spring
2 by drawing 2 free body diagrams. Now I want
tension in spring 3 and 4. What should I do?
Draw a free body diagram--
Of? That's my point. Of tension? No, that's
not the case. That's not the right thing.
I know what is behind your statement. But
I want you to say it correctly. I need to
draw the free body diagram of what?
Particle A.
Particle yes. Which particle?
A, ring A.
Ring a, that's right. That's what I want to
hear. The ring-- all the tension are at the
ring. It's not on the ceiling-- it's not.
I want you to notice something that student
eventually some of you make a mistake here.
Notice the tension at both sides of this spring
depend on what side? It is there. It is equal
and-- opposite. And they are in different
direction. Everybody understand that. I'm
going to give you an example. Which I want
you to use it in some of your homework in
future. Let's assume this. You have there--
because people make a mistake in free body
diagram. Let's assume that there is a hook
on that corner of the-- that corner. And there
is a hook in that corner. And we have a cable
connecting these 2 hook together. When the
cable is loose, there is no tension in the
cable, yes? If it is let's say this distance
from there, that corner to that corner. Let's
say it is 75 feet. Is that correct or not?
So if I get exactly a cable of 75 feet. And
connect from there to there, what's the tension
in the cable?
0.
Nothing, yes or no? Until I put one of these.
I buy one of these knuckle bolt they call
it I put that in the middle. And I tighten
the bolt, is that correct on it? The cable
is to be stretched. First of all I want you
to understand the force is in the cable, yes
or no? The length of the cable has nothing
to do to amount of force as you notice. Force
would be 0, could be 10 pounds, could be 1,000
pounds. Everybody understand that. When we
do-- I am going to draw that in this corner.
Because this is an issue I have seen in the
past. That student make mistake is this. When
you draw, this is the cable or could be a
spring, could be a rope or anything. If I
draw that between point a and b. Please write
it down in your notes. If I-- and I say tension
equal to 100 pounds. How do you show that
tension in this cable? See some of you said--
some place, some of you say this place. Some
of you saying that place. That's exactly my
point. Depends where you cut it. You see if
you make a cut here. There is a force going
this way for example. But that's not what
usually we do. We usually cut it around ring
a or ring--
B.
B. Note what happened. If I cut it here and
cut it here, actually this is not one free
body diagram. It is--
3.
3 free body diagram-- free body diagram of
ring a. The rest of the attachment I am not
showing. Don't get me wrong. There are other
item there too, yes? Free body diagram of
ring b. And free body diagram of the cable
itself, yes or no? Okay, the cable you said
it is the tension. And everybody is showing
like that, correct. So the tension means that
we use different color. The tension means
this should be going that way. And this should
be going that way. Each one with 100 pound,
yes or no? That means this cable is stretching,
yes? What should I put at hook a-- or yes
hook a? The force equal and--
Opposite.
Opposite. What should I put at hook b? The
force equal and--
Opposite.
Opposite. So from now in many of your homework,
you don't care about this part. Because you
know the cable is under tension, yes or no?
You want the forces on the end of the cable
or on the other side, yes or no? Which is
hook a and hook b. That's what I was stressing
before. Therefore look, no one gets surprised
you see that these 2 forces are. You just
told me equal and--
Opposite.
Oh that's exactly what we are doing here remember?
On this one the tension is going upward at
this end of the cable. On the other end of
the cable which was here, we are going down.
Because these are the 2 ends of the cable,
is that correct or not? So if you understand
that, please apply that. And when these 2
are way in this format. Obviously the cable
is going to be in tension. If this is changing
the direction the cable must be-- which is
not the cable actually. It should be some
wood or something. Still it would be the--
or a spring, it should be in compression,
yes or no? When we get to that-- when we get
to chapter 4 or 5. So for time being remember
that. This is the free body diagram of hook
a. Free body diagram of hook b. Tension and
compression will come in future. So don't
worry about this. So now where were we? So
we were here at this point and this is what
I said. "I need the free body diagram." Look
where I am cutting it. I am not cutting it
here and here and here. I'm cutting at exit
around the ring. A because I want forces in--
remember, force at c is going down way. Force
at b is going down but forces at a is going
upward, is that correct or not? That was my
point here. So I'm sure that you all understood.
So therefore let's go here. We don't need
this out here any longer. So when we go to
point a, this is what you see. Point a, we
already have a cut. There are 3 cuts here.
You see, free body diagram, one cut, one cut,
one cut, 3 cuts. The lower cut is going down
this way. And the value of that one was already
we calculate it to be equal to 300--
Pounds.
Pounds. Okay and then what's the angle here?
So-- now my worry is this. I have seen this
happen in the past. When I see them draw the
free body diagrams. They take this picture,
they put a force from here to there, everybody
understand. Which has nothing to do with the
length of the spring. It has nothing to do
with that. This is the action of the spring
on ring a. Which is the force-- this is actually
the length. Everybody and the length put in
inch, foot, meter, millimeter, whatever. The
force is different. So you showed the direction
of a spring. This is what I'm saying here.
And then you would show there this is unknown.
I put it in color. So you put here and you
better call that tension chord. We just called
it a tension chord in magnitude. And the angle
of that one is equal to. But this started
at a, finishes at--
B.
B, is that correct? Not all the way a, b.
That's what I'm saying here, is that correct
or not? Because this is one side. The other
side I do not care. The same format. We are
going to use the other direction which is
this one. And then use the force there, which
is tension 3. And of course to complete our
free body diagram, we should put our axis
there. Which is like that, x axis, y axis,
whatever you want. We need only x axis. And
this angle equal to 30 degrees. And that angle
equal to 60 degrees. How many unknown do I
have?
2.
Can I solve it?
Yes.
Yes or no?
Yes.
Okay, we did it already in the previous example,
right? Again, what is the idea here? Since
this is in equilibrium. T 3 plus t 4 in vector
form. You don't need to repeat that. What
we already know. T 3 plus t 4 plus this one,
whatever that was. What you call it from a
t 2 in vector form equal to--
0.
0. But that's not the point. The point is
summation of the forces along x direction
must be equal to--
0.
0. Summation of the forces in the y direction
must be equal to--
0.
0. The same technique you were using here
because this was so simple. There was no x
component that had you-- y component. That's
what we use here. Here we have to use both
of them together simultaneously, yes or no?
Now, we are-- now. In this type of problem
I do not require you to write a t 3 and t
4 and t-- this one in vector form. It is not
necessary because we already practiced that
in earlier homework. And we don't need it.
All you have to do apply the principle of
equilibrium. Which I just mention. Sigma f
x equal to 0 as sigma f y equal. If you're
writing the formula itself-- I mean the forces
itself in vector form is not required in my
classes. Although in the math some of your
instructors may ask you to do so. So don't
be surprised if they insist you to draw--
use all the forces. I'm assuming you already
know how to write the forces in vector form.
Everybody understand that. That is the first
few line of homework that I gave. When we
came to the equilibrium we go down to the
concept that I presented to you. This is the
concept. The result was-- the result of this
concept is 2 equations. Sigma f, x equal to
0 and sigma f, y equal. Okay what's sigma
f x? What's sigma f x equal to 0? This one
gives me t 3 times cosine of 60 degrees. Minus
T 3 cosine of 60 degrees. Negative because
it's a negative. This one gives me t 4 times
cosine of--
30 degrees.
30 degrees, that's right. T 4 times cosine
of 30 degrees equal to 0, yes or no? Now,
if I were you I'd immediately put the number
down there. Because cosine of 60 degree is
equal to sine of 30 degree equal to one half,
0.5. And this is 0.866. You put that in your--
from your calculator. So this is what you
get. You get 0.866 t 4 from here, equal to
0.50 for t 3. Or in other words, you are establishing
a relationship between t 3 and t 4 in the
first equation. I simplify for you this one.
Effectively t 3 becomes equal to 1.732. I
go to 4 digits. T 4-- actually 3 digits should
be sufficient. But I went one extra digit,
that doesn't hurt. As I said we need minimum
of 3 digits for all calculation. Is that correct
or not? Then, you write your sigma f y. And
sigma f y equal to 0. Now what do we have.
We have, from here, we have t 3 sine of 60
degree positive. Because it is going vertically
positive, this one gives also positive. So
plus t 4 sine of 30 degrees positive. And
then minus 300 equal to 0, exactly the way
the equation. Sum of the y component of all
the forces should be equal to 0, correct?
Now this effectively becomes-- I mean everybody
can do that. But I just did it partially for
you. That means 2 equations for 2 unknown.
You should be able to solve it. Or can you
put that number here. We resolve this issue.
Any other answer become t 4 become equal to
150 pound. And t 3 becomes equal to 260 pounds.
That's for this one. Which shows you the idea
behind the equilibrium of particle. So all
the rest of your homework are equilibrium
of particle in different format. Now, let's
practice that. This was the idea. Let's go
to one of the quizzes or couple of quizzes
that I gave to you in the past. Go to the
next page of your handout. Let's do problem
number 2, quiz-- sample quizzes. There we
are, page 11. Page 11.
Professor?
Now that's it, that's the end. There is nothing
there. You want to ask any questions go ahead.
Yes?
You could solve that last problem with triangle
rule on f of 2 right?
Yes you can.
Okay.
But do you say whether or not you want one
method or another?
Doesn't make any difference.
Okay.
But quite frankly if I were you I would always--
from now on, except the 3 problems that you
use the triangle rule and sine rule for the
problems. From now on always use components.
Okay.
Because many times this is the simple problem.
Many times it's more than 2 forces. If it
is more than 2 forces. You have 3, 4 forces
you want to add up. It's not going to be a
triangle, yes?
Yeah.
So it's not going to work. That not going
to work. Yes but your answer to your question,
obviously yes. And it's unfortunately in the
book in the solution manual. There is a lot
of triangle. I'm not against it. I just rather
use the components. Because you have to use
it for 4, for 5, for 6. And usually a structure
has more than 2 forces, you know that. This
building has 2,000 forces that run it. So
when-- we get to a real part-- a real problem.
It's not that simple. You already know that.
So you are using the j method. Use as much
as you can from probably the first line of
your homework. Use the sine rule and cosine
rule. For all the equilibrium, try to use
the component if you can. But both methods
give you the right answer. Yes, good.
If I already did the home-- that problem and
I already solved it and got the same answer
for the forces. Using the triangle method,
is that okay to keep it like that?
Yeah sure.
Okay.
I said it's okay. I said nothing wrong with
it.
Okay.
I said I prefer you to do it that way just
to get into habit of you know. Doing it with
components.
Okay.
But nothing wrong. Just leave it as it is.
Okay.
You have done it. It's fine. Actually from
now on, it's your choice still to do it either
way.
Yeah but--
Don't get-- use it. Whatever is simpler for
you. Yes question. When you do that I think
you have a question. Alright, any question?
Let's now look at-- while I'm erasing the
board you think about problem number 2 and
3. These are the 2 problems we want to do
in the next 5, 10 minutes. Until we go to
something new. Tell me how you are going to
handle this. Where you want to draw the free
body diagram? Start doing a couple of free
body diagrams. First the problem number 2
on page 11. Okay seeing what you have. What
you don't have and then proceed. Okay, I am
going to draw it here while you are working
on that one. So I'm going to show you what
happens right afterwards. You keep working
on yours. System, alright that's at 30 degree.
This is the statement of the problem in your
handout. This problem number 2, one of my
old quizzes. Okay, let's look. The picture
looks something like that. But let's see what
is given? What is not? If the weight of the
lamp is given or not?
Yes.
Yes, okay. How much is it? The weight is not
given, the--
It's the mass.
Okay, mass of the lamp7. This is the lamp
there. And that is given equal to 4 kilograms.
The idea is to finding the tension in each
rope or cable, yes or no?
Yes.
Yes, alright. Where should I start? Now notice
immediately from what I did. You recognize
here there is a tension in this cable, this
cable. Again whether you are looking at b
and a or b or c. The tension are in different
direction. We don't need all of them. We are
concentrating on the free body diagram of
b or c, yes or no? Right, which one should
I start first? B or c?
B.
B of course. Obviously because f is the goal
that we have. That's the goal. The question
is find the value of f and also find the value--
the tension in each cable. So if f is not
given, this tension is unknown. These tension
are unknown. In this free body diagram I am
going to have 3 unknown. In this free body
diagram I am going to have only--
2.
Which is similar to what I-- you see that?
I did that for class. I gave you this quiz--
this quiz. Is there any difficulty to do that?
Should it be any difficulty to do that? So
from now, what I am saying that now we are
in various stage of the study. It's very simple
problem. All of you are familiar with it.
When we get to more complex problem, it's
the same procedure. So if you pay attention
to the detail of the sub-- concept that I
present to you at the beginning of the lecture.
Then all the problem will look all the same.
Look when I did this one, this one should
be very simple for you, yes or no? It's all
you have, all the idea how to do it. Is that
correct or not? So therefore, you draw free
body diagram of b first, yes?
Yes.
So you draw this, a little circle there. Don't
forget that little circle. Everybody should
do that. Put in your notes and put it in your
handout or wherever you want to draw it. To
recognize whether these connections are at?
Remember I said disconnected body. Now we
are disconnecting the ring d from the rest
of the body. According to what you wrote.
How many connection do you have? 1, 2 and--
3.
3 and every one of them should be represented
for time being only. Because it's a rope,
rope represents the tension. That's it, nothing
else. In future if this is a piece of metal,
maybe we have more stuff in that. Is that
correct or not? So that's the same principle
we are going to apply then. But here for this
one it's very simple. So here is the ring
b, is that correct? So write in your notes,
free body diagram. Free body diagram of b.
That should be sufficient. And then here you
put w. Of course before we do that first we
have to calculate w. So let's re-do, let's
do messing around so that's w over there,
yeah? In vector form. And then we have here--
this is the same problem. As you see this
is exactly the same problem as we did before.
Except I would like to call this tension b,
a. Why b, a? Because I want to show you. But
would you think the substrate I am giving
you the direction. Start at b goes toward--
A.
A. The other tension at a actually would be
going starting from a going to--
B.
B and that's tell me what you do? Is that
correct or not? Anyhow, so at-- what's the
angle here? It's 60 degrees. And therefore
this one which is d, b-- stop, I should put
that. I have to look at my free body diagram.
This is not going that way. It is going downward
at the angle of 30 degrees. So this is equal
like that. That is d, b, c and that angle
is 30 degrees, right? Now, I didn't use it
here because I should have. Usually when you
draw the free body diagram, put all the known
forces in one color. The unknown forces in
different color. To just show me or show yourself
which is known, which is-- which forces are
known. Which forces are not-- unknown. I didn't
do it here, sorry for that. But let's make
it a double color as we know. So these 2 forces
are unknown. And the black one is already--
Known.
Known because of what's given. Anyhow, making
the story short as you see it is very simple.
So therefore, first you calculate the w again.
The w of the lamp is equal to 4 kilograms
times 9.81-- 3 digits. Don't use 9.8, 9.81.
And that becomes 39.2 newton. So that's the
weight of the lamp to be 39.2 newton. Then
he question is what is the cable? Again, I
have to write 2 equations and 2 unknown. Okay,
the 2 equations I write it here. There is
enough room. Again, this is exactly the same
problem. No difference between this one and
the one I did in the spring. So you write
TBA magnitude times cosine 60 degrees and
it is negative, yes or no? Right? X component
and x component of this one which is positive.
Plus TBC cosine of 30 degrees equal to--
0.
0. Summation of x component, correct? Then
summation of y component. And we get plus
TBA sine of 60 degrees. Because y component
here is positive. The other 2 are--
Negative.
Negative minus TBC. Sine of 30 degrees minus
39.2 equal to 0. Again, 2 equations for 2
unknown. Can you solve it? Of course you can.
So I'm not going to bother you with that part.
And then we end up with the following scenario.
TBC becomes equal to 39.2 newton by accident.
And TBA becomes equal to 67.9 newton. This
is step 1 which was the free body diagram
of the b-- I better put b here. That would
complete our free body diagram. Now next move
on to the particle and do the same thing there.
It's-- actually that one is very simple. Let's
do it. This is-- write it down in your notes.
Free body diagram of ring, other ring is ring
c of ring c. And then this is force f vertical
and I don't have the direction of that. This
force we already have calculated. Now look
whatever this side the force going this way.
This side the force going upward, correct?
With the same magnitude, correct? Therefore,
if I want to put this is c going toward b
so TBC. But in magnitude already will calculate
it to be equal to 39.2 newton, is that correct
or not? Yes? Alright and the angle is 30 degree.
And then I have this force here which is the
angle of 30 degrees. But that force is tension
in cable c, d, yes or no? Yes, we can put
it in magnitude this time. Let's put it all
in magnitude. We can use all magnitude or
all weight so you have to be consistent, yes
or no? Do I need to solve this problem or
do you know the answer? What's the TCD should
be equal to without even doing any calculation?
Any answer? I don't hear you.
39.2.
Yes, 39 point--
2.
2, Ooviously, why?
It's the same.
It's symmetrical. Because x component-- if
x component of that must be equal to x component
of that. Both of them are 30 degree, must
be they're equal, yes or no? So I don't even
to solve it. This is quiz. I'm using your
ideas, not your time trying to calculate that.
Let's get this straight. In my quizzes there
is a concept. There is the calculation. Calculation
I try to put as much minimum effort of people
going into that. And the maximum part of your
quiz will be about the-- concept that you
are using. Of course these are very simple.
You already have seen it in physics. You will
repeat in a way, is that correct or not? But
when we get to the bulk of study the same
principle will be applied. So please be careful
that your procedure has to be correct. And
even you make sometimes mathematical error.
That doesn't bother me. By grading a little
bit difficult but I give you the maximum grade.
So if you're procedure is correct. You make
mathematical error. Still you get 80, 90,
95% of your grade because you know how to
solve the problem because of lack of time.
We're 50 minute duration. As I said before
you make a mistake here and there. That's
not for me,that's not that important. However,
as an engineer you should not make any mistakes.
You should not send a rover to the-- some
star and then forget to convert inch to the
millimeter. Remember this episode that they
sent $100 million rover somewhere. And they
forgot, one of the engineer-- this is the
rule. Always use the unit in every-- do this
in every problem. And I do, I always use newton,
newton, newton. Unfortunately many of you
don't do that. And carry that into your homework.
So each time that you are writing an equation.
Remember all the units, see that that matches
with what you are doing. And then you should
be okay. Okay the next question is problem
number 3. How do you handle problem number
3? Quickly. Is there any difference between
problem number 2 or 3? Notice problem number
1 and 2 are exactly the same, is that correct
or not? One is lamp, the other one is sand
or whatever I need that from there, yes or
no? Problem number 3 is a little bit different.
Sometime I have given this in the past but
student sometimes make a little bit of mistake
there. They shouldn't because you have seen
also that one in physics classes. Okay I am
going to do that on the middle board now.
Okay? How do we start on that one? What do
we have? And what you don't have? Please write
it down. Again, so you want to make a note
of all of this material. Alright, okay. First
of all it is an inclined surface there. And
there is a pulley here. There is a rope going
over the pulley. And there is a box here on
the slope. Slope are assumed, the contact
are assumed to be frictionless. They are in
the contact-- the boxes are in the contact
with the big glide surfaces. And then this
angle-- this is not to scale. It's given,
40 degrees. Apparently this is angle beta
which is not given. And then this is box 1.
And this is box 2. M 1 is given equal to 200
kilogram. M 2 is given equal to 300 kilogram.
They want you to find the angle beta and the
tension in the rope. Where should we start
guys? Assuming all the contact with the n
slide surface here is frictionless. The contact
with the pulley rope and pulley is frictionless.
When we get to chapter 8, we talk about the
friction in detail. Actually the whole chapter
is about the friction. We are going to talk
about it include the friction in every problem.
But for time being assuming everything is
frictionless, a smooth surface. Is that correct?
It's not a fact, is that correct or not? Yes?
How do we start? Where should I start? How
do it? First of course we must calculate the
weight. That is very simple. So weight of
w 1 in a format that we want to show is 200
times 9.81. All equal to 1,960 newton. And
weight of box 2 which is 300, you will do
the same thing. 9.81 all equal to 2,943 newton,
alright. Which box should I start? Of course
this time we do not need to draw the free
body diagram of the pulley. Because from physics
you know that if there is the rope. The tension
is in the pulley. If this side and that side
both tension will stay the same if there is
no friction. Is that correct or no? So this
is no use to draw the free body diagram of
the pulley. We have to use the free body diagram
of the--
Box.
Box, okay. So which one should I start first?
Obviously I should start here. Because this
one 40 degrees is given. And this one-- which
makes it very similar to the previous problem,
yes or no? So if in one class I gave you the
other quiz. Probably the second class I give
you this quiz. They never think they are the
same but actually they are--
The same.
The same. Everybody understand because I am
using the same concept, right? Alright so
box. How do we do that? Okay let's pull the
box there. Okay this box is attached, there
is a connection here, yes or no? Correct as
I said it. We are separating the box from
all its--
Connection.
Connection. So this time let's put all the
unknown in color. So you have here a tension
in the cable. I'm throwing in the magnitude,
yes or no? Yes?
Yes.
Okay what else? There is the--
Weight.
Weight. So this is as I said this a particle
remember? All the weight is the box. I'm showing
a box but all the forces going. Go concurrent
to the point of center of gravity of the box,
yes or no? Which is at the middle hopefully,
is that correct or not? I mean it's been material
are all the same. See so that everybody the
same. So therefore we have here the weight
of w. And the weight of w is how much? The
weight of w hanging there is 1,962. What else
should I have there? Contact force, contact
force, yes or no? Which is normal-- which
is normal to what?
The surface.
No because I have seen a few people make a
mistake sometimes. Don't make the mistake
that normal means vertical. That's not what
we mean normal. All of you are familiar normal
to the direction of the--
Surface.
Surface, is that correct or no? That's why
it's-- perpendicular to that surface. So wherever
that surface is, so I need to use the blue
color. Because that's an unknown. So this
is n. So magnitude of n is unknown, magnitude
of 2 is unknown. And what are the angles?
Let's see that. This angle here is 40 degrees,
yes? Now okay? This is the same problem. Look
exactly more or less the same. It has 2 unknown
and 1 known forces. As you said you can use
the triangle if you want. Or you use the component,
yes or no?
Yes.
What's the best way to do this? Actually the
best way-- the reason I am doing this problem
is this. Because to show you all the little
trick here which you have seen it in the past
by the way. Here I can chose the x, y axis
as usual, yes or no? But that's not necessary.
I can change my x, y axis for this problem.
Isn't that better to use x and y parallel
to the slope or perpendicular to the slope?
So let's do that. But I have seen people shifted
this which I don't like because the weight
should be always vertical, is that correct
or not? So changes your pages a little bit.
By choosing this system of coordinates x and
y. Notice t is in the direction of the x.
N is in the direction of y. The only first
which is not in the direction of the 2 axes.
It will be the w, okay? Now, based on this
system of coordinates. You write sigma f,
x equal to 0 sigma f, x. Becomes equal to
here you have t going to the right. N does
not have any-- see this is the y there. N--
oh before I do that, what is this angle guys?
Because I need that angle, don't I? Yes? 40.
Good. Everybody agree? Now I-- am happy if
all of you say 40 and you know that. But the
problem is in the quizzes. One third of people
put that angle wrong. I don't know why. Everybody
happy with that 40? How did you calculate
that to get that point? I show you here because
this is going to be repeated so many times.
So many equations although you said it correctly.
But I just for the benefit of the few who
didn't say anything. So I'm going to do that
one more time. How we calculate that. This
is usually typical of what you have-- to do
in many problems. If this is alpha, obviously
the angle of this line with the horizontal
line is alpha. Alpha could be any number,
is that correct or not? Yes? You are looking
for the angle of this one with a vertical,
yes or no? But this line is perpendicular
to the-- first line. This is line a, b. And
this is the line normal to a, b, is that correct
or not? Yes? I ask you what is this angle?
Is that correct or not? That's what I did
there. Is that correct? I asked you what's
that? Many of you immediately said 40. I love
it. Is that correct or not? Everybody is happy
with the answer. Now how did we get that?
Notice this is how you do it. If you extend
this line to here, this is alpha. So this
angle becomes how much? That angle, because
this is 90 degrees, that's the total. If in
this triangle-- if in this triangle a, c,
d. This is 90 degrees because these 2 lines
are perpendicular to each other, yes or no?
So if this is alpha this angle is how much?
90 minus alpha. Then you go to this little
triangle. If this is 90 minus alpha, this
must be--
Alpha.
Alpha. That's why you immediately come up
to that answer. Because you already have done
it once or twice, is that correct or not?
It's not a mystery. Just by doing it a couple
of times you do that. So everybody should
be practicing this. Because I tell you that
in many, many homework in future. We are going
to be immediately use that. I don't want any
mistake. Everybody should have the right answer.
It's very simple, you do it. It's that correct
or not? Yes? So actually there's another way
of looking at it. If this has an angle with
the-- angle of this line a, b. Which is horizontal
with alpha, this line cannot have the same
angle with a horizontal line. Must be with
the vertical, yes or no? Yes? 2 line perpendicular
to each other cannot have the same angle with
the horizontal, yes or no? One is alpha. The
other one is 90 minus alpha, is that correct
or not? So therefore that's your answer, absolutely,
very good. So sigma f, x equal to what? T
and then n doesn't contribute anything. Minus
because the y component-- the x component
going negative with that way. So minus 1,962
times sine of 40 degree. Because this is cosine,
the other one is sine. So we have sine equal
to 0, equal to 0, sorry for that. So immediately
we calculate t. T become equal to 1,261 newton.
And if you want the n we are running out of
time. So I don't know what you want to write
n? You write sigma f, y equal to 0, is that
correct? So my point is finish here because
I wanted to show you that. I wanted to show
you that. But let me put the answer here for
you. And n, this n becomes equal to by writing
sigma f, y which is this and that. This does
not contribute. So n becomes equal to 1, 503
newton. Now after you finish this box obviously
you go to the other box, yes or no? Yes? And
in the other box, let me draw the free body
diagram. But there it is-- I don't have time
to finish it. But it is-- actually it is very
simple. You can do it at home. Because we
want to move on to the other subject. So on
the other box, so you write the box there
again like this. And then you draw the weight
which was give, 2,943. That's the weight of
the box 2. So you put it there. And then you
have a tension here. The tension in the cable
also was given. Because tension we calculated
to be 1, 261. So 1,261 and then there is a
normal force. Notice what will happen here.
There is a-- because these are all going to
intersect in there. There is the normal force
here. And that angle of course is equal to--
Beta.
Beta. I just showed you there. Is that correct
or not? Yes? So how many unknown to you have
here in this scenario? This is known. This
is known. Your unknown is n and beta. Which
is I did it already for you in the other problem,
yes or no? You write the equation in terms
of n and beta. You divide by itself or where--
right? Okay, we are done for this part. Next
time we are going to talk about the pulley.
[ Music ]
