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Let’s look at another problem here. Let
me just explain the problem first and then
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we will first draw the free body and seek
to solve the problem. I will just explain
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to you, how you will go about solving the
entire problem. There is a particular rod
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here A B, let me call this as E. So A E B
that is pinned at A to the fixed frame, as
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you can see here it is hashed. This is a fixed
frame. It is pinned to a body which is L in
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shape, B C D pinned to A B at B and it is
pinned to a wheel which is resting on A E
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B at E and this wheel is
just standing at E. Now this B C D is prevented
from moving at C and this is the problem.
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We wished to find out all the unknowns related
to this. When we say we have to solve the
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problem, we need to solve for the reactions
from the fixed supports and the reactions
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that appear as an interaction between the
rigid bodies. How do I go about doing this?
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The first exercise is to understand that there
are three rigid bodies in this.
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One rigid body is this A E B. The other rigid
body is B C D and the third rigid body is
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this particular wheel which is hinged about
D. There are three rigid bodies, this will
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help me draw the free body diagrams appropriately.
One is A B, you can either call it as A E
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B or A B. The other rigid body is B C D and
the wheel D, the three rigid bodies that we
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have. What is the first exercise that we need
to carry out? We need to remove the restrains
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that appear due to fixed frame of reference.
So let’s do that particular exercise to
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start with. Now I am going to just shift to
simple line diagram. I am just going to draw
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this A B as a line, B C D as an L and a circle
for a wheel. This is a point A B, we have
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like this which is BCD and you have a wheel.
Now what all have we removed here? We have
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removed the support that restrains motion
in the horizontal direction which means I
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should have a reaction from the fixed frame
on to this system of planar rigid bodies.
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So let’s call this as C or Cx along the
x direction. So we are going to take this
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as x direction and this as y direction for
our own convenience.
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At K it is pinned here which means the fixed
frame of reference doesn’t let it move with
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respect to the fixed frame of reference at
A, both in the vertical and the horizontal
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direction which means there will be two forces
that will be needed by the fixed frame of
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reference in order to restrain the motion.
Let’s call this as Ay and Ax. So we have
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3 unknowns Ax, Ay and Cx which are to be found
out from equilibrium equations. In addition
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in this particular problem it is given that
there is a moment acting here. Think of some
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motor that is applying a torque of 120 Newton
meter. I need to transfer that over here.
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There is a 120 Newton meter that is acting
external to this. Assume all the rigid bodies
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are negligible mass compared to the 120 Newton
meter that is applied. The first free body
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concerns this entire system without the fixed
frame of reference and it contains all the
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three. Then we seek to draw the free body
of each one of them separately.
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Let’s start with the wheel because that
seems to be the simplest one. This is the
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wheel D that we seek to draw. What happens
here as we have discussed earlier, in order
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to draw the free body of this wheel, I need
to make sure that I fix A E B and B C D the
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two other rigid bodies fixed to the fixed
frame of reference which means A E B and B
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C D are not moving at all. In which case this
point D of B C D will restrain this wheel
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from moving in x as well as y direction which
means I will have a reaction that appears
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on the wheel as Dy and Dx that is responsible
for preventing motion from x and y. Apart
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from that you already notice here that the
wheel is resting on A E B which means there
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is a normal reaction equal to, let’s just
call this as N E or since it is in y direction,
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we will just call it as Ey. We have a question
here. Will I have a horizontal reaction over
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here? I don’t know. Let’s say it is not
frictionless, it has friction. Let’s just
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introduce Ex for now.
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There is a resistance in the x direction.
we will come back to simplification as far
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as this wheel is concerned in a while. Have
you missed out anything else? the mass of
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this particular wheel is negligible which
means there are only two points that i have
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to consider. One point of restrain coming
from B C D and the other was one is due to
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resting at B. so this is complete as far as
wheel D is concerned. Let’s seek to draw
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the other two free bodies. I am going to use
line diagrams in order to make it simple.
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Now let’s just look at this particular B
C D. If you look at B C D, we have to fix
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A E B and the wheel to the fixed frame of
reference and have a look at what will be
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the restrains that will be offered. There
are three points at which restrains will be
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offered on this B C D, one is at B by A E
B. The other is at C due to fixed frame of
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reference and the other one is at D due to
this wheel remaining un moveable and therefore
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I will have a reaction.
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For now since we already know that we have
drawn Dx and Dy in this direction. We will
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take equal and opposite directions for this
point D, this is Dy Dx. We have finished this.
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Looking at this there is a restrain offered
here. So we will just stick to that. At B
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we will have a restrain in the vertical direction
and therefore a force appears. There is a
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restrain in the horizontal direction which
means there is another force Bx that appears.
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Are we complete? The answer is yes. The mass
we have here is negligible which means we
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have completed the free body of B C D.
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This is x direction, this is y direction.
What remains now is A E B. Again I am going
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to resort with line diagram, it makes it simple
for drawing. At B remember I am going to fix
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B C D as well as the wheel E D which means
at this particular point, I will have an equal
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and opposite reaction which means I will have
By here and Bx like this. Simple. At E there
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is an interaction with the wheel and therefore
I should take the opposite signs of this Ex
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and Ey and insert those as the forces, so
Ex and Ey. This way we can avoid adding more
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number of unknowns, we take equal and opposite
reactions automatically in this. At A we have
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already drawn for the system of rigid body.
I will have Ax and Ay appearing. We have drawn
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all the free body diagrams here that of wheel,
that of B C D and that of A E B.
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The next exercise is to find out what all
unknowns do we have to solve. Let’s look
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at that. The simplest one is this. I am going
to start with this particular wheel. Let me
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draw that again here. Now we are looking at
the unknowns. For the wheel we have Dx and
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Dy, Ex and Ey. We already accounted for this,
we have Cx that needs to be solved and we
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have Bx and By.
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For this body we have to solve for Ax and
Ay, we already accounted for Ex, Ey, Bx, By
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which means Ax and Ay. In all how many? 2
plus 2, 4 plus 4, 5 plus 4 is 9 unknowns.
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Can I write down 9 equations? The answer is
yes. Because I have 1, 2, 3 rigid bodies.
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Each can generate 3 equations and therefore
I can solve for the 9 unknowns. Now if I do
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it this way, it will be a long procedure.
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Let me look for certain simplifications that
I can do. So simplification one. If we examine
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the wheel, let me draw it over here. You have
Dx, Dy, Ex, Ey. These are the only forces
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that are acting. If I take moment about this
particular point and immediate result is that
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Ey is along the vertical direction that passes
through D. Dx and Dy don’t take part in
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that moment equilibrium which means only Ex
takes part in the moment equilibrium. Therefore
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if I take moment about D lets say positive,
there is only one force that results in that.
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Let’s say this radius is r, we already know
that this is 0.2 meters. We get it is anticlockwise
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in its motion Ex times 0.2 should be equal
to 0. It immediately gives me Ex equals zero.
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There is no horizontal force that appears
on this. This is clear? Now if Ex is equal
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to 0, sigma Fx equals 0 will immediately tell
me that Dx is equal to 0. You get this? I
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am just going to erase it to make it simple,
this is not there, this is not there. Like
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what we did earlier, for a rigid body where
there are only two forces two points at which
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forces are acting, immediate answer that we
get here is Dy and Ey are opposite to each
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other.
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So sigma Fy equals 0 gives me Ey equals minus
Dy. I need not now look at all these four.
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I just need one unknown, either Ey or Dy and
that will reduce the unknowns in the other
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free body diagrams. You have Dx and Dy here,
you have Ex and Ey over here. So let’s seek
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to now remove those so that we simplify. Please
remember these simplifications will help us
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solve in a much better way. I am going to
remove this, I am going to call this as Ey
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equal and opposite. So I am going to remove
this and shall we just take this instead of
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Ey, we will just put it as normal reaction.
That’s the reaction that we see from the
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ground for a wheel. There will be N acting
on this. Here also I don’t have this Ex,
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this is nothing but the N acting from the
wheel. Now we have reduced it.
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We don’t have this anymore, this anymore,
instead it is replaced by a single unknown
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N. Now instead of 9 equations, we have 1,
2, 3, 4, 5, 6 equations to be solved. Mind
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you in doing that we have already used up
all the three equations necessary for the
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wheel. Going back to this, there are 1, 2,
3, 4 unknowns present in this. 1, 2, 3, 4,
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5 unknowns present in this. In order to solve
for this, what would be the points we take
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in order to write down the moment equations.
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Is there a simplification possible in this?
There are simplifications that we can think
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of but mind you, I don’t have to look at
the wheel anymore. That’s already taken
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care of. How do I solve for the other unknowns?
We already saw 5 here, 5 there. If I take
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moment about this point, what all unknowns
can I avoid? I can avoid Ax, Ay as well as
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Bx which means I will have an equation that
is in terms of N and By. Is that clear? So
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N and By will take part, if I have equal zero.
This will have N1 and By taking part. Go to
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the other free body, what is the equation
that gives me a relationship between N and
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By only? That’s not very difficult, it is
the vertical force equilibrium.
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So far this, I will do this. I will do sigma
Fy equal zero that it will not involve Bx
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as well as Cx, it will involve only these
two N and By. This involved N and By, this
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involved N and By, I can solve using these
two equations N and By. Once I have solved
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for N and By, I am now left with Ax, Ay, Bx.
in this case Bx and Cx. If I take the vertical
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equilibrium here, Ay can be solved for because
N and By are already known. So I can solve
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for that by simply taking sigma Fy equals
zero here.
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I am left with the horizontal equilibrium
there that will relate Ax and Bx. If I take
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the horizontal equilibrium there, it will
relate Bx and Cx. I am sorry, I should have
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inserted 120 Newton meter here. Taking moment
about A, I should involve this 120 Newton
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meter. So Ax equals Bx, Cx equals Bx are the
two equations I get. I need three equations
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because I have Ax, Bx and Cx. How do I get
the third one? I will go back to this guide
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and if I take moment about this, I avoid Ax
and Ay. There is only Cx that needs to be
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solved here.
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Remember this system of planar bodies free
body will help me solve for Cx directly. This
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is an important point you have to note. If
I do sigma MA equals zero here, I will get
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this directly. Having got on this directly
I can find out Bx, having found out Bx I can
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found out Ax which means I have solved for
all the unknowns in this particular problem.
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Please use this kind of an approach in order
to solve the problem, else you will end up
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with 9 equations with 9 unknowns that need
to be solved and it will be too cumbersome
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for you to solve.
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It is important to look at simplifications
that you can do, you will always have to look
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at, is there a simple way to solve for one
unknown or the other. If you do use this approach,
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it will be possible for you to solve many
problems in a very simple way. You can draw,
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you can make these strips in order to understand
the problem that we discussed. This is that
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A B, let me draw this point of interaction
as E.
Now this is a point fixed to the fixed frame,
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this is the point A fixed to the fixed frame
and remember this is not fixed to the fixed
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frame but it fixes B C D and A E B. Now we
will remove each restrain and insert a force.
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As before this has x direction this has y
direction.
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This restrain essentially does not allow the
points C from moving in this direction. I
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am going to remove this and insert a force
like this. Nice to do this kind of an exercise,
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this gives us understanding. In order to make
this body free from the fixed frame, I need
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to remove A. What does A do? It doesn’t
allow the point A to move either in vertical
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or in horizontal direction. I am going to
insert the two restrains Ax and Ay. Now this
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is a free body that you have for the entire
system. Supposing I have to remove this particular
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planar body from the system, I need to fix
A B and fix B C D. The pin at point D does
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not allow it to move. So what I am going to
do is I am going to remove this and insert
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two forces that prevented movement at D. In
addition at this point E, there is horizontal
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and vertical restrains so I can add those
two. An equal and opposite reaction has to
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be inserted over here.
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I am just going to remove this and insert
it over here. There is a restrain this way
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equal and opposite restrain as inserted over
here, let me just move it over here like this.
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There is one more like this that I have to
insert, an equal and opposite reaction at
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D. Here we have made a simplification to find
out that these two are zero. We might as well
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remove them which mean this is also zero.
You can do this kind of simplification in
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order to understand, how to draw the free
body diagram.
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Now if you look at it, I have a force over
here let me retain that force. There is another
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force on this. If I have to remove A E B now,
I have to remove the restrain here. When I
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remove the restrain here, I need to add the
two restrains related to that and this is
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the free body diagram of B C D.
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In a very similar way you can do the same
thing. This point is restrained by B C D and
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therefore let me just take this out. It will
have a reaction like this. A is already restrained
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So let me just take those two. These are the
two other restraints. At E also I have one
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restraint that comes from the wheel. I have
used something like this, so this is like
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this. This is E upward equal and opposite
reaction occurs here and this is the free
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body apart from the moment that occurs at
A. This should be very clear to you. Try this
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exercise a few times to get a physical feel
of how to draw the free body. Once you become
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a master in free body diagrams then lot of
things can be accomplished in a very simple
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way.
Thank you.
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