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Today we will see on systems of rigid bodies,
supposing there are forces and moments acting.
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How do I find out equations of equilibrium?
I will focus mainly on planar rigid bodies.
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I am going to give a few tricks that you can
use in order to solve problems. From a problem
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solving angle only, I am going to look at
this particular session. To start with let’s
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examine on planar rigid body subjected to
various forces let’s say F1, F2, F3, F4.
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Let’s say there is a couple here MA and
so on. One thing that we already know is we
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can convert this in to an equivalent rigid
body, equivalent body or this body with…
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We know that we can write an equivalent of
these four systems on the same rigid body
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as resultant force acting, so let me call
this as F. All these are vectors and resultant
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moment lets say this is point o, Mo. Now over
this we have to apply the condition of what
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is happening to this particular body. Is it
moving or is it stationary?
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Let’s look at the first question of it being
stationary. If it is being stationary, we
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say that the acceleration is equal to 0. It
is a principle of static’s that we used
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right now. It is not rotating which means
the angular acceleration is 0, it is not moving
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in the x or y direction which means those
2 accelerations are equal to 0. If we apply
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the principle F equals m a, mind you that
I am writing this as vector. This m is the
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mass of this particular rigid body. If this
is equal to 0 we get F equal to 0 and this
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is one equation.
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The other equation has to do with the moment
is equal to I times alpha where alpha is the
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angular acceleration of this rigid body. I
am just writing it as scalar. It is something
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that is about an axis that is perpendicular
to the board and this is also equal to 0 in
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a condition of static’s which means we get
these two, mind you these two vector equations.
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This is only one particular direction and
therefore we just we are going to make it
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as scalar equation. This will have two components,
the moment we give an x direction and a y
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direction.
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Therefore this will simplify to
F equals zero will imply its first component
Fx equal to zero and the component along y
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direction being equal to zero and the moment
about o equal to zero. Fx and Fy are components
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of this vector F which is a resultant of all
these forces put together which you have already
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seen. This is a simple procedure, we can find
the relationship between the forces and moments
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by solving these equations. We can solve for
these values, once we know the forces acting
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on the body.
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Let’s look at a simple example. Supposing
I have a body which has a force F acting like
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this, along horizontal and there is a particular
point here
which is hinged. There is another point over
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here which is a roller support. I wish to
find out what will be the resultant force,
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resultant reaction that will come on to this
particular rigid body. Very simple problem.
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The first exercise I have to do here is I
have to now represent all these constraints
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that I have here which is the hinged constraint
and the roller constraint has forces that
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take part in the constraint. We call those
as reactions.
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In this particular rigid body this hinge will
offer two reactions. I am going to represent
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this as x and this as y. So I will have one
reaction like this, let say this is A, this
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is B. This is point C which will have a horizontal
component of this so let me call that as Ax.
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There is a vertical component of this reaction
Ay and there is a particular direction to
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this, so let me just represent the direction.
Let’s say this angle is theta with respect
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to vertical. I can represent this as a force
B. Mind you in this particular case, I know
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at what angle the force is acting which means
only the magnitude is an unknown. Here the
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two unknowns are Ax and Ay. So if I complete
it, this shows a free body diagram. This free
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body diagram is essential in order to solve
for unknowns using system of equations. Let’s
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write it down.
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I am looking at static’s, I know that each
of these resultants components have to be
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equal to zero. I have to convert this in to
same rigid body this is the same rigid body
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I am drawing. I have force here, force here,
force here, force here and a force over here.
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Let’s say the resultant force is acting
like this. This is vector F. I am taking a
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point over here which is let’s say o. We
will come to what point to take at a later
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stage and a moment m at o. If this is stationary,
the force should be equal to 0. The vector
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itself should be equal to zero or Fx Fy should
be equal to 0. I mean just change to P so
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that we don’t confuse between this F and
that F. This M0 which is the moment about
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an axis that is coming out is also equal to
0.
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How do we write F? F is nothing but we have
these forces Ax acting along this direction
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plus Ay acting along this direction plus B
which is acting along the direction theta
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two vertical plus p that is acting along horizontal
direction. Mind you it’s a vector equation.
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I have to now separate this in to horizontal
components and vertical components. The other
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equation, so if it is stationary I have to
say this is equal to 0. The other equation
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tells me that M0 should be equal to 0, M0
I can find out by lets just do that exercise.
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I have force P over here, if have to find
out the moment of this p about this o, what
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is a relationship? Let me call this point
as one and let say this is a vector r. Then
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the moment about this o power of force p can
be given as M is equal to r cross p.
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In this particular case since it is just a
planar problem, if I find out the perpendicular
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distance here let’s say I am going to call
this as d1 then this is simply, the direction
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is going to be like this. If I hold some object
and pull it this way, it is going to rotate
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like this, as an understanding that I can
have and therefore I can say this is minus
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d1 times p. p is the value of this force.
I know p is along the horizontal direction.
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So primarily x component of p. So like this
I can find out, this is Mo for the first point.
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We will have to do that for the other force
B, the force Ay and the force Ax and then
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insert it over here, in order to get relationships
between these forces.
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One more important thing to understand here
is I wish to tell clearly which is a direction
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that is assumed to be positive. If you go
back, we have used x positive direction to
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be like this, y positive direction to be this
way and if I take vector outward to the board,
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it’s a positive direction. If I have a vector
outward, the moment is going to be anticlockwise
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which is taken to be positive. One way of
doing this is I can write Fx or the component
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of this force along this x direction which
is this positive equal to zero. This is a
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positive quantity equal to 0. I can do this
same thing over here, M at o taking anticlockwise
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to be positive is equal to 0.
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By doing this we will not miss out on the
sign conventions. For example in this particular
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case, this particular force p is trying to
rotate about this point o in a clockwise direction
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which is negative of anticlockwise and you
can see here d1 is a positive number, p magnitude
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is a positive number which means we will have
a negative sign over here indicating that
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it is actually a positive clockwise value
represented by anticlockwise. You can do the
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same exercise for each one of these. One of
the advantages of planar problem is all I
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need to do is to find out the distance of
the vector direction to a particular point
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in question.
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For example if this is the force B, I can
extend this line. Find out what would be the
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perpendicular distance of o with respect to
this B? Let me call this as d1. Again if you
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notice, the force is such that acting to rotate
the body in the clockwise direction. I will
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get a negative value when I insert it in this
particular thing and this is how I can solve
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the problem. In the examples we will have
a few problems that you will tackle. But before
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ending there is one important trick or principle
that we should know. Remember whether the
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force is over here or the force is over here.
A moment of this force about o will remain
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the same.
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Even if I had put the force over here, let’s
say this point is two, I would have gotten
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the same moment. So it’s only the perpendicular
distance of the force with respect to particular
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point that is important to us. In a similar
way here, it is simpler if I write it as Ax
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and Ay because the distance that I can find
out for this.
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For example I am just going to erase this
and draw fresh so that you can understand,
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how easy it is if I write it in terms of components.
If this is the body, if I take this force
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p, I know this is a horizontal force. All
I need to know is to find out the vertical
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coordinate difference between this point o
and this point one. This y coordinate one
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with respect to o will immediately tell me
that is the lever arm that I have to use to
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multiply with the magnitude of p.
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Similarly if I have to split this reaction
in to Ax and Ay, it becomes easy because all
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I need to do is I had to find out the difference
between the y coordinates of this point. Let
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me call this as three between the y coordinate
of three and y coordinate of o or in other
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words this is y3 minus yo. Then I take this
magnitude, multiply with Ax. Again to make
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it very simple, I will physically look at
what the direction is. If this is the positive
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distance and this is the way, the forces acting,
it is going to go anticlockwise which means
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in this particular equation I will have it
as positive.
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So in that sense if I write, this will turn
out to be Ax times, let me call this as y0
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3 distance between o and 3. This is a positive
sign because it is going anticlockwise. On
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the other hand if I take this one, it is going
to go clockwise so I will have a minus Ay
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times… right now I have to look at the distance
in the horizontal direction which means this
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is o, this is x o three. Here I am not going
to bother about whether 3 o or o 3. I will
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just take the magnitude of the distance. This
negative sign will take care of all the other
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list.
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The notion of directions that we have used
is here have to be retained. Let’s complete
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this by adding these. There is this force
B that is acting, let’s say that is acting
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like this at an angle theta, again I can find
out the perpendicular distance from it and
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use that particular thing. Alternatively I
can also write this force as two components
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like this. This is By and Bx. In this particular
case given B, I can write By should be equal
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to B cos theta and this to be B sine theta
where theta is already known. By writing this,
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if I take this coordinate let’s say this
is 4 I can use the coordinate and write the
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moment with respect to By separately and Bx
separately. So I will add that particular
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part to it and then I will take. I have completed
finding out moment for everything and if I
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said this is equal to 0, I will get a relationship,
an additional relationship. How many unknowns
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do I have here?
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Going back to this particular diagram, this
is something I don’t know. This is something
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I don’t know. The magnitude of this is something
that I don’t know. 1 2 3, I should have
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3 equations here, 3 independent equations
that I should be able to solve. This force
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equation which translates in to two component
equations will give me two equations and the
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moment equation together will give three equations
which will help solve these three unknowns.
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This is all to it solving a problem with a
single rigid body. If it is a multiple rigid
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body or a system of planar rigid bodies, we
will have to do a few more amendments I am
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going to talk about it.
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