Now it’s time to give you a bag of tricks
that you can use in order to solve the problems
easily. I am only going to focus on static’s
of systems of planar rigid bodies. Let’s
go to the first trick. I am going to call
it as first trick, trick number one. Let’s
say I have a particular rigid body and I have
forces all around and there is a moment. Let
me just draw so that it is easy. Let’s say
F1, F2, F3. There is a 
moment here. I have just drawn some moments
that may appear on this. I wish to find out
the resultant force and moment. One of the
important things to know is one of the problems
that we will face is which point should I
take in order to find the resultant. The answer
is any particular point it can take. There
is no difference as such because every point
of the rigid body will be stationary. I can
arbitrarily choose a point A, find the resultant
force F and the resultant moment MA so that
I can write these two equations.
Now some of the points to remember here is
can I instead of looking at this, this is
force at this particular point is equal to
zero and this is a resultant force and the
moment about this particular point A is equal
to zero. Now the question arises, supposing
I take some other point and let’s say I
found out the resultant moment and that is
let say MB. Can I add the other equation and
say MB equal to zero. Remember there are two
equations that I can find from here which
is x component and y component of resultant
force equal to zero, MA equal to zero. Supposing
I say, I am going to take another point here.
I know that this point is also stationary
and therefore the resultant force is going
to be the same. But moment will be different,
MB; can I add that say I will have the fourth
equation. Can I do this that’s the equation?
The answer to this is very simple. If I look
at this particular point B and try to find
out the moment given that this is the resultant
of forces at B, it is nothing but MB is equal
to MA plus the moment of this force with respect
to B.
So let’s say this is rA. This is equal to
rA cross F. We already have F is equal to
zero which means this will go to zero implies
this is equal to zero. This is equal to MA,
I already have MA equal to zero. This means
this is equal to zero or in other words given
that MA equal to zero and F equal to zero,
natural result is MB is equal to zero. This
cannot be an independent equation compared
to these two. This is one thing that we have
to understand which automatically means that
I will be able to get only three independent
equations that I can solve.
Now let say I have a force on a body and I
seek to find out the three equations that
are needed in order to solve this problem.
Let say these are the forces that are acting
on this. Like before we can choose a particular
point and go about doing it. We will get F
equal to zero and let say there is a moment
over here. We have MA equal to zero. Now if
I have to ask this question, can I take lets
say MA equal to zero, I will choose another
point B and say MB equal to zero. These two
together, if I take only these two equations
is this going to be valid. The answer is not
fully, I need to take one more equation, lets
say point here c and then find out the moment
of the resultant force and then said each
one of them equal to zero. Can I reproduce
the solutions that I get for this or in other
words, let me just repeat this. Can I rewrite
these set of equations in to this set of equations?
The answer is yes provided if I just connect,
let me just erase all this MA, MB, Mc so that
it is easy for you to understand, provided
these points A B C that I choose are not co-linear
and that this resultant force is non-zero.
If it is zero then what will happen is some
of these equations will give zero equal zero
which is the same that you will get here also.
If this resultant force is equal to zero,
we will get zero equal to zero or in other
words, if I have to replace an equivalent
set of equations, it is possible to use only
moment equations provided you choose three
points that are not lying on a single line
like this. Suppose take a point here C prime
which is co-linear, I will find that I will
reproduce the equation that I got for B and
C, a combination of A and B. As long as you
do this it’s easier. Sometimes it may be
beneficial to use these equations compared
to these equations. I will show those in the
examples.
Let’s take a simple example meaning simply
supported beam let say. Simply supported usually
is given by this. There is a pin support here
or hinge support there and a roller support
here. Let say this length is L, there is a
force acting on this let call this as P. The
aim is to find out the reactions say at A
and at B. We will use this system of equations
that we get from equilibrium equation in order
to solve this problem. The first exercise
that we have to do is draw the free body.
Remember this is a single rigid body that
we have here. So it is not going to be difficult
for us to draw free body of this. Let say
this distance is 2 L by 3 and this distance
is L by 3, removing this I will have two reactions.
One reaction like this and the other reaction
is like this.
Let call this Ay and this as Ax. Another reaction
over here, since it is a roller support it
does not restrain moment alone x direction
and therefore let say this is B or By and
this is a force p.
Let’s examine the number of unknowns. We
have Ax is an unknown, Ay is an unknown, By
is an unknown. Given p we need to find out
this. How many equations can we generate?
There are three equations we can generate
which means that we will be able to solve
this problem completely. Let’s try to solve
this problem. The first exercise is if you
look at this carefully there are three vertical
forces. The next thing that we do usually
is examine the types of forces that are present
in this particular rigid body. We have three
vertical forces, there is only one single
horizontal force. The first exercise I can
do is I can write the summation of all components
on the rigid body along x direction is equal
to zero that implies Ax equal to zero. This
is the immediate result that I can get.
What is the next step? I could probably go
for the net force in the y direction is equal
to zero that will involve Ay and B y. This
is where we use a simple trick. Instead of
going for vertical component being equal to
zero of the net force on this, let’s look
at any other option. We have the moment equation
that is possible. I can take a moment about
this particular point which means it will
involve Ay and By because they have a distance
from this particular point. But instead if
I take a point here on A or on B, I know that
the moment of Ay at A is equal to zero or
if I had taken this particular point, the
moment of by is equal to zero at this point.
So I will use either of these points and write
the moment equation. Let me write it over
here.
I am going to write sigma or summation of
all the moments at this vertical point A is
equal to zero. It automatically eliminates
use of Ay as well as Ax, only P and By will
come in to picture. Let me take anti clockwise
to be positive. This helps in writing it out
clearly. Again this exercise is simple. I
just hold this, look at what kind of moment
this will create. p will try to push it down
and therefore will introduce a clockwise rotation.
Anticlockwise is positive and this distance
I already know is 2 L by 3 and therefore it
will be minus P times 2 L by 3 is the contribution
of P for moment at A. The other one is to
do with By and that is at a distance L. I
hold again pivot at this particular point
A and c, what is the direction contribution
of By? It is anti-clockwise like this and
therefore this is plus By times L and that
is equal to zero because of the condition
that the rigid body is stationary. Remember
in this particular case this is known, these
are known which means I can immediately find
out what is By. L will get cancelled, it is
nothing but 2 P by 3.
The next step is to find out Ay. I can do
using vertical forces equal to zero or moment
about B equal to zero either one of them will
give the answer. But the key point that I
am trying to make out here is try and find
out a point about which you can take a moment
in such a way that you have the least number
of unknowns that take part in the equation.
If you do that then it is easier to solve
for reactions and that is very simple. If
this is 2 P by 3 and this is p, vertical equilibrium
will automatically give you Ay equals P by
3 because Ay plus B y equal to p. Thank you.
Now let’s look at one more trick or tip
that you can use before solving for systems
of rigid bodies. Supposing I have rigid body,
I just make it a rigid body of this sort and
lets say these are the two points this is
A and B on which some forces are acting. Let
say this is F1, this is F2.
Now one of the immediate simplifications that
can be done as far as this particular rigid
body is concerned is like we did earlier,
if you look at it we have force here and force
here and if we want to find out the relationships
or to solve for let say F2 or F1, we should
take moment about A or moment about B and
solve the problem. In this particular case
let’s say moment about A is equal to zero,
if this has to be stationary. Then what do
we get here? We are going to just change this
problem a little bit and see whether we can
get some simplification. Instead of taking
this let me take a component of this which
is F2x and component like this which is F2y.
Let’s look at the moment contribution of
these two components of this force. The immediate
thing that we find is since this x, let say
x is along this direction, if x where along
AB if I take AB and set that to be a x direction
then I know if I take moment about this, this
particular F2x will not take part in moment.
The only force that will take part in the
moment is F2y.
Therefore this will result in let say this
length is AB, this will result in F2y times
AB equals 0 where AB is the length. This automatically
means that this is equal to 0. By similar
argument if I take moment about this, it is
possible to show that the horizontal component
or the component along AB which is F1x and
F1y are the two components that we can split
and we can immediately say that this is equal
to 0 or in other words I can simplify this
as a problem where if this is A, this is B,
I will have forces acting on this way and
this way let say this is F1, this is F.
If I use horizontal equilibrium along AB,
I can write this to be minus 1. I have used
all the three equations in order to come up
with this particular conclusion. In other
words this is equivalent to saying that if
I have a body where forces are acting only
on two particular point AB then the forces
will act along 
that line AB and let’s call this as FAB.
Now this body that I am talking about need
not be one single direction like this, one
straight member like this.
Let me draw it over here, I can also have
something like this, let say this is point
B, this is point A. If I know that the forces
are acting only at A and B, I can just connect
these two and say that the resultant forces
on this particular rigid body will be acting
along this direction. If I have straight member
like this, straight body like this, we call
this as an axial member. If we have something
like this, the conclusion is that it is along
AB that the force resultant will be finally
acting. This result is very useful when we
come to systems of rigid bodies and examples
of system of rigid bodies being trusses, beams
and so on.
