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Chapter 6 : Equilibrium of single rigid bodies
 6.1 Introduction
 6.2 Equilibrium equations and degrees of freedom
 6.3 Types of supports
 6.4 Procedure to analyze problems
 6.5 Selftest
 6.6 Problems, 2D
 6.7 Problems, 3D
In the preceeding chapters we have considered various aspects as to
how we can use a basic law of physics ( sum of the forces equal to zero)
to determine unknown forces in connection with individual particles where
rotational aspects were not of interest.
For single rigid bodies (like a bridge) we found in the preceeding
chapter that these considerations are not sufficient but that
the ability of forces to rotate a body has to be taken into account.
This lead to the definition of the moment of a force and to the
statement that the sum of the moment of ALL forces has to be zero in order
for the body not to rotate.
In this chapter I summarize these results and the procedures of analysis
as applied to determining forces acting on rigid bodies.
After briefly talking about several types of supports for rigid bodies
I will bring several workedout
examples first for singlebody and then for multibody problem.
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We have two types of equations resulting from :
 Vectorial sum of the forces equals zero
(6.2a)

or in component form :
 Vectorial sum of the moments equals zero
The sum of the moments around ANY point of ALL the
forces acting on a rigid body has to be ZERO.
(5.7a)

or in component form :
where the moment vector is defined by :
The vector
is defined as a vector going from the point of rotation to
anywhere on the line of action of the force
vector .
The point of rotation (the point around which you consider
the sum of the moments) can be chosen arbitrarily. Some choices
will require less computations than others.
Look here if you want to
know why your choice for the location of the point of
rotation is arbitrary.
In the case of 2dimensional problems (where we normally choose
as coordinates the x and yaxes ) forces have zero zcomponent.
Hence, the sum of zcomponents of all forces is automatically
satisfied and we have to consider only the sum of the forces
in the x and in the ydirection.
Furthermore, in 2dimensional problems, rotation takes place
only around an axis perpendicular to the display area (which is
usually the zaxis). Or in other words, none of the forces
has a moment around the x and the yaxis. Therefore only
the sum of the moments around the zaxis has to be considered.
Often we speak in this case of a point of rotation lying in the
plane of the display area.
Hence, as far as the equilibrium of a
single rigid body
is concerned the number of algebraic equations available is :
3 equations for 2dimensional problems
6 equations for 3dimensional problems
This puts restrictions on the types and number of forces which support
a single rigid body (system). If the number of unknowns ( = the number
of degrees of freedom ) resulting from the support forces is equal to
the number of available
equilibrium equations, we speak of a statically determinate system,
if the number of unknowns exceeds the number of available equilibrium equations, the system is called statically indeterminate. In such a case the
body's deformation for example (even if it is only minute) is needed to
solve for all the unknown support forces. If the number of unknowns is less
than the number of equations the body is said to be underconstrained,
the body will simply move and/or rotate under influence of the given
external loads.
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Discussing the different ways as to how a body can be supported is
important because the type of support tells you something about the
support force itself and the number of unknowns resulting from it.
Twodimensional supports 
Symbols/text used 
Active Force  # of unknowns 

 1 

 1 

 2 
Threedimensional supports 
Symbols/text used 
Active Force  # of unknowns 

 1 

 2 

Ball/socket 

 3 
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 Sketch the FreeBody Diagram of the rigid
body carefully. Interior aspects of the body usually are
not of interest. The F.B.D. should show
ALL forces acting on
the body from the outside.
This means : all given loads and/or
field forces ( gravity etc. ) and the unknown forces (often
called the support forces). Quite often it is important to
recognize that the problem statement includes information
about for example the direction of a support force
either in the text or in the figures.
In many problems the weight of some or all items
can be neglected. Reading the problem statement carefully will
tell you whether or not that is the case.
In real life you will have to err always on the
safe side.
 Choose a coordinate system if not given in the
problem statement. The location and orientation
of your system do not effect the principal outcome of
your calculations but influence the amount of work
you will have to do in order to arrive at a solution.
Look for dominant directions and align these with
one of the coordinate axis. By in large only experience
will teach you to find solutions with the least amount
of work.
Label uniquely all forces ad relevant distances as far as not
given already in the problem statement. The same label must
not be used for two different items.
 Set up the relevant equilibrium equations using the symbols
you defined previously yourself and which were given in the
problem statement.
If the problem is 2dimensional make use of the different
ways you can determine the moment of a force.
 Solve the derived equations for the unknown forces and
summarize your results.
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Sorry, none available. Humor me though and try those available
in previous chapters.
Chapter 2
Chapter 3
Chapter 4
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Problems in 2dimensions are distinct in that for a single rigid body you have
to consider :
Two equations for the sum of the forces, one for each direction.
The sum of the moment around an axis perpendicular to the plane in
which the force lie.

for a total of three algebraic equations. It is often of advantage to
substitute additional moment equation(s) for one or both sumoftheforces
equation(s).
Problem 6.6a : Crate in equilibrium
Problem 6.6b : Ladder against wall, I
Problem 6.6c : Ladder against wall, II
Problem 6.6d : Beam on incline
Problem 6.6e : Supports for a truss
Problem 6.6f : Underconstrained plate
Problem 6.6g : Balancing Act I
Problem 6.6h : Balancing Act II
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Problem 6.7a : Hooked beam
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Zig Herzog, hgn@psu.edu
Last revised: 01/12/07