Equilibrium of CONCURRENT
FORCES
If
all the forces on a body go through one point of intersection, they
are CONCURRENT. When this happens the forces cannot make the
body
rotate. This is the simplest situation for a Free Body Diagram in two
dimensions.

Lecture Notes : Concurrent-Forces.pdf Concurrent-Forces.one

Video: Concurrent Forces

VIDEO

What is Equilibrium?
Equilibrium
means "no acceleration".

Since
a force is a "push" or "pull" exerted on a body, equilibrium
means
that the total of all forces acting on a body must be zero.

According
to Newton's second law,

F
= m * a If

a
= 0 then

F
must be zero.

(Remember! In Newton's second law

F is the TOTAL force
on the body)

Since we are studying STATICS, from now on we
assume every body is in equilibrium.

Equilibrium
of Concurrent
Forces Concurrent
means that the forces intersect through a single point.

If
forces are concurrent, we can add them together as vectors to get the
resultant.

If the body is not accelerating, it must be
in equilibrium, so that means the resultant is zero.

For
concurrent forces, the body is a point.

So for concurrent
forces in equilibrium, the forces should all add up to give zero.

If a body is not accelerating is in equilibrium , so
resultant of all forces = 0.

A
typical concurrent force situation is a lifting eye. The
pulling
forces in any cables must pass through the centre of
the eye. If there is only one eyebolt (correctly positioned
over
the centre of gravity) and the load is suspended, the
bolt
force must pass through the same centre. Hence all forces pass through
one point (the centre of the eye), so we have concurrent equilibrium.

All
forces on a suspended load are concurrent. (Assuming the load remains
level when lifted). It is possible to maintain equilibrium
even
when the cables are at different angles. In the example below, Cable B
must have less tension than Cable A; Diagrams 1.
The Space Diagram (SD) The
initial problem is usually sketched. This illustration or picture shows
the layout and dimensions. If this diagram was drawn to scale, the
units would be length (mm, m etc). It is nice to be accurate,
but it does not have
to be to scale.2. The
Free Body Diagram (FBD) The Free Body Diagram is a strict
diagram that isolates the body for study. See Free Body Diagrams
for more information. The idea of the FBD is to focus on one particular
part or group of parts (called the body) and replace every external
member with the force they would apply.1. Isolate the body. (An
outline is best because we are supposed to forget about the inside of
the body)2.
Locate border crossings. Identify the contact points where
forces are crossing the boundary. Gravity acts through centre.3. Line of Action.
Some types of connections have a known direction. E.g. Cables have
force running through the centreline.4. "To the Body". Since
Newton's 3rd law has every action with an opposite reaction, we
must eliminate half the forces. Identify those forces that are applied
"to the body", and eliminate those done "by the body". If
the FBD were drawn to scale, the body might be length (mm, m etc) and
the forces might be another scale (N, kN etc). Warning! Do
not get Linear dimensions and Force dimensions mixed up. You cannot add
metres and Newtons together!3. The Force Polygon
(FP) The force polygon must be drawn strictly to
scale, and everything is a Force. The only information coming from the
FBD is;Force magnitudes Force
Angles Warning!
Do not attempt to bring any FBD Lengths into the FP. There are
no metres in the Force Polygon. In
some cases the Free Body diagram does not even look like the original.
This is most obvious for concurrent forces. Since all forces go through
one point, we can treat the body as a DOT!Cable
connection in a structure, specially designed to make the centreline of
every cables intersect at one point.

Example Diagrams. These
cranes are not accelerating, so they are in equilibrium. Therefore all
the forces on any body
should add up to zero. The body is actually the connection point which
is probably a lifting eye of a hook. The
FBD shows as much as we know from the Space diagram - in this case
angles are known but only one magnitude. The force polygon should form
a closed loop (since resultant = 0), so this defines the lengths (and
hence the magnitudes) of F1 and F2. CAD programs are very
helpful when working with force polygons.(Ivanoff)
Special Contact Points When
drawing the Free
Body Diagram we must include all the forces that cross the boundary
(outline) of the body. Some of these contact points have
special clues about the direction of the force and location of
the
force.Cable
Joint : The force must run through the centreline of the
cableFrictionless
Joint: The force must be perpendicular to the
surface.Wheels
and Rollers:
The force must run through the centre of the axle. Free running wheels
are frictionless so force is perpendicular to the surface and all
forces pass through the centre of the axle.Pulleys: The tension
in the cable is the same on each side of the pulley and all forces pass
can be made to go through the centre of the axle.Friction : The force
can be in any directionPin Joint: The force
can be in any direction Contact points are
also called Support Reactions; Here is a table (Ignore the
last one at this stage).Number
of Forces acting on a Body One Force This
is impossible for equilibrium. The forces are supposed to add up to
zero (unless the body is accelerating. E.g. A falling rock).Two
Forces If
a body has only 2 forces, they must be co-linear. E.g. A linkage
between 2 pivot pins must have the force running through the line of
the pins. (This assumes gravity force is ignored, otherwise you have
three forces)Three
Forces If
a body has exactly 3 forces, they must be concurrent.
This is called the Three Force Principle. This can be very handy in
solving problems because many mechanisms have bodies with 2 or
3
forces.Four or more Forces... We
cannot assume the forces will be concurrent, unless specially
made
that way. (Like the five-way cable connection below). When
forces
are not concurrent they can create rotations, which we deal with in a
later chapter. (Non Concurrent forces)Five deliberately concurrent
forces

The
Equilibrium Equations Equilibrium simply says the resultant
is zero. Mathematically, this can be stated that the Fx and Fy
components are zero. So, for concurrent forces in 2 dimensions
(planar), equilibrium means that... Very
often we know the angle of the forces but not the magnitudes. When
solving mathematically, this means we will need to use simultaneous
equations. (See example below)Worked Example 1
Using Goal Seek in Excel Goal
seek can be
used here. We are trying to find two variables (magnitudes of Fa and
Fb), but Goal Seek can only do one variable at a time. So
here's the trick... Guess
one of them first, then put the other force in terms of the guessed
variable, then
Goal Seek the original guess to give the answer you want.
Spreadsheet
(Excel) Method 1. Set up the problem as usual. Calculate Fg
etc. 2. Since we know the angles, get the sin() &
cos() to save time later. 3. Guess a value for Fa. 4.
Now write Fb in terms of Fa. for example, we know
Fax + Fbx = 0, so Fbx=-Fax. i.e. - 0.5
* Fb = - 0.258819 * Fa so
Fb = 517.64 N This
is wrong, but we don't know that yet. Now we check... Does Fy
= 0? Check if Fay + Fax + Fg = ? where... Fay
= 1000 * sin(75) = 965.93 N Fby = 517.64 * sin(120) =
448.29 N So get total Fy 965.93 + 448.29 - 2354.4 =
-940.186 N Whoops,
missed. But that's OK, we were only guessing Fa. Goal Seek We
want Excel to keep guessing Fa (cell B7) so that the final total (cell
B12) equals zero. Go to Tools > Goal Seek...
Follow the instructions. (In Excel 2007 it is in What if?) Excel
finds the answer for you! It seeks Fa to give 1664.81 N, and
so Fb is 861.77 N

Questions:
Concurrent Forces.pdf
(Ivanoff old edition) Ivanoff_concurrent.pdf (New version with answers)
Homework Assignment: Do
every question graphically (CAD) 4.1 to 4.6. Show FBD and
Force Polygon. Do these questions in Excel: 4.1,
4.2, 4.3 Do this question mathematically by simultaneous
equations: 4.1 Note: Excel is not permitted during this test (Tester: Exam mode).