MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING 

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


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.
  1. Cable Joint: The force must run through the centreline of the cable
  2. Frictionless Joint: The force must be perpendicular to the surface.
  3. 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.
  4. 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.
  5. Friction: The force can be in any direction
  6. Pin 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

Example: A Lifting Eye

Two ropes are attached to this lifting eye. Force A is at 75o, and Force B is at 60o from the horizontal. 

If the load is 240kg, what are the tensions in cable A and cable B?

The Free Body Diagram
1. Isolate: Take the eye as a dot
2. Border crossings: There are 3 forces
3. Line of action: Must run along cable centrelines
4. "To the body": Cables always pull


Notes:
  • You should convert to 360o notation in the FBD.
  • A FBD is almost always compulsory.
Graphical (CAD) method
Start with what you know.
We know the weight (gravity force) of 240kg load.
Fg = 240 * 9.81 = 2354.4N

See instruction for How to add forces in Solid Edge.

1. Start somewhere on the page (draw a small circle to show the start point).
2. Draw the gravity force: 2354.4 at 270o
3. Now draw Fa as a line from the end of the last one. Use an unknown length (take 1000 for starters) at 75o.
4. Go back to origin and draw Fb as another unknown length (say 1000) at 120o.
5. Now trim-corner Fa & Fb, Fb & Fg, into a triangle.
6. Add dimensions.
Mathematical (components) method
Step 1: Convert angles to 360 Notation:
Fa = Fa N at 75o
Fb = Fb N at 120o
Fg = 240 * 9.81 = 2354.4N at 270o

Step 2: Get X and Y components:
Fax = Fa * cos(75) = 0.2588*Fa
Fay = Fa * sin (75) = 0.9659*Fa
Fbx = Fb * cos(120) = -0.5*Fb
Fby = Fb * sin (120) = 0.8660*Fb
Fgx = 2354.4 * cos(270) = 0 N
Fgy = 2354.4 * sin (270) = -2354.4 N

Step 3: Write Equilibrium equations;
Fx = 0;
Fax + Fbx + Fgx = 0
0.2588*Fa  -  0.5*Fb + 0 = 0                    (eqn 1)
Fy = 0;
Fay + Fby + Fgy = 0
0.9659*Fa  +  0.8660*Fb - 2354.4 = 0        (eqn 2)

Step 4: Solve equations;
These are simultaneous equations that can be solved by substitution (Or matrices for many variables)
From eqn 1:  Fb = 0.5176*Fa            
Substitute this into eqn 2...
0.9659*Fa  +  0.8660*0.5176*Fa - 2354.4 = 0
Now we have one variable so we can solve it:
Fa = 2354.4 / 1.4142 = 1664.8 N
Now subs back into eqn 1..
Fb = 0.5176 * Fa = 861.77 N


Using CAD to check the maths. The Horiz/Vertical dimensions are added to check the Fx and Fy components of each force.
Mathematical (Triangle Geometry) Method
Since there are only 2 forces, the Force Polygon is a triangle. However, it is not a right-angle triangle, so we need the Cosine rule or the Sine rule.

Find the angles from geometry; 
B = 90 - 75 = 15o
C = 75 + (180-120) = 135o
A = 180 - B - C = 30o

So we can use the Sine Rule: (Whew!)
c/sin(C) = 3329.6
Amazing eh? This is the sine number for this triangle...
OK now, since
b/sin(B) = 3329.6
then
b = 3329.6*sin(15) = 861.77 N
Also
a = 3329.6 * sin(30) = 1664.81 N

Most maths books use capital letters for angles and lower case for the length of the opposite sides. Therefore;
  • COSINE RULE: a2 = b2 + c2 - 2*b*c*Cos (A)
  • SINE RULE: a/sin(A) = b/sin(B) = c/sin(C)

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).