This document discusses equilibrium of particles and free body diagrams (FBD) in statics. It begins by defining equilibrium of a particle as having zero net external force. A particle is a model of a real body where all forces act at a single point. The document then discusses how to draw FBDs by showing all forces and moments acting on a body. It provides examples of drawing FBDs for various systems involving spheres, rings, and cables. It also discusses applying the equations of equilibrium to solve for unknown forces using the FBD approach.

1. Project inrodused to
dr.ahmad alshorman
Prepared by :
Osamah zahi melhem
Ahmad bassab alkhazaleh
Salah tawalbeh

2. Equilibrium of a particle
A particle is in equilibrium if the vector
sum of the external forces acting on it
is zero. Hence a particle is in
equilibrium if:
1. It is at rest and remains at rest −
Static.Equilibrium
2. It moves with constant velocity −
Dynamic.Equilibrium

3. What’s the particle
The particle is a model of a real body.
The word "particle" does not imply
that the particle is a small body.
Modelling a body as particle is
equivalent to the assumption that all
forces applied on body act at the
same point. This assumption is
acceptable in many.practical
engineering applications

4. Forces affect on Particles

5. Statics (MET 2214)
Prof. Simin Nasseri
Free Body Diagram (FBD)
How to draw a Free Body
Diagram:
Draw outlined shape - Imagine the
particle isolated or cut “free” from
its surroundings
Show all forces and moments -
Include “active forces” and
“reactive forces”. Place each
force and couple at the point that
it is applied.

6. Statics (MET 2214)
Prof. Simin Nasseri
FBD

7. Statics (MET 2214)
Prof. Simin Nasseri
FBD F.B.D of the ring A:

8. Equilibrium equations in
component form
1-in 2-d
2-in 3 -d

9. Example of 2-D

10. Example of 3-D

11. Statics (MET 2214)
Prof. Simin Nasseri
Example 1
The sphere has a mass of 6 kg and is supported
as shown. Draw a free-body diagram of the
sphere, cord CE, and the knot at C.

12. Statics (MET 2214)
Prof. Simin Nasseri
Sphere
There are two forces acting on the sphere. These are its weight
and the force of cord CE.
The weight is: W = 6 kg (9.81 m/s2) = 58.9 N.

13. Statics (MET 2214)
Prof. Simin Nasseri
FBD of sphere
This is the way we show the FBD of the sphere:
FCE
58.9 N

14. Statics (MET 2214)
Prof. Simin Nasseri
Cord CE
There are two forces acting on the cord. These are the
force of the sphere, and the force of the knot. A cord
is a tension only member. Newton’s third law
applies.

15. Statics (MET 2214)
Prof. Simin Nasseri
Knot at C
There are three forces acting on the knot at C. These
are the force of the cord CBA, and the force of the
cord CE, and the force of the spring CD.

16. Statics (MET 2214)
Prof. Simin Nasseri
FBD of the knot at C
FCE
FCBA
FCD
C
60o

17. Statics (MET 2214)
Prof. Simin Nasseri
Example 2
Draw the FBD diagram of the ring A:
W= 2.452 KN

18. Statics (MET 2214)
Prof. Simin Nasseri
FBD of the ring A
Is this the FBD of A?
No! this is not the free
body diagram of A!

19. Statics (MET 2214)
Prof. Simin Nasseri
FBD of the ring A

20. Statics (MET 2214)
Prof. Simin Nasseri
Example 3
Draw the free body diagrams of C and E and the cable CE:

21. Statics (MET 2214)
Prof. Simin Nasseri
FBD of E

22. Statics (MET 2214)
Prof. Simin Nasseri
FBD of C

23. Statics (MET 2214)
Prof. Simin Nasseri
FBD of cable EC

24. Statics (MET 2214)
Prof. Simin Nasseri
Example 4
Draw the FBD of ring A.
W=78.5 N

25. Statics (MET 2214)
Prof. Simin Nasseri
FBD of A

26. Statics (MET 2214)
Prof. Simin Nasseri
Part 2
Applying the Equilibrium Equations

27. Statics (MET 2214)
Prof. Simin Nasseri
FBD
Draw the free body diagrams:
W
N
W
N
f
Normal force = The force you have when there is a contact between
surfaces
(the ball is in contact with the ground).
Friction force = You have this when the surface in contact is not
frictionless and the friction prevents the motion of the object.
30

28. Statics (MET 2214)
Prof. Simin Nasseri
FBD
0
0
x
y
F
F




W
N
W
N
f
30
0
0
x
y
F
F




x
y
x
y

29. Statics (MET 2214)
Prof. Simin Nasseri
y
FBD
W
N
N
f
N = W
N - W.cos 30 = 0
f - W.sin 30 = 0
W.cos30
W.sin30
x
x
y

30. Statics (MET 2214)
Prof. Simin Nasseri
Example 2:
Determine the tension in cables AB and AD for
equilibrium of the 250 kg engine.
FBD of the ring A

31. B
0, cos30 0
0, sin30 2.452 0
Solving for T :
sin30 2.452 , 4.90
Subsituting into the first equation:
4.25
x B D
y B
B B
D
F T T
F T kN
T kN T kN
T kN
  
  
 



Solution of Example 2
According to the free body diagram of the ring A, we have
three forces acting on the ring. The forces TB and TD have
unknown magnitudes but known directions. Cable AC
exerts a downward force on A equal to:
W = (250kg)(9.81m/s2) = 2452N = 2.245KN
TBcos30
TBsin30

32. Moment
In statics, moments are effects (of a force) that cause
rotation. When computing equilibrium, you must be
able to calculate a moment for every force on your
free-body diagram. To determine a force's moment,
you use one of two different calculations, as you can
see in the following list.

33. calculations
Scalar calculation (for two dimensions): To
calculate the moment about a Point O in scalar
calculations, you need the magnitude of the force
and the perpendicular distance from Point O to
the line of action of the ForceF.
Vector calculation (for two or three
dimensions): To compute the moment vector
about a Point O in vector calculations, you must
determine the Force F in Cartesian vector form
and the position vector from Point O to the line of
action of the Force F.

34. Figures of moment

36. Moment of a couple
If two opposite moments act to cause an object
to rotate, such as when your two hand are at
the 'quarter-past-three' position on a car
steering wheel, it is called a couple. The
moment of a couple is called the torque. It is
quite often said of engines and applys to the
ability of the engine to turn the wheels, or
wrongly by Jeremy Clarkson from 'Top
Gear' as in, "This engine has a lot of torques."

37. an example of a moment of a couple