1. Engineering Mechanics: Statics
Chapter 3:
Equilibrium of a Particle
Engineering Mechanics: Statics

2. Chapter Objectives
To introduce the concept of the free-body diagram for a particle.
To show how to solve particle equilibrium problems using the equations of equilibrium.

3. Chapter Outline Condition for the Equilibrium of a Particle
The Free-Body Diagram
Coplanar Systems
Three-Dimensional Force Systems

4. 3.1 Condition for the Equilibrium of a Particle
Particle at equilibrium if
- At rest
- Moving at constant a constant velocity
Newton’s first law of motion
∑F = 0
where ∑F is the vector sum of all the forces acting on the particle

5. 3.1 Condition for the Equilibrium of a Particle
Newton’s second law of motion
∑F = ma
When the force fulfill Newton's first law of motion,
ma = 0
a = 0
therefore, the particle is moving in constant velocity or at rest

6. 3.2 The Free-Body Diagram
Best representation of all the unknown forces (∑F) which acts on a body
A sketch showing the particle “free” from the surroundings with all the forces acting on it
Consider two common connections in this subject – Spring
– Cables and Pulleys

7. 3.2 The Free-Body Diagram Spring
- Linear elastic spring: change in length is directly proportional to the force acting on it
- spring constant or stiffness k:
defines the elasticity of
the spring
- Magnitude of force when spring
is elongated or compressed
F = ks

8. 3.2 The Free-Body Diagram Spring
where s is determined from the difference in spring’s deformed length l and its undeformed length lo
s = l - lo
- If s is positive, F “pull”
onto the spring
- If s is negative, F “push”

9. 3.2 The Free-Body Diagram Example Given lo = 0.4m and k = 500N/m
To stretch it until l = 0.6m, A force, F = ks
=(500N/m)(0.6m – 0.4m) = 100N is needed
To compress it until l = 0.2m,
A force, F = ks
=(500N/m)(0.2m – 0.4m)
= -100N is needed

10. 3.2 The Free-Body Diagram Cables and Pulley
- Cables (or cords) are assumed to have negligible weight and they cannot stretch
- A cable only support tension or pulling force
- Tension always acts in the
direction of the cable
- Tension force in a continuous
cable must have a constant
magnitude for equilibrium

11. 3.2 The Free-Body Diagram Cables and Pulley
- For any angle θ, the cable is subjected to
a constant tension T
throughout its length

12. 3.2 The Free-Body Diagram Procedure for Drawing a FBD
1. Draw outlined shape
- Isolate particle from its surroundings
2. Show all the forces
- Indicate all the forces
- Active forces: set the particle in motion
- Reactive forces: result of constraints and supports that tend to prevent motion

13. 3.2 The Free-Body Diagram Procedure for Drawing a FBD
3. Identify each forces
- Known forces should be labeled with proper magnitude and direction
- Letters are used to represent magnitude and directions of unknown forces

14. 3.2 The Free-Body Diagram
A spool is having a weight W which is suspended from the crane bottom
Consider FBD at A since these forces act on the ring
Cables AD exert a resultant force of W on the ring
Condition of equilibrium is used to obtained TB and TC

15. 3.2 The Free-Body Diagram
The bucket is held in equilibrium by the cable
Force in the cable = weight of the bucket
Isolate the bucket for FBD
Two forces acting on the bucket, weight W and force T of the cable
Resultant of forces = 0
W = T

16. 3.2 The Free-Body Diagram Example 3.1
The sphere has a mass of 6kg and is
supported. Draw a free-body diagram of the
sphere, the cord
CE and the knot at C.

17. 3.2 The Free-Body Diagram Solution FBD at Sphere
View Free Body Diagram
Solution
FBD at Sphere
Two forces acting, weight and the force on cord CE.
Weight of 6kg (9.81m/s2) = 58.9N

18. 3.2 The Free-Body Diagram Solution Cord CE
Two forces acting, force of the sphere and force of the knot
Newton’s Third Law: FCE is equal but opposite
FCE and FEC pull the cord in tension
For equilibrium, FCE = FEC

19. 3.2 The Free-Body Diagram Solution FBD at Knot
Three forces acting, force by cord CBA, cord CE and spring CD
Important to know that
the weight of the sphere
does not act directly on
the knot but subjected to
by the cord CE

20. 3.3 Coplanar Systems
A particle is subjected to coplanar forces in the x-y plane
Resolve into i and j components for equilibrium
∑Fx = 0
∑Fy = 0
Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal o zero

21. 3.3 Coplanar Systems Scalar Notation
- Sense of direction = an algebraic sign that corresponds to the arrowhead direction of the component along each axis
- For unknown magnitude, assume arrowhead sense of the force
- Since magnitude of the force is always positive, if the scalar is negative, the force is acting in the opposite direction

22. 3.3 Coplanar Systems Example Consider the free-body diagram of the
particle subjected to two forces
Assume unknown force F acts to the right for equilibrium
∑Fx = 0 ; + F + 10N = 0
F = -10N
Force F acts towards the left for equilibrium

23. 3.3 Coplanar Systems The chain exerts three forces on the ring at A.
The ring will not move, or will move with constant velocity, provided the summation of the forces along the y axis is zero
With any force known, the magnitude of other two forces are found by equations of equilibrium

24. 3.3 Coplanar Systems Procedure for Analysis 1. Free-Body Diagram
- Establish the x, y axes in any suitable orientation
- Label all the unknown and known forces magnitudes and directions
- Sense of the unknown force can be assumed

25. 3.3 Coplanar Systems Procedure for Analysis
2) Equations of Equilibrium
- Given two unknown with a spring, apply
F = ks
to find spring force using deformation of spring
- If the solution yields a negative result, the sense of force is the reserve of that shown in the free-body diagram

26. 3.3 Coplanar Systems Procedure for Analysis
2) Equations of Equilibrium
- Apply the equations of equilibrium
∑Fx = 0 ∑Fy = 0
- Components are positive if they are directed along the positive negative axis and negative, if directed along the negative axis

27. 3.3 Coplanar Systems Example 3.2 Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.

28. 3.3 Coplanar Systems Solution FBD at Point A
Initially, two forces acting, forces
of cables AB and AD
- Engine Weight
= (250kg)(9.81m/s2)
= 2.452kN supported by cable CA
Finally, three forces acting, forces
TB and TD and engine weight
on cable CA

29. 3.3 Coplanar Systems Solution +→ ∑Fx = 0; TBcos30° - TD = 0
+↑ ∑Fy = 0; TBsin30° kN = 0
Solving,
TB = 4.90kN
TD = 4.25kN
*Note: Neglect the weights of the cables since they are small compared to the weight of the engine

30. 3.3 Coplanar Systems Example 3.3 If the sack at A has a weight
of 20N (≈ 2kg), determine
the weight of the sack at B
and the force in each cord
needed to hold the system in
the equilibrium position
shown.

31. 3.3 Coplanar Systems Solution FBD at Point E
- Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA

32. 3.3 Coplanar Systems Solution +→ ∑Fx = 0; TEGsin30° - TECcos45° = 0
+↑ ∑Fy = 0; TEGcos30° - TECsin45° - 20N = 0
Solving,
TEC = 38.6kN
TEG = 54.6kN
*Note: use equilibrium at the ring to determine tension in CD and weight of B with TEC known

33. 3.3 Coplanar Systems Solution FBD at Point C
Three forces acting, forces by cable CD
and EC (known) and
weight of sack B on
cable CB

34. 3.3 Coplanar Systems Solution +→ ∑Fx = 0; 38.6cos30° - (4/5)TCD = 0
+↑ ∑Fy = 0; (3/5)TCD – 38.6sin45°N – WB = 0
Solving,
TCD = 34.1kN
WB = 47.8kN
*Note: components of TCD are proportional to the slope of the cord by the triangle

35. 3.4 Three-Dimensional Force Systems
For particle equilibrium
∑F = 0
Resolving into i, j, k components
∑Fxi + ∑Fyj + ∑Fzk = 0
Three scalar equations representing algebraic sums of the x, y, z forces
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0

36. 3.4 Three-Dimensional Force Systems
Make use of the three scalar equations to solve for unknowns such as angles or magnitudes of forces

37. 3.4 Three-Dimensional Force Systems
Ring at A subjected to force from hook and forces from each of the three chains
Hook force = weight of the electromagnet and the load, denoted as W
Three scalars equations applied to FBD to determine FB, FC and FD

38. 3.4 Three-Dimensional Force Systems
Procedure for Analysis
Free-body Diagram
Establish the z, y, z axes in any suitable orientation
Label all known and unknown force magnitudes and directions
Sense of a force with unknown magnitude can be assumed

39. 3.4 Three-Dimensional Force Systems
Procedure for Analysis
Equations of Equilibrium
Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 when forces can be easily resolved into x, y, z components
When geometry appears difficult, express each force as a Cartesian vector. Substitute vectors into ∑F = 0 and set i, j, k components = 0
Negative results indicate that the sense of the force is opposite to that shown in the FBD.

40. 3.4 Three-Dimensional Force Systems
Example 3.5
A 90N load is suspended from the hook. The
load is supported by two cables and a spring
having a stiffness k = 500N/m.
Determine the force in the
cables and the stretch of the
spring for equilibrium. Cable
AD lies in the x-y plane and
cable AC lies in the x-z plane.

41. 3.4 Three-Dimensional Force Systems
Solution
FBD at Point A
- Point A chosen as the forces are concurrent at this point

42. 3.4 Three-Dimensional Force Systems
Solution
Equations of Equilibrium,
∑Fx = 0; FDsin30° - (4/5)FC = 0
∑Fy = 0; -FDcos30° + FB = 0
∑Fz = 0; (3/5)FC – 90N = 0
Solving,
FC = 150N
FD = 240N
FB = 208N

43. 3.4 Three-Dimensional Force Systems
Solution
For the stretch of the spring,
FB = ksAB
208N = 500N/m(sAB)
sAB = 0.416m

44. 3.4 Three-Dimensional Force Systems
Example 3.6
Determine the magnitude
and coordinate direction
angles of force F that are
required for equilibrium of
the particle O.

45. 3.4 Three-Dimensional Force Systems
Solution
FBD at Point O
Four forces acting on
particle O

46. 3.4 Three-Dimensional Force Systems
Solution
Equations of Equilibrium
Expressing each forces in Cartesian vectors,
F1 = {400j} N
F2 = {-800k} N
F3 = F3(rB / rB)
= {-200i – 300j + 600k } N
F = Fxi + Fyj + Fzk

47. 3.4 Three-Dimensional Force Systems
Solution
For equilibrium,
∑F = 0; F1 + F2 + F3 + F = 0
400j - 800k - 200i – 300j + 600k
+ Fxi + Fyj + Fzk = 0
∑Fx = 0; Fx = 0 Fx = 200N
∑Fy = 0; 400 – Fy = 0 Fy = -100N
∑Fz = 0; Fz = 0 Fz = 200N

48. 3.4 Three-Dimensional Force Systems
Solution

49. 3.4 Three-Dimensional Force Systems
Example 3.7
Determine the force
developed in each cable
used to support the 40kN
(≈ 4 tonne) crate.

50. 3.4 Three-Dimensional Force Systems
Solution
FBD at Point A
To expose all three unknown forces in the cables

51. 3.4 Three-Dimensional Force Systems
Solution
Equations of Equilibrium
Expressing each forces in Cartesian vectors,
FB = FB(rB / rB)
= FBi – 0.424FBj FBk FC = FC (rC / rC)
= FCi – 0.424FCj FCk FD = FDi
W = -40k

52. 3.4 Three-Dimensional Force Systems
Solution
For equilibrium,
∑F = 0; FB + FC + FD + W = 0
-0.318FBi – 0.424FBj FBk FCi
– 0.424FCj FCk + FDi - 40k = 0
∑Fx = 0; FB FC + FD = 0
∑Fy = 0; – 0.424FB – 0.424FC = 0
∑Fz = 0; FB FC - 40 = 0

53. 3.4 Three-Dimensional Force Systems
Solution
Solving,
FB = FC = 23.6kN
FD = 15.0kN

54. Chapter Summary Equilibrium
A particle at rest or moving in constant velocity is in equilibrium
All forces acting on a particle = 0
To account for all the forces, draw a free-body diagram
FBD shows all the forces, with known or unknown magnitudes and directions

55. Chapter Summary Two Dimensional
Two scalars equations ∑Fx = 0 and ∑Fy = 0 is applied to x, y coordinate system
If the solution is negative, the sense of the force is opposite to that shown on the FBD.
If problem involves a linear spring, stretch or compression can be related to force by
F = ks

56. Chapter Summary Three Dimensions
For 3D problems, equilibrium equation ∑F = 0 should be applied using a Cartesian vector analysis
First, express forces on the FBD as Cartesian vectors
Second, sum the forces and equate the i, j and k components = 0 so that ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0

57. Chapter Review

58. Chapter Review

59. Chapter Review

60. Chapter Review