Concept of Particles and Free Body Diagram Why FBD diagrams are used during the analysis? It enables us to check the body for equilibrium. By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body. It helps to identify the forces and ensures the correct use of equation of equilibrium. Note: Reactions on two contacting bodies are equal and opposite on account of Newton's III Law. The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces. Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable. Physical Meaning of Equilibrium and its essence in Structural Application The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium. A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium. The rigid body in equilibrium means the body is stable. Equilibrium means net force and net moment acting on the body is zero. Essence in Structural Engineering To find the unknown parameters such as reaction forces and moments induced by the body. In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters. For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.

1. Equilibrium of Rigid Bodies

2. FORCES ARE VECTORS
THEREFORE WE NEED
TO USE THE
TECHNIQUES OF
VECTOR ALGEBRA

3. Catagories
2 D FORCE SYSTEMS

4. EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS

5. Only ONE unknownOnly ONE unknown (Force component)(Force component) can be foundcan be found
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS

6. Example:Example:
P
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
Determine the value of the force P so as toDetermine the value of the force P so as to
satisfy the equilibrium?satisfy the equilibrium?
F 0 -350+250-80+P=0 P=180 kN
x
+
→ = →∑

7. Example
Consider 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
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS

8. F1
Example:
If the stepped bar is in equilibrium find the force F1.
Resultant of Collinear Forces
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS

9. EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS

10. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y
plane its equilibrium condition equation can be written as

 ΣFx i + ΣFy j = 0
Both of these vector equations above to be valid, implies that
both the x and the y components should be equal to zero. Hence,
 +→ ΣFx = 0 F1x + F2x + ….. = 0
 +↑ ΣFy = 0 F1y + F2y + ….. = 0
Both algebraic sums equal to zero.
∑ = 0F

Only TWO unknowns can be foundOnly TWO unknowns can be found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
‘‘CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces’’

11. ∑ = 0F

0=∑ xF 0=∑ yF
0=+ ∑∑ jFiF yx

andand
TwoTwo Force componentForce component unknownunknownss cancan
be foundbe found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

12. • Resolve the given forces into i and j
components and apply the equilibrium
+→ ∑F∑Fxx = 0= 0
+↑ ∑F∑Fyy = 0= 0
• Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal to zero.
Only TWO unknowns can be foundOnly TWO unknowns can be found!!
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

13. Determine the magnitudes of F1 and F2 for
equilibrium. Set θ=60°.
Example:
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

14. FF11=1.827 kN F=1.827 kN F22=9.596 kN=9.596 kN
Only TWO unknowns can be foundOnly TWO unknowns can be found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

15. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example: (T)
53°
24°

16. Example
Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

17. SINCE the mass of the
engine is given i.e. unit is
‘kg’ (scalar) and not the
weight (FORCE)
the calculations should be
corrected to a vector having
a unit of Newton.
(mass * gravity )
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

18. 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
assummed
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

19. 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 axis and
negative, if directed along the negative
axis
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

20. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

21. Solution
FBD at Point A
- Initially, two forces acting, forces
of cables AB and AD
- Engine Weight [W=m.g]
= (250kg)(9.81m/s2
)
= 2.452 kN supported by cable CA
- Finally, three forces acting, forces
TB and TD and engine weight
on cable CA
FBD of the ring AFBD of the ring A
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

22. Solution
+→ ∑Fx = 0; TB cos30° - TD = 0
+↑ ∑Fy = 0; TB sin30° - 2.452 = 0
Solving,
TB = 4.904 kN
TD = 4.247 kN
*Note: Neglect the weights of the cables since they
are small compared to the weight of the engine
FBD of the ring AFBD of the ring A
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

23. Example
If the sack at A has a weight
of 20 N , determine
the weight of the sack at B
and the force in each cord
needed to hold the system in
the equilibrium position shown.
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

24. Solution
TEC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
TEC

25. FBD of the ring EFBD of the ring E
FBD of the ring CFBD of the ring C
TEC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

26. Solution
FBD at Point E.
Three forces acting,
forces of cables EG
and EC and the weight
of the sack on cable EA
FBD of the ring EFBD of the ring E
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

27. Solution
Use equilibrium at the ring to determine tension in
CD and weight of B with TEC known
+→ ∑Fx = 0; TEG sin30° - TECcos45° = 0
+↑ ∑Fy = 0; TEG cos30° - TECsin45° - 20 = 0
Solving,
TEC = 38.637 N
TEG = 54.641 N
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

28. FBD of the ring EFBD of the ring EFBD of the ring CFBD of the ring C
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

29. Solution
FBD at Point C
Three forces acting, forces by cable CD
and EC (known) and
weight of sack B on
cable CB.
FBD of the ringFBD of the ring CC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

30. Solution
+→ ∑Fx = 0; 38.637cos45° - (4/5)TCD = 0
+↑ ∑Fy = 0; (3/5)TCD + 38.637sin45° – WB = 0
Solving,
TCD = 34.151 N
WB = 47.811 N
*Note: components of TCD are proportional to the slope
of the cord by the 3-4-5 triangle
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

31. Example: (T)
The 50-kg homogenous smooth sphere rests on the 30°
incline A and bears against the smooth vertical wall B.
Calculate the contact forces at A and B?
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces

32. FBD of the sphereFBD of the sphere
30°
A
B
30°
A
B
30°
RRAA
RRBB
CC
WW
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example:

33. y A A
+
x B B
W 50x9.81 490.5
F 0 F cos30 -490.5= 0 F 566.381 N
(assumed direction correct)
F 0 566.381sin30 -F 0 F 283.191 N
+
= =
↑ = ° ⇒ =
→ = ° = ⇒ =
∑
∑
(assumed direction correct)
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example:
N

34. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS

35. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
3 unknowns3 unknowns
5 unknowns5 unknowns
3 unknowns3 unknowns
6 unknowns6 unknowns

36. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y-z
axes, its equilibrium condition equation can be written as

 ΣFx i + ΣFy j + ΣFz k = 0
Both of these vector equations above to be valid, implies that the
x, the y and the z components be equal to zero separately.
Hence,
 +→ ΣFx = 0 F1x + F2x + ….. = 0
 +↑ ΣFy = 0 F1y + F2y + ….. = 0
 + ΣFz = 0 F1z + F2z + ….. = 0
∑ = 0F

Only TOnly THREEHREE unknowns can be foundunknowns can be found
Three-D Force Systems
Concurrent at a point

37. When the system of external 3 dimensional
forces acting on an object in equilibrium:
Σ F = (ΣFx) i + (ΣFy) j + (ΣFz) k = 0
so each component of this equation must
be determined separately:
ΣΣFFxx =0,=0, ΣΣFFyy =0=0,, ΣΣFFzz =0.=0.
Three-D Force Systems
Concurrent at a point

38. • Resolve the given forces into i, j and k
components and apply the equilibrium
+→ ∑F∑Fxx = 0= 0
+↑ ∑F∑Fyy = 0= 0
+ ∑F∑Fzz = 0= 0
• Equations of equilibrium require that the algebraic
sum of x, y and z components must be equal to
zero.
TTHREEHREE unknowns can be foundunknowns can be found!!
Three-D Force Systems
Concurrent at a point

39. The 100-kg cylinder is
suspended from the
ceiling by cables
attached at points B, C
and D.
What are the tensions in
cables AB, AC & AD ?
Note that:
the gravity effect is in –the gravity effect is in –veve
y direction.y direction.
Example:
Three-D Force Systems
Concurrent at a point

40. Solution Strategy:
•Isolate the part of the cable system near point A,
•Obtain a free-body diagram subjected to forces due to the
tensions in the cables.
•Because the sums of the external forces in the x, y, and z
directions must IN BALANCE, obtain 3 INDEPENDENT
equations for the three unknown cables that are in tension.
•To do so, express the forces exerted by the tensions in
terms of their components.
Three-D Force Systems
Concurrent at a point

41. Drawing the Free-Body Diagram and Applying the Equations
Three-D Force Systems
Concurrent at a point

42. • Isolating the part of the cable system near point A and
show the forces exerted by the tensions in the cables.
The sum of the forces must equal zero:
Σ F = TAB + TAC + TAD − (981 N)j = 0
• Writing the Forces in Terms of their Components
• Obtain a unit vector that has the same direction as the
force TAB by dividing the position vector rAB from point
A to point B by its magnitude.
rAB = (xB − xA)i + (yB − yA)j + (zB − zA)k
= 4i + 4j +2k (m)
Three-D Force Systems
Concurrent at a point

43. Solution
kji
r
r
e 333.0667.0667.0 ++==
AB
AB
ABλ AB
Three-D Force Systems
Concurrent at a point

44. Expressing the force TAB in terms of its components by
writing it as the product of the tension TAB in cable AB
and the unit vector eAB...
TAB = TABeAB == TAB (0.667 i + 0. 667 j + 0.333 k)
Express the forces TAC and TAD in terms of their
components using the same procedure.
TAC = TAC (−0.408 i + 0.816 j − 0.408 k)
TAD = TAD (−0.514 i + 0.686 j + 0.514k )
λAB =
λAB
Three-D Force Systems
Concurrent at a point

45. Substituting these expressions into the equilibrium equation
TAB + TAC + TAD − (981 N)j = 0
Because the i, j, and k components must each equal to
zero, this results in three equations of:
i-component: 0.667TAB − 0.408TAC − 0.514TAD = 0
j-component: 0.667TAB + 0.816TAC + 0.686TAD = 981
k-component: 0.333TAB − 0.408TAC + 0.514TAD = 0
Solving these 3 equations successively, the tensions are:
TAB = 519 N
TAC = 636 N
Three-D Force Systems
Concurrent at a point

46. Three-D Force Systems
Concurrent at a point
Example: