statics

1. Engineering Mechanics: Statics
Fifteenth Edition
Chapter 3
Equilibrium of a Particle
Section 3.1 Condition for the
Equilibrium of a Particle
Section 3.2 The Free-Body
Diagram
Section 3.3 Coplanar Force
Systems

2. Equilibrium of A Particle, The Free-
body Diagram & Coplanar Force
Systems
Objectives:
a. Draw a free body diagram
(FBD)
b. Apply equations of equilibrium
to solve a 2-D problem.

3. Reading Quiz (1 of 2)
1. When a particle is in equilibrium, the sum of forces acting
on it equals ______ . (Choose the most appropriate
answer)
A. A constant
B. A positive number
C. Zero
D. A negative number
E. An integer

4. Reading Quiz (2 of 2)
2. For a frictionless pulley and cable, tensions in the cable
(T1 and T2) are related as _______.
A. 
1 2
T T
B. 
1 2
T T
C. 
1 2
T T
D. 1 2
T T sin
 

5. Applications (1 of 3)
The crane is lifting a load. To decide if the straps holding the
load to the crane hook will fail, you need to know forces in the
straps. How could you find those forces?

6. Applications (2 of 3)
For a spool of given weight, how would you find the forces in
cables AB and AC? If designing a spreader bar like the one
being used here, you need to know the forces to make sure the
rigging doesn’t fail.

7. Applications (3 of 3)
For a given force exerted on the boat’s towing pendant, what
are the forces in the bridle cables? What size of cable must
you use?

8. Section 3.3: Coplanar Force Systems
This is an example of a 2-D or
coplanar force system.
If the whole assembly is in
equilibrium, then particle A is
also in equilibrium.
To determine the tensions in
the cables for a given weight of
cylinder, you need to learn how
to draw a free-body diagram
and apply the equations of
equilibrium.

9. The What, Why and How of A Free
Body Diagram (FB D) (1 Of 2)
Free-body diagrams are one of the most important things
for you to know how to draw and use for statics!
What? - It is a drawing (sketch) that shows all external forces
acting on the particle (or body).
Why? - It is key to being able to write the equations of
equilibrium—which are used to solve for the unknowns (usually
forces or angles).

10. The What, Why and How of A Free
Body Diagram (FBD) (2 of 2)
How?
1. Imagine the particle to be isolated
(or cut free) from its surroundings.
2. Show all the forces that act on
the particle.
Active forces: They want to move
the particle.
Reactive forces: They tend to
resist the motion.
3. Identify each force and show
all known magnitudes and
directions. Show all unknown
magnitudes and / or directions as
variables Note: Given cylinder mass = 40Kg

11. Equations of 2-D Equilibrium (1 of 2)
Particle A is in equilibrium, therefore the net force at A is zero.
So   0
B C D
F F F
or 
 0
F
In general, for a particle in
equilibrium,

 0 or
F
   
 
X y
F F 0 0 0
i j i j (a vector equation)
Or, written in a scalar form,
 
X Y
F 0 and F 0
 
These are two scalar equations of equilibrium (E-of-E). They can be used
to solve for up to two unknowns.

12. Equations of 2-D Equilibrium (2 of 2)
Write the scalar E-of-E:
o
X B D
F F cos 30 F 0
    

o
Y B
F F Sin30 392.4N 0
   

Solving the second equation gives: 
B
F 785N

From the first equation, we get: 
D
F 680N

Question: Suppose you chose the opposite direction for ?
How does this change the answer?

13. Simple Springs
Spring Forces = spring constant X deformation of spring
F= k s

14. Cables and Pulleys
With a frictionless pulley and cable
1 2
T T
 
Cable can support only a tension or “pulling” force, and this
force always acts in the direction of the cable.
Cable is in tension

15. Smooth Contact
If an object rests on a smooth
surface, then the surface will exert a
force on the object that is normal to
the surface at the point of contact.
In addition to this normal force N, the
cylinder is also subjected to its weight
W and the force T of the cord.
Since these three forces are concurrent
at the center of the cylinder, we can
apply the equation of equilibrium to this
“particle,” which is the same as
applying it to the cylinder.

16. Example I (1 of 2)
Given: The box weighs 550 lb and
geometry is as shown.
Find: The forces in the ropes AB
and AC.
Plan:
1. Draw a FB D for point A.
2. Apply the E-of-E to solve for the
forces in ropes AB and AC.

17. Example I (2 of 2)
Applying the scalar E-of-E at A, we get;
o
X B C
4
F F cos30 F 0
5
 
   
 
 

o
Y B C
3
F F sin30 F 550 lb 0
5
 
    
 
 

Solving the above equations, we get;
𝑭 𝑩=𝟒𝟕𝟖 𝒍𝒃 𝑭 𝑪=𝟓𝟏𝟖𝒍𝒃

18. Example Ⅱ (1 of 2)
Given: The mass of cylinder C
is 40 kgrand geometry
is as shown.
Find: The tensions in cables D
E, EA, and E B.
Plan:
1. Draw a FBD for point E.
2. Apply the E-of-E to solve for the forces
in cables DE, EA, and EB.
What are the implied assumptions?

19. Example Ⅱ (2 of 2)
Applying the scalar E-of-E at E, we get;
+→ ∑ 𝐹 𝑥=− T ED + ¿
+ ↑ ∑ 𝐹 𝑦 =¿
Solving the above equations, we get;
TED=340 N ←∧TEA=196 N ↓

20. Concept Quiz
1. Assuming you know the geometry of the ropes, in which
system above can you NO T determine forces in the
cables?
2. Why?
A. The weight is too heavy.
B. The cables are too thin.
C. There are more unknowns than equations.
D. There are too few cables for a 1000 lb weight.

21. Group Problem Solving (1 of 3)
Given: The mass of lamp is 20kgr
and geometry is as shown.
Find: The force in each cable.
Plan:
1. Draw a FB D for Point D.
2. Apply E-of-E at Point D to solve for the unknowns (FCD & FDE).
3. Knowing FCD, repeat this process at point C.

22. Group Problem Solving (2 of 3)
Applying the scalar E-of-E at D, we get;
 
o
Y DE
F F sin30 20 9.81 0
   

o
X DE CD
F F cos30 F 0
   

Solving the above equations, we get:
and
𝑭 𝑫𝑬=𝟑𝟗𝟐 𝑵 𝑭 𝑪𝑫=𝟑𝟒𝟎 𝑵

23. Group Problem Solving (3 of 3)
Applying the scalar E-of-E at C, we get;
o
X BC AC
3
F 340 F sin45 F 0
5
 
    
 
 

o
Y AC BC
4
F F F cos 45 0
5
 
   
 
 

Solving the above equations, we get;
and
𝑭𝑩𝑪=𝟐𝟕𝟓 𝑵 𝑭 𝑨𝑪=𝟐𝟒𝟑𝑵

24. Attention Quiz (1 of 2)
1. Select the correct FB D of particle A.
A. B.
C. D.

25. Attention Quiz (2 of 2)
2. Using this FBD of Point C, the sum of forces in the x-direction
 
X
F
 Is ______.
Use a sign convention of +  .
A.
o
2
F sin50 20 0
 
B. o
2
F cos50 20 0
 
C.
o
2 1
F sin50 F 0
 
D.
o
2
F cos50 20 0
 

26. Three-dimensional Force Systems
Objectives:
• Drawing a 3-D free body diagram
• Applying the three scalar equations (based on one
vector equation) of equilibrium.

27. Reading Quiz (1 of 1)
1. Particle P is in equilibrium with five (5) forces acting on it in
3-D space. How many scalar equations of equilibrium can
be written for point P?
A. 2
B. 3
C. 4
D. 5
E. 6

28. Applications (1 of 1)
You know the weight of the electromagnet and its load.
You need to know the forces in the chains to see if it is a safe
assembly. How would you do this?

29. The Equations of 3-D Equilibrium (1 of 2)
When a particle is in equilibrium, the
vector sum of all the forces acting on it
must be zero
This equation can be written in terms of its
x, y and z components.
∑Fx i+∑Fy j+∑Fz k=0

30. The Equations of 3-D Equilibrium (2 of 2)
This vector equation will be satisfied only when
These equations are the three scalar equations of equilibrium. They are
valid for any point in equilibrium and allow you to solve for up to three
unknowns.

31. Example I (1 of 3)
Given: The four forces and geometry
shown.
Find: The tension developed in cables
AB, AC, and AD.
Plan:
1. Draw a FBD of particle A.
2. Write the unknown cable forces TB, TC , and TD in Cartesian
vector form.
3. Apply the three equilibrium equations to solve for the
tension in cables.

32. Example I (2 of 3)
Solution:
TD=T D(−0.5i−0.5 j+0.7071k)
W =−300 k
TD=T D cos120°i+TD cos120° j+TD cos 45°k
In component form, the forces are:

33. Example Ⅰ (3 of 3)
Applying equilibrium equations:
Equating the respective i, j, k components to zero, we have
Using equations for and ,
we can determine
Substituting TC and TD into , we find

34. Example Ⅱ (1 of 3)
Given: A 600 N load is supported
by three cords with the
geometry as shown.
Find: The tension in cords AB, AC
and AD.
Plan:
1. Draw a free body diagram of Point A. Let the unknown
force magnitudes be FB, FC, FD.
2. Represent each force in its Cartesian vector form.
3. Apply equilibrium equations to solve for the three
unknowns.

35. Example Ⅱ (2 of 3)
FC=– FC i N
FB=FB (sin 30 °i+cos3 0 ° j )={0.5 FB i+0.866 FB j}N
¿ 𝐹 𝐷
(1 𝒊 −2 𝒋+2 𝒌
(1
2
+2
2
+2
2
)
1/ 2
)=0.333 𝐹 𝐷 𝒊−0.667 𝐹 𝐷 𝒋+0.667 𝐹 𝐷 𝒌

36. Example Ⅱ (3 of 3)
Now equate the respective i, j, and k
components to zero.
Solving the three simultaneous equations
yields
FC=646 N (since it is positive, it is as assumed, e.g., in tension)
FD =900 N
FB=693 N
∑ Fx =0.5 FB – FC +0.333 FD=0
∑ F y=0.866 FB – 0.667 FD=0
∑ Fz =0 .667 F D – 600=0

37. Concept Quiz (1 of 2)
1. In 3-D, when you know the direction of a force but not its
magnitude, how many unknowns corresponding to that
force remain?
A. One
B. Two
C. Three
D. Four

38. Concept Quiz (2 of 2)
2. If a particle has 3-D forces acting on it and is in static
equilibrium, the components of the resultant force
(∑ Fx ,∑ Fy ,∧∑ Fz) ________.
A. have to sum to zero, e.g., −5 i + 3 j + 2 k
B. have to equal zero, e.g., 0 i + 0 j + 0 k
C. have to be positive, e.g., 5 i + 5 j + 5 k
D. have to be negative, e.g., −5 i −5 j − 5 k

39. Group Problem Solving (1 of 4)
Given: A 400 lb crate, as shown, is in equilibrium and supported by
two cables and a strut AD.
Find: Magnitude of the tension in each of the cables and the force
developed along strut AD.
Plan:
1. Draw a free body diagram of Point A.
Let the unknown force magnitudes be
FB, FC, F D.
2. Represent each force in the Cartesian
vector form.
3. Apply equilibrium equations to solve for
the three unknowns.

40. Group Problem Solving (2 of 4)
 
weight of crate 400 lb
W k
FB =FB (r AB
r AB
)=FB {(− 4 i− 12 j+3 k)
(13) }lb
F𝐶=F𝐶 (r A 𝐶
r A 𝐶
)=F𝐶 {(2i −6 j+3 k )
(7) }lb
F𝐷 =F𝐷 (r𝐷 A
r𝐷 A
)=F 𝐷 {(12 j+5 k)
(13) }lb

41. Group Problem Solving (3 of 4)
The particle A is in equilibrium, hence
FB +FC +F D +W =0
Now equate the respective i, j, k components to zero
(i.e., apply the three scalar equations of equilibrium).
∑ Fx =−( 4
13 )FB +(2
7 )FC= 0
∑ F y=−(12
13 )FB −(6
7 )FC+(12
13 )F D=0
∑ Fz =( 3
13 )FB +(3
7 )FC +( 5
13 )FD – 400=0

42. Group Problem Solving (4 of 4)
Solving the three simultaneous equations gives the forces

43. Attention Quiz (1 of 2)
1. ​
​
Four forces act at point A and point A is in equilibrium.
Select the correct force vector P.
A.  
20 10 –1 l
{ }
0 b
i j k
B. 
 
10 20 1
{ 0 lb
}
i j k
C.   
20 10 10 } l
{ b
i j k
D. None of the above

44. Attention Quiz (2 of 2)
2. In 3-D, when you don’t know the direction or the magnitude
of a force, how many unknowns do you have
corresponding to that force?
A. One
B. Two
C. Three
D. Four