3 EQUILIBRIUM
System isolation and
equilibrium conditions
1
Objectives Students must be able to #1
Course Objective
• State the conditions of equilibrium, draw free body
diagrams (FBDs), analyse and solve problems
involving rigid bodies in equilibrium.
Chapter Objectives
• Analyse objects (particles and rigid bodies) in
equilibriums
• Classify problems in equilibrium into SD and SI
categories
2
Equilibrium Definition
stationary
An object is in equilibrium when it is
stationary or in steady translation
relative to an “inertial reference frame”.
moving with Constant
Whether object in stationary
(moving in steady translation) or not,
depends on “reference frame”.
velocity
really
equilibrium?
Equilibrium
Centrifugal
acceleration
Earth
Newtonian Mechanics
“Inertial Reference Frame”
- Earth Frame
- Central Universe Frame
3
F1
3/1 Introduction
A
F2
F3
- Equilibrium is the most important subject in statics.
C
- In statics, we deal primarily with bodies at rest.
A 
R
(i.e. they are in the state of “equilibrium”).
- More precisely, when a body is in
equilibrium, the wrench resultant of all
forces acting on it is zero; i.e.
R  F  0
(  M A )
C 0
-These requirements are necessary and sufficient conditions for
equilibrium; i.e.
If is true
body in equilibrium
If body in equilibrium
is true
- From the Newton’s second law of motion, a body that moves with constant
velocity, rotates with constant angular velocity; i.e. “zero acceleration”,
can also be treated as in a state of equilibrium.
4
Equilibrium in 2D
- All physical bodies are inherently 3D, but many may be treated as
2D; e.g. when all forces are on the same plane.
3/2 Mechanical system isolation
 F 0
M  M 0
R
- Before we apply the equilibrium conditions
i
i
we need to know what force or couple are involved.
Isolate body
F F
F
F
- Draw Free Body Diagram (FBD)
FBD is used to isolate body (or bodies / system of bodies)
so that force/couple acting on it can be identified.
5
Free Body Diagram (FBD)
• FBD is the sketch of the body under consideration
that is isolated from all other bodies or
surroundings.
• The isolation of body clearly separate cause and
effects of loads on the body.
• A thorough understanding of FBD is most vital for
solving problems.
6
body in interest
Construction of FBD
1) Pick body/combination of
bodies to be isolated
2) Isolating the body. Draw
“complete external boundary”
of the isolated body
3) Add all forces and moments
(including that are applied by
the removed surrounding)
Free Body Diagram
F
y
150 cm
mg
50 cm
x
N
4) Indicate a coordinate system
5) Indicate necessary dimensions
Most important step is solving problems in mechanics.
*** If an FBD is not drawn (when it is needed), you will get
no credit ( 0 point ) for the whole problem!!!! ***
F
y
150 cm
mg
50 cm
x
N
7
Note on drawing FBD
Free Body Diagram
• Establish the x, y, z axes in any suitable
orientation.
• Label all the known and unknown force
magnitudes and directions on the
diagram
• The sense of a force having an
unknown magnitude can be assumed.
Use different colours in diagrams
• Body outline
• Load (force and couple)
• Miscellaneous (dimension, angle, etc.)
F
y
150 cm
mg
50 cm
x
N
- blue
- red
- black
8
Equilibrium Solving Procedure
• Formulate problems from physical situations.
 F 0
M  M 0
R
i
i
(Simplify problems by making appropriate assumptions)
• Draw the free body diagram (FBD) of objects under
consideration
• State the condition of equilibrium
• Substitute variables from the FBD into the
equilibrium equations
• Substitute the numbers and solve for solutions
– Delay substitute numbers
– Use appropriate significant figures
• Technical judgment and engineering sense
– Try to predict the answers
– Is the answer reasonable?
9
Equilibrium Free Body Diagram (FBD)
• FBD is the sketch of the body under consideration that is
isolated from all other bodies or surroundings.
• The isolation of body clearly separate cause and effects of
loads on the body.
• A thorough understanding of FBD is most vital for solving
problems.
10
Equilibrium
Equilibrium
FBD Construction
•
•
•
•
Select the body to be isolated
Draw boundary of isolated body, excluding supports
Indicate a coordinate system by drawing axes
Add all applied loads (forces and couples) on the isolated
body.
• Add all to support reactions (forces and couples) represent
the supports that were removed.
• Beware of loads or support reactions with specific directions due
to physical meanings
• Add dimensions and other information that are required in
the equilibrium equation
No FBDs  Cannot apply equilibrium conditions
 Stop...........
11
Equilibrium Help on FBD

Establish the x, y axes in any suitable orientation.
 Label all the known and unknown applied load
and support reaction magnitudes
 Beware of loads or support reactions with
specific directions due to physical meanings
 Otherwise, directions of unknown loads and
support reactions can be assumed.
12
Equilibrium
Equilibrium On FBD Analyses
Objective: To find support reactions
• Apply the equations of equilibrium
F  0
M
O
0
or
or
F
M
x
O, x
F
 0,  M
 0,
y
F  0
 0,  M  0
 0,
O,y
z
O ,z
• Load components are positive if they are directed along a
positive direction, and vice versa
• It is possible to assume positive directions for unknown
forces and moments. If the solution yields a negative
result, the actual load direction is opposite of that shown in
the FBD.
13
Contents
• Equilibrium of Objects
– Particles (2D & 3D)
– 2D Rigid Bodies
– 3D Rigid Bodies
the heart ()
of Statics
 F 0
M  M 0
R
i
i
Rigid Bodies
R
• SD and SI Problems
Particle
 F 0
i
Ideal particle
can not rotate.
(no couple
acting on it)
14
Particles
FBD construction
To construct a complete FBD of a particle
• Select the particle to be isolated
• Draw the particle as a point
• Indicate a coordinate system
• Add all active forces/moments (weight, etc.)
• Add all support reactions (e.g. tension in the tangential
direction of a cable, tensile and compressive forces in a
compressed and stretched springs)
15
Particles
Equilibrium Analyses #2
Equations of Equilibrium
• Apply the equations of equilibrium
F  0
(  F ) iˆ  (  F
x
F
x
 0,
y
) jˆ  (  F z ) kˆ  0
F
y
 0,
F
z
0
• Components are positive if they are directed along a
positive axis, and negative if they are directed along a
negative axis.
• Assume the directions of unknown forces in the positive x,
and y axes. If the solution yields a negative result, this
indicates the sense of the force is the reverse of that
shown on the FBD.
16
Particle Equilibrium
Particles
Equilibrium in 3D
z
F3
F2
F
F
F
x
0
y
0
z
0
W
y
x
F1
FB
FD
FC
17
H-Ex3-1#1) The sphere has a mass of 6 kg and is supported as
shown.Draw a free-body diagram of the sphere, the cord CE, and
the knot at C.
18
How many
unknowns?
How many
Equations?
  Fx  0 
FCBA cos 60o  FCD  0
  Fy  0 
 0
FCBA sin 60o  FCE

58.9FCE
N

FCE
58.9

sin 60o sin 60o
 cos 60o  58.9 cot 60o
 FCE
FCBA 
FCD

FCE
  Fy  0 
  Fy  0 
y
x
FCE   58.9   0
FCE   58.9 N
 0
FCE  FCE
  FCE  58.9 N
FCE
Action-reaction pair
- use same symbol
- opposite direction
Determine the tension in cables AB and AD for equilibrium of
the 250-kg engine shown.
FBD of A
0.25 g kN
g  9.807 m/s2
20
FBD of A
0.25 g kN
Particle A is in equilibrium.
  Fx  0   
TB cos 30  TD  0
(1)
  Fy  0   
TB sin30  0.25g kN  0
(2)
Solve (2) and subst. into (1) TB  4.9035 kN, TD  4.2466 kN
TB  4.90 kN, TD  4.25 kN
#
21
Example Hibbeler Ex 3-3 #1
2unknown,
2 Eqs
(at this stage)
3unknown,
2 Eqs
(at this stage)
If the sack at A has a weight of 20 lb,
determine the weight of the sack at B
and the force in each cord needed to
hold the system in the equilibrium
position shown.
=?
Equilibrium of FBD of E
  Fx  0 
TEG sin30  TEC cos 45  0
  Fy  0 
TEG cos 30  TEC sin 45  20 lb  0
Solve (1) & (2) TEC  38.637 lb, TEG  54.641 lb
TEC  38.6 lb, TEG  54.6 lb
(1)
(2)
#
FBD of E
22
Recommended
TEC
 38.637 lb
FBD of E
FBD of C
Equilibrium of FBD of C
  Fx  0 
TEC cos 45 lb  (4 5)TCD  0
  Fy  0 
(3 5)TCD  TEC sin 45 lb  WB  0
Solve (3) & (4) TCD  34.151 lb, WB  47.811 lb
TCD  34.2 lb, WB  47.8 lb
(3)
TEC  38.637 lb
(4)
TEG  54.641 lb
#
23
Example Hibbeler Ex 3-5 #1
A 90-N load is suspended from the
hook. The load is supported by two
cables and a spring having a
stiffness k = 500 N/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.
24
3D particle Equilibrium
How many unknowns,
how many equations?
  Fx  0 
  Fy  0 
  Fz  0 
Equilibrium of FBD of A
  Fx  0 
FD sin30  (4 5)FC  0
(1)
  Fy  0 
FD cos30  FB  0
(2)
 Fz  0
(3 5)FC  90  0
(3)
Solve (3),
FC  150 N
#
Solve (1),
FD  240 N
#
Solve (2),
FB  207.85 N  208 N
#
Fs  ks 
FB  ksAB  500sAB
sAB  207.85 / 500
(no FBD, no score)
perfect answer sheet
sAB  0.41569  0.416 m #
25
Example Hibbeler Ex 3-7 #1
3D particle
Equilibrium
Determine the force
developed in each cable
used to support the 40kN (4 tonne) crate shown.
F  0
(  F ) iˆ  (  F
ˆ  (  F ) kˆ  0
)
j
y
z
F
y
x
x
 0,
F
 0,
F
z
0
26
Particle Equilibrium
Hibbeler Ex 3-7 #2
Position vectors
rAB  3iˆ  4 jˆ  8kˆ m
r  3iˆ  4 jˆ  8kˆ m
AC
Force vectors
rAB
1
 FB
( 3iˆ  4 jˆ  8kˆ )
rAB
89
rAC
1
FC  FC
 FC
( 3iˆ  4 jˆ  8kˆ )
rAC
89
FD  FD iˆ
W  40kˆ kN
FB  FB
27
Equilibrium of A
  F  0  FB  FC  FD  W  0


FC
FB
ˆ
ˆ
ˆ
( 3i  4 j  8k ) 
( 3iˆ  4 jˆ  8kˆ )  (FD iˆ)  ( 40kˆ )  0
89
89
8FC
3FB 3FC
4FB 4FC ˆ
8FB
ˆ
(

 FD ) i  (

) j (

 40) kˆ  0
89
89
89
89
89
89
28
Particle Equilibrium
Example Hibbeler Ex 3-7 #4
Equilibrium of A
  Fx  0 
  Fy  0 
  Fz  0 
3FB
89
4FB
89
8FB
89



3FC
89
4FC
89
8FC
89
 FD  0 (1)
0
 40  0
(2)
(3)
From (2), FB  FC
From (3),
From (1),
16FC
89
6FC
89
 40  0 or FC  23.585 kN,
 FD  0 or FD  15.0 kN,
FC  FB  23.6 kN #
FD  15.0 kN
#
29
2D Equilibrium
Equilibrium of 2D Rigid Bodies
• Use similar analyses as the particles
• Additional consideration
– Action forces in supports/constraints
– Free-body diagram (FBD) of 2D rigid bodies
– Equilibrium equations (scalar form) for rigid bodies
– Two-force and three-force members
31
Force Reaction (2D)
To write an FBD, first, you will need to know what kind of force
we will get when eliminating the environment/surrounding.
1) Flexible cable, belt, chain, or rope


T

always away from the body

tangent to the cable
T
32
2) Smooth surfaces
N
only this direction
N
3) Rough surfaces
Not always
this direction
- Contact force at
contact point normal
to the surface/contact
plane
- always compressive
-A rough surface can
produce a tangential force
(F, friction) as well as a
normal force (N)
F
R
N
only this direction
- direction of F depend on
situations (chapter 6)
33
4) Roller supports
only this direction
N
- Roller, rocker, or ball
transmits a compressive
force normal to the
supporting surface
N
5) Freely sliding guide
not always
this direction
The vector N may be up or down
depend on problem. If not known,
you may assume any of the two.
After further calculation, if N is +,
correct sense was assumed. If
negative, N goes the other way.
N
M
N
M existence due to its
bending resistance
34
6) Pin connection
Pin not free to turn
Pin free to turn
Rx
not always
this direction
Rx
M
Ry
Ry
As a general rule, if a support prevents translation of a body in a given
direction, then a force is developed on the body in the opposite direction.
Similarly, if rotation is prevented, a couple is exerted on the body.
7) Built-in or fixed support
F
M
not always
this direction
V
A
A
Weld
A
35
8) Gravitational attraction
m
G
W=mg
Resultant of the
gravitational attraction is
the weight W = mg and act
toward center of the earth
passing through the center
mass G
9) Spring action
F
x
x is positive
F
x is negative
F
Normal
distance
For linear springs , F = kx
36
Equilibrium
You may assume either case.
The sign will indicate its sense of
direction later.
37
2D Equilibrium
Equilibrium construction of a FBD
To construct a complete FBD for a 2D rigid bodies
• determine which body is to be isolated
• draw external boundary of isolated body
• indicate a coordinate system (axes)
• add all loads (forces and couples, be they applied or
support)
No FBDs


Cannot apply equilibrium conditions
NO SCORES
38
F1
F3
F2
V
y F
F1
M
x
mass m
W=mg
A
y
M
mass m
P
M
x P
A
F
B
P
F3
F2
No internal force
is shown in FBD
V
P
N
W=mg
y
x
T
m
A
B
Correct?
N
F
V
**Write FBD before allowing any movements of vectors (sliding/free)**
not
always
“mg”
39
y
3/A Correct the incomplete FBD
x
for all figures below
F
Ax
Ay
40
3/B Correct the wrong or
incomplete FBD
y
x
f
for all figures below
mg
If rough surface
F
Ox
41
3/B Correct the wrong or
incomplete FBD
y
x
for all figures below
D
C
Ax
C
42
We will choose problem 1 and 4 as samples
Equilibrium
System Isolation & FBD
43
A
y
mg
x
f
By
M
mg
Ax
y
x
Ay
m0g
N
|_ with
the incline
44
2D Equilibrium
Example Hibbeler Ex 5-1 #1
Draw the free-body diagram of the uniform beam shown.
The beam has a mass of 100 kg.
45
2D Equilibrium
Example Hibbeler Ex 5-1 #2
46
2D Equilibrium
Example Hibbeler Ex 5-3 #1
Two smooth pipes, each having a mass of 300 kg, are
supported by the forks of the tractor. Draw the FBDs for each
pipe and both pipes together.
47
2D Equilibrium
48
2D Equilibrium
Example Hibbeler Ex 5-3 #2
49
2D Equilibrium
Example Hibbeler Ex 5-4 #1
Draw the FBD of the unloaded platform that is suspended off the edge
of the oil rig shown. The platform has a mass of 200 kg.
50
2D Equilibrium
Example Hibbeler Ex 5-4 #2
51
52
3/3 Equilibrium Conditions
- We can find the wrench resultant force

F1

F2
P
G

R
C
P
0
G
“equilibrium”
R0
MP  0
M
P

F3
“equilibrium”
F  0
F  0
P
C 0
M
AllPoint
0 ?
- Equilibrium conditions (3D) of the body is equivalents to
F  0
M
P
0
(for some point P )
We can prove this!
F  0
M
O
0
(for all point O)
- For co-planer forces only (2D), eq conditions is equivalents to
Fx= 0
Fy = 0
Mo = 0
(for all point O) (2D)
54
Equilibrium and its Independent Equations
ABC
Equilibrium
F1
A
A  B  C and C  A
(no tendency to initiate translation,
no tendency to initiate rotation)
F2
F3
A 
C
R
equilibrium
A 
C
R
R  F  0
(  M A )
C 0
A
any O
(trivially easy)
R  F  0
M
A
R  F  0
0
(some point: A)
F  0
M
 (r
G
A
B
A
i
M
o
0
(any point: O)
Important Meaning
0
 Fi )  Bi  0
2D
F
F
x
0
x
0

MP  0

MQ  0
Only at most 3 independent Equations57
2D Equilibrium
Scalar Form
Equilibrium Eqn. for 2D Rigid Bodies
F
F
M
x
0
y
0
O
0
Particle
Rigid Body
The sum of the moment about any point O is zero.
58
2D Equilibrium
2D Equilibrium Procedure #1
Free-Body Diagram
• Establish the x, y axes in any suitable orientation.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on
the body.
• Label all the loadings and specify their directions
relative to the x, y axes. The sense of a force or
couple moment having an unknown magnitude but
known line of action can be assumed.
• Indicate the dimensions of the body necessary for
computing the moments of forces.
59
2D Equilibrium
2D Equilibrium Procedure #2
Equations of Equilibrium
• When applying the force equilibrium equations,
F
x
 0 and
F
y
0
orient the x and y axes along lines that will provide the
simplest resolution of the forces into their x and y
components.
• If the solution of the equilibrium equations yields a
negative scalar for a force or couple moment magnitude,
this indicate that the sense is opposite to that which was
assumed on the FBD.
60
2D Equilibrium
2D Equilibrium Procedure #3
Equations of Equilibrium
• Apply the moment equation of equilibrium about a point O
that lies at the intersection of the lines of action of two
unknown forces.
M
O
0
In this way, the moments of these unknowns are zero
about O, and a direct solution for the third unknown can be
determined.
61
H5.7) Find the tension in the cord at C and
the horizontal and vertical component of
reactions at pin A. The pulley is frictionless.
The body is in equilibrium.
Fx = 0 Fy = 0 MAnyPoint = 0
T are the same
at both side?
+
  Fx  0
T sin   Ax  0
  Fy  0 
T cos   500  Ay  0
  M A  0
500(0.2)  T (0.2)  0
T  500 N
Ax  T sin   250 N
Ay  500  T cos   933 N
If couple also applies at pully,T are the same in both side?
If pulley rotates, T are the same in both side ? with constant angular velocity?
63

F1

F2
y
x

F3
Body in equilibrium
Fx= 0 Fy = 0
Mo = 0
(for some point O)
Fx= 0 Fy = 0
(for all point O)
Mo = 0
Independent Equations
** Whatever 2D-equilibrium conditions you used,
at most 3 unknowns (scalar) may be found.
Hey , why you cannot get anymore,
you can take another points to take
moment, and get new equation?
not
independent
equations!
** However, sometimes you can not get
those 3 unknowns in 2D problem.
64
H5.7) Find the tension in the cord at C and
the horizontal and vertical component of
reactions at pin A. The pulley is frictionless.
The body is in equilibrium.
Fx = 0 Fy = 0 MAnyPoint = 0
  Fx  0
+
T sin   Ax  0
  Fy  0 
T cos   500  Ay  0
  M A  0
500(0.2)  T (0.2)  0
  M D  0 
Not
independent
Equations!
(0.2)   Fy  0
   M A  0
   M D  0
 Ay (0.2)  (T cos  )(0.2  0.2sec  )  0
T  500 N
Ax  T sin   250 N
Ay  500  T cos   933 N
65
2D Equilibrium
Example Hibbeler Ex 5-6 #1
Determine the horizontal and vertical components of
reaction for the beam loaded as shown. Neglect the
weight of the beam in the calculation.
66
2D Equilibrium
Example Hibbeler Ex 5-6 #2
Find: Ay, Bx, By
FBD of ADB
67
2D Equilibrium
Example Hibbeler Ex 5-6 #3
Equilibrium of ADB
  Fx  0   
600 cos 45 N  Bx  0
Bx  424.26  424 N
  MB  0
 
#
(100 N)(2 m)  (600 sin 45 N)(5 m) 
(600 cos 45 N)(0.2 m)  Ay (7 m)  0
Ay  319.50  319 N
#
68
2D Equilibrium
Example Hibbeler Ex 5-6 #4
Equilibrium of FBD of ADB
  Fy  0   
Ay  600 sin 45 N  100 N  200 N  By  0
By  404.76  405 N
#
69
2D Equilibrium
Example Hibbeler Ex 5-8 #1
The link shown is pin-connected at A and rests against a
smooth support at B. Compute the horizontal and vertical
components of reactions at the pin A.
70
2D Equilibrium
Example Hibbeler Ex 5-8 #2
FBD of link BA
71
2D Equilibrium
Example Hibbeler Ex 5-8 #3
Equilibrium of FBD of link BA
  M A  0
 
90 N  m  (60 N)(1 m)  NB (0.75 m)  0
NB  200 N
#
  Fx  0   
Ax  NB sin30 N  0, Ax  100 N
#
  Fy  0   
Ay  NB cos 30 N  60 N  0
Ay  233.21  233 N
#
72
3/57 200-kg beam has a C.G. at G. The 80-kg man is exerted a
3 unknowns, 3 Equations.
300-N force on the rope. Calculate the force reaction
at Ait.which
You can solve
y
is a weld pin.
mg =
80(9.81)
+
300 N
Ay
200(9.81) N =mg?
C
MA
N
300N
Ax
80(9.81)
Ay
200(9.81)
MA
Ax
N is internal force.
300 N
T =300N
300 N T
mg
x
T = 300N?
T=300N
80(9.81)
Ay
200(9.81)
4 unknowns, 3 Eq.
You can’t solve it?
MA
You can solve it !
Use FBD of rod
Ax
tension is internal force.
C
73
3/57 200-kg beam has a C.G. at G. The 80-kg man is exerted a
300-N force on the rope. Calculate the force reaction at A which
is a weld pin.
80(9.81)
80(9.81)
Ay
200(9.81)
Sun’s
Gravitational
force
MA
Ax
tension is internal force.
C
74
Draw the FBD of the foot lever shown.
why?
Two-Force Member In Equilibrium
Body in
Equilibrium
P
Fx= 0
Fy = 0
Mo = 0
To keep the body in equillibrium,
the second force must …
- direction of opposite side.
P
body in
equilibrium
- has the same manitude
- be co-linear
bodies in
equilibrium
77
Fx= 0
Three-Force Member for Equilibrium
Fy = 0
Mo = 0
To keep the body in equillibrium,
the second force must …
1)

F  0
F2
F1
F3
2)
4 forces need to
be concurrent
or parallel
to make body
in equilibrium?
(co-planer)
Concurrent OR Parallel (with proper moment arm)
Parallel
concurrent
(with proper moment arm)
F2
O
F3

F2

F1
F1
M = 0

F3
78
Draw the FBD of the foot lever shown.
why?
FB
FB
Two force
in Equilibrium
You should write the FBD of 2-force
(in equilibrium) like this.
85
The lever ABC is pin-supported at A and connected to a short
link BD as shown. If the weight of the members is negligible,
determine the force of the pin on the lever at A.
4 unknowns, 3 eq.
Hibbeler Ex 5-13
Ax  Bx  400  0
Ay  By  0
400(0.5)  Ay (0.1)  Ax (0.2)  0
By
y
Ay
+
Ax
x
Bx
1600
2800
Ay 
N
3
3
2800
Dx  D y  Bx  B y  
N
3
AX 
Bx
2 unknowns, 3 eq.
 By  Dy  0
 Bx  Dx  0
Dy
By
Dx
Two Force
Members
 D y (0.2)  Dx (0.2)  0
D x  D y  Bx  B y
86
Solve by using 3-force member Concept
Three Forces are
concurrent.
Hibbeler
Ex 5-13
O
y
45o
0.5 m
+
x
0.5+0.2
=0.7 m
45o
Assume
Directions
F
FA

2 force
members
0.5-0.1
=0.4 m
3-Unknowns, 3-Eqs
  tan1
0.7
 60.26
0.4
The lever ABC is pin-supported at A and connected to a short
link BD as shown. If the weight of the members is negligible,
determine the force of the pin on the lever at A.
[  Fx  0]
2 force
members
FA cos 60.26  F cos 45  400 N  0
[  Fy  0]
0.5 m
y
+
FA sin60.26  F sin 45  0
O
x
FA  1074.9 N
F  1319.9
N
Ans
F
FA
  60.26
88
Any two-force members or three-force members here?
89
3/27 In a procedure to evaluate
the strength of the triceps muscle,
a person pushed down on a load
cell with the palm of his hand as
indicated in the figure.
If the load-cell reading is 160 N,
determine the vertical tensile force
F generated by the triceps muscle.
The mass of the lower arm is 1.5
kg with mass center at G.
91
y
If what you want
o know is Oy ?
If what you want
o know is Ox ?
Free Body Diagram
x
T
Ox
0.15
0.25
Oy
0.15
160N
1.5(9.81)
3 Unknown,
3 Equations. You can solve it.
By easier way.
Because what we want to know is the force T of triceps.
Taking summation of moments at point O will eliminate Ox and Oy
which are unknowns out, and the calculation will be easier.
+
M
O
0
0.25 T  0.15(1.5)(9.81)   0.15  0.15 160  0
T  1832 N
Ans
92
Technique: Finding unknown force only in 1 step
Concrete slab with mass of 25,000 kg is pulled by cable of
tension P. Determine the tension T in the horizontal cable using
only 1 equilibrium equation.
Given:  =
60°
y
P
x
30
D
W = mg
30
A
6m
Key
60
T
R
Here we have 3 unknowns (P,R,T). To use only 1equilibrium eq.
To determine T, we need to take moment at the point where
the other unknown forces (R,P) passes.
93
+
P
4m
30
D
W = mg
30
A
 MD  0
(T) (length AD)  (W) (4 cos 60 )  0
(T )(6 tan 30 )  (25000)(9.8)(4cos60 )  0
60
T
6m
R
y
x
(T) (3.464)  490500
T  141,600
N
ANS
94
Alternative Equilibrium equations
- Recall: In general, you have these three equilibrium equations:
Fx= 0
(for some point O)
Fy = 0
Mo = 0
(E)
- Alternatively, you may use either
MA = 0
MB = 0
Fx= 0
line AB not |_ to the x direction
MA = 0
A B C not on the same straight line
R=0
not |_
B
R=0
x
A
Mc = 0
MB = 0
B
C
A
96
Calculate the magnitude of the force at pin A.
400 N
Ax
y
500 N
Ay
80 N-m
By
D
x
Free Body Diagram
How many unknowns ?
M
We have 3 unknowns of
Ax
Ay
By
M
A
M
F
x
0
F
0
M
B
0
F
0
M
0
M
A
A
0
B
y
0
x
0
D
0
97
400 N
Calculate the magnitude of the force at pin A.
Ax
we can determine Ay from  MB  0
+
 (A y ) (.15)  (500)(. 125)  (400)(. 25)  80  0
Ay
500 N
By
y
80 N-m
x
Free Body Diagram
A y   783.3 N
we can determine Ax from  Fx  0 +
A x  500  0
A y   500 N
Thus, the resultant of force at pin A:
R 
(500) 2  (783.3) 2
 929.3 N
ANS
98
Constraints and Statical determinacy
** need to know when a problem can be solved or what force/couple can be found.
1) Statically Determinate
Improper Constraint
- Not enough constraint
Unknown 3 ,
Independent eq 2
Unknown 3 ,
Independent eq 3
2) Statically Indeterminate
- can’t get all unknowns
Unknown 4 ,
Independent eq 3
- Not in equilibrium
No solution: (cant maintain moment)
unknown 3 ,
independent 2
99

F1
y
x

F3
Body in equilibrium
Fx= 0 Fy = 0
Mo = 0
(for some point O)
Fx= 0 Fy = 0
Independent Equations

F2
** Whatever 2D-equilibrium conditions you used,
at most 3 unknowns (scalar) may be found.
Hey , why you cannot get anymore,
you can take another points to take
moment, and get new equation?
not
independent
equations!
** However, sometimes you can not get
those 3 unknowns in 2D problem.
Mo = 0
(for all point O)
Independent Equations on categories of 2D problem.
Equilibrium

1) Co-linear
x
F

F2

F1
3
Choose this
direction as x axis
Fx= 0
Fy = 0
Depend on xaxis selection
Mo = 0
(Choose point O on the line of action) 100
2) Concurrent at a point

F2

F3
O

F1
Fx = 0
y
Fy = 0
x
Mo = 0
Fx = 0
3) Parallel 
F1

F3

F2
Fy = 0
y
4) General

F1

M

F4

F3
Depend on xaxis selection
Mo = 0
x

F2
concurrent at
point O
Fx = 0
y
Fy = 0
x
Mz = 0
101
Note on Solving Problems
1. If we have more unknowns than the number of independent
equations, then we have a statically indeterminate situation.
We cannot solve these problems using just statics.
2. The point (axis) which we check for the moment,
affects the simplicity of the solution. Choosing
appropriate one is the key for fast problem solving.
3. The order in which we apply equations may affect
the simplicity of the solution. For example, solving
 FX = O first allows us to find the first unknown x
quickly.
4. If the answer for an unknown comes out as
negative number, then the sense (direction) of
the unknown force is opposite to that assumed
when starting the problem.
N
If you get N<0,103
something wrong!
3/48 The small crane is mounted on one side of the bed
of a pickup truck. For the position 45, determine the
magnitude of the force supported by the pin at O and the
oil pressure P against the 50-mm-diameter piston of the
hydraulic cylinder BC.
Body in interest:
AOBC?
AOC?
Because We also want to find
force P exerted by BC to the
object AOC
104
45
y
FBD
Find P ,O
Ignore mass of
link CB
x
W
C
Ox
C
which
dirction?
B
Oy
2 Force Member
// with CB
Direction // CB
0.11
F

C G
E D
H
Can you find  ?
CH
tan


0.34
EO
O


F 0.36
B
CE  OB

EO

( FD  FG )  OB
EO

( FD  FG )  OB
ED  DO
(0.34sin   0.11cos  )  0.36
tan  
0.11sin   0.34sin 
  56.2
 MO  0
To find C:
MW  120(9.8)[(0.785  0.34)cos45o ]
M F  ( F cos  )(0.36)
MW  M F  0
o
+
P
F
d
  
2
2

F  5060 N
5060
0.05 
 

 2 
N-m
2
 2.58 MPa
105
Ans
45
Find
P ,O
A
y
0.785
W
0.11
x
F
C
F
0.34

O
  56.2
To find reactions at O
x
  F cos   Ox  0
 Ox  2820 N
F
y
o
Ox
Oy

F
FDB
B
 W  F sin   O y  0
 O y  3030 N
| O | Ox2  Oy2  4140 N (
)
Ans
106
3/46 It is desired that a person be able to begin closing the van
hatch from the open position shown with a 40-N vertical force P.
As a desired exercise, determine the necessary force in each of the
two gas-pressurized struts AB. The mass center of the 40-kg door
is 37.5 mm directly below point A. Treat the problem as twodimensional.
Why direction: CB?
1125 mm
A
550 mm
O
2F
W=mg
600 mm
P= 40N
Unknowns:
Oy
30
Ox
175 mm
B
F, Ox, Oy
Equilibrium eqs:
107
1752  5502  6002  2(550)(600) cos  )
Find F
   16.787o
1125 mm
Oy
A
2F
550 mm
G
O
x
G
Ox

O
30
W=mg
B
P = 40

175 mm
30
  30o  
A
600 mm
W=(40)(9.81)
2F
y
M P : easy but M 2 F  ? M W  ?
30
B
M P  (40)(1.125)
+
 MO  0
M 2 F  (2 F )(0.55sin  )
M W  W (0.55sin  )
P(1.125)  2 F (0.55sin  )  W (0.55cos(30   ))  0
45000  210106.6115  317.696 F  0
F  802.989  803
N
ANS
108
Find appropriate L to make
Bodies in equilibrium.
: known
=
2T cos  (b)
x
mg
: known
2T cos
x=?
110
111
3/52 The rubber-tired tractor shown has a mass of 13.5 Mg
with the center of mass at G and is used for pusshing or
pulling heavy loads. Determine the load P which the tractor
can pull at a constant speed of 5 km/h up the 15-percent grade
if the driving force exerted by the ground on each of its four
wheel is 80% of the normal force under the wheel. Also find
the total normal reaction N under the rear pair of wheels at B.
Normal reaction at A and B are equal?
If not, which one should be larger?
112
y
x
825
1200 600
0.8 NA
NA
FBD
tan   0.15    8.531o
W=mg=13.5*1000*9.81=132400 N
F
F
y
A
0.8NB
P
NB
 0.8N A  W sin   0.8N B  P  0
x
M
W
 N A  W cos  N B  0
 W cos 1200  W sin  825  N B 1800  P  600  0
N A  6.3 kN
N B  124.7 kN
P  85.1 kN
Ans
113
Recommended Problem
3/34 3/49 3/35 3/54 3/57
114