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5.3 Equations of Equilibrium5.3 Equations of Equilibrium
� For equilibrium of a rigid body in 2D,
∑Fx = 0; ∑Fy = 0; ∑MO = 0
� ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on x and y components of all the forces acting on the body
� ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
� For coplanar equilibrium problems, ∑Fx = 0; ∑Fy
= 0; ∑MO = 0 can be used
� Two alternative sets of three independent � Two alternative sets of three independent equilibrium equations may also be used
∑Fa = 0; ∑MA = 0; ∑MB = 0
When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
� Consider FBD of an arbitrarily shaped body
� All the forces on FBD may be
replaced by an equivalent replaced by an equivalent
resultant force
FR = ∑F acting at point A and a
resultant moment MRA = ∑MA
� If ∑MA = 0 is satisfied, MRA = 0
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations
� If FR satisfies ∑Fa = 0, there is no
component along the a axis and
its line of axis is perpendicular its line of axis is perpendicular
to the a axis
� If ∑MB = 0 where B does not lies
on the line of action of FR, FR = 0
� Since ∑F = 0 and ∑MA = 0, the
body is in equilibrium
Alternative Sets of Equilibrium Equations
� A second set of alternative equations is
∑MA = 0; ∑MB = 0; ∑MC = 0
� Points A, B and C do not lie on the
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
� Points A, B and C do not lie on the
same line
� Consider FBD, if If ∑MA = 0, MRA = 0
� ∑MA = 0 is satisfied if line of action of FR passes through point B
� ∑MC = 0 where C does not lie on line AB
� FR = 0 and the body is in equilibrium
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisFree-Body Diagram
� Establish the x, y, z coordinates axes in any suitable orientationsuitable 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for AnalysisFree-Body Diagram
� The sense of a force or couple moment having an unknown magnitude but 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
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for Analysis
Equations of Equilibrium
� Apply the moment equation of equilibrium ∑M = 0 about a point O that lies on the ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces
� The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Procedure for Analysis
Equations of Equilibrium
� When applying the force equilibrium ∑Fx = 0 and ∑F = 0, orient the x and y axes along and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components
� If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.6
Determine the horizontal and vertical
components of reaction for the beam loaded.
Neglect the weight of the beam in the Neglect the weight of the beam in the
calculations.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
FBD
� 600N force is represented by its x and y components
� 200N force acts on the beam at B and is � 200N force acts on the beam at B and is
independent of the
force components
Bx and By, which
represent the effect of
the pin on the beam
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
M B ;0=∑→+
� A direct solution of Ay can be obtained by applying ∑MB = 0 about point B
� Forces 200N, Bx and By all create zero moment about B
NB
BN
x
x
424
045cos600
=
=−o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NA
mAmNmNmN
M
y
B
319
0)7()2.0)(45cos600()5)(45sin600()2(100
;0
=
=−−+
=∑oo
NB
BNNNN
F
NA
y
y
y
y
405
020010045sin600319
;0
319
=
=+−−−
=∑↑+
=
o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Checking,
M A ;0=∑
NB
mBmN
mNmNmN
M
y
y
A
405
0)7()7)(200(
)5)(100()2.0)(45cos600()2)(45sin600(
;0
=
=+−
−−−
=∑oo
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.7
The cord supports a force of 500N and wraps over
the frictionless pulley. Determine the tension in the
cord at C and the horizontal cord at C and the horizontal
and vertical components at
pin A.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
� Principle of action: equal but opposite reaction observed in the FBDobserved in the FBD
� Cord exerts an unknown load
distribution p along part of
the pulley’s surface
� Pulley exerts an equal but
opposite effect on the cord
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
FBD of the cord and pulley
� Easier to combine the FBD of the pulley and
contracting portion of the cord so that the contracting portion of the cord so that the
distributed load becomes internal to the system
and is eliminated from the
analysis
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
mTmN
M A
0)2.0()2.0(500
;0
=+
=∑
Tension remains constant as cord
passes over the pulley (true for
any angle at which the cord is
directed and for any radius of
the pulley
NT
mTmN
500
0)2.0()2.0(500
=
=+
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NA
Fx
030sin500
;0
=+−
=∑→+o
NA
NNA
F
NAx
NA
y
y
y
x
933
030cos500500
;0
250
030sin500
=
=−−
=∑↑+
=
=+−
o
o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.8
The link is pin-connected at a and rest a
smooth support at B. Compute the horizontal smooth support at B. Compute the horizontal
and vertical components of reactions at pin A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
FBD
� Reaction NB is perpendicular to the link at BB
� Horizontal and vertical
components of reaction
are represented at A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
M A
0)75.0()1(60.90
;0
=+−−
=∑
NAx
NA
F
NN
mNmNmN
x
x
B
B
100
030sin200
;0
200
0)75.0()1(60.90
=
=−
=∑→+
=
=+−−
o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NNA
Fy
030cos20060
;0
=−−
=∑↑+o
NA
NNA
y
y
233
030cos20060
=
=−−
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.9
The box wrench is used to tighten the bolt at
A. If the wrench does not turn when the load
is applied to the handle, determine the is applied to the handle, determine the
torque or moment applied to the bolt and the
force of the wrench on the bolt.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
FBD
� Bolt acts as a “fixed support” it exerts force � Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of EquilibriumM A
12
;0
=∑
NAx
NNA
F
mNM
mNmNM
x
x
A
A
00.5
060cos3013
552
;0
.6.32
0)7.0)(60sin30()3.0(13
1252
=
=+
−
=∑→+
=
=−
−
o
o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
NNA
F
y
y
060sin3013
1252
;0
=−
−
=∑↑+
o
( ) ( )( )
NmM
mNmNM
MCCW
NA
NNA
A
A
y
y
y
6.32
07.060sin303.013
1252
;0
0.74
060sin3013
52
=
=°−
−
=∑↑
=
=−
−
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are Ay pass through this point and these forces are not included in the moment summation
� MA must be included
� Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the boltbolt
� For resultant force on the wrench,
� For directional sense,
� FA acts in the opposite direction on the bolt
( ) ( )
o1.8600.5
0.74tan
1.740.7400.5
1
22
==
=+=
−
N
N
NFA
θ
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Checking,
;0=∑MC
0.8.51.6.32.2.19
0)7.0(0.74.6.32)4.0(1312
52
=−+
=−+
mNmNmN
mNmNmN
Example 5.10
Placement of concrete from the
truck is accomplished using the
chute. Determine the force that the
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
chute. Determine the force that the
hydraulic cylinder and the truck
frame exert on the chute to hold it in
position. The chute and the wet
concrete contained along its length
have a uniform weight of 560N/m.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Idealized model of the chute
� Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a at A and the hydraulic cylinder BC acts as a short link
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD� Since chute has a length of 4m, total supported
weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, Gassumed to act at its midpoint, G
� The hydraulic cylinder exerts a horizontal force FBC on the chute
Equations of Equilibrium� A direct solution of FBC is obtained by the
summation about the pin at A
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
mNmF
M
BC
A
)2(30cos2240)5.0(
;0
+−
=∑o
NAx
NA
F
NF
mN
x
x
BC
7900
07900
;0
7900
0)0625.0(30sin2240
=
=+−
=∑→+
=
=+ o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
2240
02240
;0
=
=−
=∑↑+
NA
NA
F
y
y
Checking,
0)0625.0(30sin2240)1(30cos2240
)30cos1(2240)5.0(7900
;0
2240
=+
++−
=∑
=
mNmN
mNmN
M
NA
B
y
oo
o
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.11
The uniform smooth rod is subjected to a force and
couple moment. If the rod is supported at A by a
smooth wall and at B and C either at the top or smooth wall and at B and C either at the top or
bottom by rollers, determine
the reactions at these supports.
Neglect the weight of the rod.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
SolutionFBD� All the support reactions act normal to the
surface of contact since the contracting surfaces are smoothare smooth
� Reactions at B and C are acting in the positive y’ direction
� Assume only the rollers located on the bottom of the rod are used for support
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
;0=∑ AM
030cos30cos300
;0
030sin30sin
;0
0)8)(30cos300()6(.4000)2(
''
''
''
=++−
=∑↑+
=−+
=∑→+
=+−+−
oo
oo
o
yy
y
xyy
x
yy
BCN
F
ABC
F
mNmCmNmB
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Note that the line of action of the force component passes through point A and this force is not included in the moment equation force is not included in the moment equation
� Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD
kNNC
kNNB
y
y
35.14.1346
10.1000
'
'
==
−=−=
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Top roller at B serves as the support rather than the bottom one
NA
ANN
x
x
173
030sin0.100030sin4.1346
=
=−− oo
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Example 5.12
The uniform truck ramp has a weight of
1600N ( ≈ 160kg ) and is pinned to the body 1600N ( ≈ 160kg ) and is pinned to the body
of the truck at each end and held in position
by two side cables.
Determine the tension
in the cables.
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
� Idealized model of the ramp
� Center of gravity located at the midpoint since the ramp is approximately uniformramp is approximately uniform
FBD of the Ramp
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Equations of Equilibrium
TmT
M A
)30cos2(20sin)30sin2(20cos
;0
+−
=∑oooo
By the principle of transmissibility, locate T at C
md
md
NT
N
TmT
0154.120sin
2
10sin
5985
0)30cos5.1(1600
)30cos2(20sin)30sin2(20cos
=
=
=
=+
+−
oo
o
oooo
5.3 Equations of Equilibrium5.3 Equations of Equilibrium
Solution
Since there are two cables supporting the
ramp,
T’ = T/2 = 2992.5NT’ = T/2 = 2992.5N