Engineering Mechanics: Statics
Equilibrium of a Rigid Body
Chapter Objectives
• Revising equations of equilibrium of a rigid
body in 2D and 3D for the general case.
• To introduce the concept of the free-body
diagram FBD for a rigid body.
• Showing how to solve rigid-body equilibrium
problems.
THREE DIMENSIONAL FORCE SYSTEMS
• Remember M = r (vector) X F (vector)
• Find the length ‘r’ Vectorially for each Force ‘F’
• Also if not given, find the vectorial representation of ‘F’
Finding the length ‘r’
• Read the coordinates of the two ends of the line
• Get the difference of the coordinates from the tail of the force coordinate to the
point of interest.
Moment of a 3D Force Review
zyx
zyx
FFF
rrr
kji
rM
== F X
In determinant form:
Sarrus‟ Rule
Moment of a 3D Force Review
zyx
zyx
zyx
zyx
FFF
rrr
kji
FFF
rrr
kji
F Xr =
+ + + - - -
= + (ryFz –rzFy) i + (rzFx -rxFz) j + (rxFy – ryFx) k
Moment of a 3D Force Review
A man exerts a 485 N pull on the
rope which is looped around the
branch at point A. Determine the
moment about point C of this force
that exerts on the branch of the tree
and state the magnitude of this
moment.
Example:
Moment of a Force AT A POINT ‘3D’
CAr
Example:
ABT
Moment of a Force AT A POINT ‘3D’
Example:
Three forces act on the rod. Determine the resultant moment they create about the flange at O and determine the coordinate direction angles of the moment axis.
Moment of a Force AT A POINT ‘3D’
Solution Position vectors are directed from point
O to each force
rOA = {5j} m and
rOB = {4i + 5j - 2k} m
For resultant moment about O,
Ro A 1 B 2 C 3M (r X F) r X F + r X F + r X F
i j k i j k i j k
0 5 0 0 5 0 4 5 2 30i 40 j 60k N.m
60 40 20 0 50 0 80 40 30
Moment of a Force AT A POINT ‘3D’
OA OA OB
Solution
For magnitude
For unit vector defining the direction of moment axis,
2 2 2
Ro
Ro
Ro
M (30) ( 40) (60) 78.103 N.m
M 30i 40 j 60ku
78.103M
0.3841i 0.5121j 0.7682k
Moment of a Force AT A POINT ‘3D’
λ
Solution
For the coordinate angles of the moment axis,
cos 0.3841; 67.4121
cos 0.5121; 120.8038
cos 0.7682; 39.806
Moment of a Force AT A POINT ‘3D’
Example (T):
a) Determine the moment of the force F about the point P:
b) Find the perpendicular distance D acting on the line of action of
the given force that causes the same moment.
The vector from P to the point of application of F is:
r = (12 3) i + (6 4) j + (5 1) k
= 9 i + 2 j 6 k (m)
N
m
m m
m
Moment of a Force AT A POINT ‘3D’
The moment is:
The magnitude of MP :
m)-(N 288738
744
629 kji
kji
FrM +==×=P
m-N 980.98288738222PM
Example (T):
Moment of a Force AT A POINT ‘3D’
The perpendicular distance D acting on the line of action of the given force that causes the same moment is
The direction of MP tells us both the orientation of the plane of the plane containing D and F and the direction of the moment.
m 998.10N 9
m-N 980.98===
F
MPD
Example (T):
Moment of a Force AT A POINT ‘3D’
Example (T):
The cable tension has a magnitude of
34.185 kN. Calculate the moment of this
cable T, which produces about the base
‘O’ of the construction crane.
Moment of a Force AT A POINT ‘3D’
ABT
EXAMPLE (T):
A force is applied to the tool, to open a gas valve. Find the magnitude of the moment
of this force about the point A.
A
B
Moment of a Force AT A POINT ‘3D’
A
B
rAB = {0.25 sin 30° i + 0.25 cos30° j} m
= {0.125 i + 0.2165 j} m
F = {-60 i + 20 j + 15 k} N
Mz = (rAB F)
0 0 1 0.125 0.2165 0 -60 20 15
Mz =
= 1{0.125(20) – 0.2165(-60)} N·m
= 15.490 N·m
Moment of a Force AT A POINT ‘3D’
i j k
Example (T): 3 cables are attached to a bracket . Replace the forces exerted by the cables with:
an equivalent resultant force and couple moment acting at point A.
Moment of a Force About A LINE ‘3D’
Further Reduction of Forces and Couples System
Further Reduction of Forces and Couples System
Further Reduction of Forces and Couples System
Further Reduction of Forces and Couples System
Further Reduction of Forces and Couples System
Further Reduction of Forces and Couples System
Solution: General system of forces and moments
Further Reduction of Forces and Couples System
z
x
When an object acted upon by a system of forces & moments is in equilibrium, the following conditions are satisfied:
1. The sum of the forces is zero:
2. The sum of the moments about any point is zero:
∑ 0=F
0∑ int =poanyM
o 1
F
d1
d2 d3
2F
3F
1M
2M
Equilibrium Conditions Coplanar (2D) General case
Equilibrium Conditions Coplanar (2D)General case
o 1
F
d1
d2 d3
2F
3F
1M
2M
Body in Equilibrium
Fundamental Equations/Conditions for Statics !
Necessary
Sufficient
∑∑∑ 0,0,0 === oyx MFF
∑∑∑ 0,0,0 === oyx MFF+ +
+
1 unknown
2 unknowns
2 unknowns
3 unknowns
STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
EQ
UILIB
RIU
M
EQ
UA
TIO
NS
CO
ND
ITIO
NS
OF
EQ
UILIB
RIU
M
3 unknowns
5 unknowns
3 unknowns
6 unknowns
There are 2 sets of Independent Equilibrium Equations
Set 1: Equilibrium of Forces (3 components)
Set 2: Equilibrium of Moments (3 components)
Equilibrium Conditions 3D General case
∑∑∑
∑∑∑
0 ,0 ,0
0 ,0 ,0
===
===
zyx
Zyx
MMM
FFF+ i + j
+ i + j + k
The Free-Body Diagram „FBD‟
It is the sketch of the body that shows all the applied known and unknown forces and moments acting on this body.
This sketch, shows the particle “free” (isolated) from its surroundings with all the external forces acting on it.
Note that; while showing the externally applied forces and moments, their directions should be indicated.
If the directions are not given, ASSUME:
Force: +ve for all directions (x, y and z)
(i.e. Same direction with the defined axis directions)
Moments: +ve for i, j, k components.
The Free-Body Diagram FBD
To apply equilibrium equations we must account for all known and unknown forces and moments acting on the object by drawing a free-body diagram FBD of the body.
Force Types Active Forces : tend to set the body in motion. Reactive Forces : result from constraints or supports and tend to prevent motion
Procedure for Analysis Equilibrium Problems
1) Draw Free-Body Diagram FBD
Establish the x, y (for 2D) axes and x, y and z (for 3D)
axes in any suitable orientation.
Label all known and unknown force and moment
magnitudes and directions on the FBD.
The sense of an unknown force and moment may be
assume.
2) Apply equations of equilibrium.
Components of the force and moments are positive if
directed along a positive axis and negative if directed
along a negative axis.
3) If solution yields a negative result the force or the
moment is in the opposite sense of that was assumed
and showed on the FBD.
Supports
Forces & couples exerted on an object by its
supports are called reactions.
E.g. a bridge is held up by the reactions exerted
by its supports.
support
support
Support Reactions
Reactive Forces General Rule:
If a support prevents the translation of a
body in a given direction, then a force
(REACTION) is developed on the body in that
direction.
Likewise, if the rotation is prevented, a
couple moment (REACTION) is exerted on
the body.
2 D SUPPORTS
• Pin Support:
– Figure a: pin support
a bracket to which an object is attached by a smooth
pin that passes through the bracket & the object
– Figure b: side view
The arrows indicate the directions of
the reactions Ax and Ay in 2D.
SYMBOL USED IN NOTEBOOK
Pin Support Details:
The arrow indicate the directions of the reaction A
SYMBOL USED IN NOTEBOOK
• Roller Support: It can move freely in the direction parallel to the surface on which it rolls, it can’t exert a force parallel to the surface but can exert a force normal (perpendicular) to this surface.
Fixed Support
– The fixed support shows the supported object
literally built into a wall (built-in).
SYMBOL USED IN NOTEBOOK
A fixed support can exert
2 components of force
AX AY
and a couple
MA
Different Support Types in 2D
Different
Different Support Types in 2D
Different Support Types in 2D
Unknow
n is d
irecte
d
alo
ng
th
e a
xis
of th
e
sh
ort
lin
k
Different Support Types in 2 D
Different Support Types in 2D
Different Support Types in 2D
Different Support Types in 2D
Different Support Types
(a) Pin in a slot (b) Slider in a slot (c) Slider on a shaft
Different Support Types in 2D
Different 2D Support Types in Practice
Example:
For the given supports and loadings of the body draw its FBD?
Free Body Diagram
FBD
2D
Example:
Free Body Diagram
FBD
2D
Example: Free Body Diagram
FBD
2D
Example:
Free Body Diagram
FBD 2D
30°
Free Body Diagram FBD 2D
Free Body Diagram FBD 2D
PULLEYS
When a cable (cord) wraps over a frictionless pulley, definitely
each and every portion of this cables (cords) is under
TENSION. To satisfy the static equilibrium, the magnitudes of
the tension T1 = T2 should be the SAME.
r =15 cm
7.8 kN 7.8 kN 7.8 kN
Free Body Diagram FBD 2D
PULLEYS
Example:
• The object in this figure has a fixed support at the left end
(point A).
• A cable passing over a pulley is attached to the object at 2
points.
• Isolate it from its supports & complete the free-body by
showing the reactions at the fixed support & the forces
exerted by the cable.
Free Body Diagram FBD 2D
Don’t forget the couple at the fixed support,
Since the tension in the cable is assumed, on both
sides of the pulley, the 2 same forces exerted by
the cable having the magnitude T.
Once the free-body diagram of an object is obtained, by
identifying the loads & reactions acting on it, for
equilibrium, the equilibrium equations can be applied.
Free Body Diagram FBD 2D
Ay
Ax
Free Body Diagram FBD 2D
Example:
Free Body Diagram FBD 2D
Example:
x
y
y
Free Body Diagram FBD 2D
Free Body Diagram FBD 2D
Example:
Solving rigid-body equilibrium problems
2D CASES
∑∑ 00 int == poanyforcesofdirection MF+ +
Only TWO unknowns can be found
G H Find the tensions in cables BG and DH
if P = 18.75 kN and Q = 12.5 kN ?
x
y
Example:
Only THREE unknowns can be found
∑∑∑ 0,0,0 === oyx MFF+ +
+
Example:
Find the reactions at the supports A and D
ϴ = 57 o
Note that the support at A is PIN type
and the support at D is ROLLER type
Example:
Find the reactions at the supports, if P = 2500 N
25 kNm
Example:
Example:
25 kNm
30o
Example:
25 kNm
6 sin 30 kN
6 cos 30 kN
0.8 cos 60 m
0.8 sin 60 m
Bx
By
Ay 1.2 sin 60 m
0.6+ 0.6+ 1.2 cos 60 m
EXAMPLE: For the given compound beam find the reactions at supports A
and B (ignore the size of the supports).
EXAMPLE: For the given compound beam find the reactions at supports A
and B (ignore the size of the supports).
Note: Seperate the given system into 2 parts as system CD and system AB.
Find the reactions at Supports C and D.
For system AB apply EQUAL MAGNITUDE BUT OPPOSITE DIRECTION
of the reaction values obtained for Support C (due to action-reaction).
Example:
For the given system, determine the reactions at A, B, C and D
(ignore the size of the supports)?
100 N
A SIMPLE PULLEY
100 N 100 N
100 N 100 N
MULTIPLE PULLIES
C
B
A
T = P
2T=P
4T=P
P /4 P
P /2
MULTIPLE PULLIES
Example: Calculate the tension T in the cable which supports 50 kN with the
given pulley arregement shown. Note that each pulley is free to rotate about its
bearing and has negligible weight compared with the carried load. Ignore the
weight of ther cable as well.
C
B
A
Solution:
C
B
A
25 kN
25 kN
12.5 kN 12.5 kN
= 12.5 kN
Example:
Relationship between pulled length ‘s’ of the rope and the
elevated height ‘h’ due to different pulley combinations.
Neglecting friction and the radius of the pulley, determine:
a) The tension in cable ADB
b) Reaction at C.
80 mm 80 mm 200 mm
150 mm
A B C
D
120 N
Example:
80 mm 80 mm 200 mm
150 mm
A B C
120 N
Example:
x
y
R1
R2 T T
FBD of the beam ABC
The mass of 700 kg is suspended from a trolley which moves along the
crane rail d = 3.5 m. Determine the force along short link BC and the
magnitude of the reaction force at pin A.
Example (T):
Different Support Types in 3D
1 unknown
1 unknown
1 unknown
3 unknowns
4 unknowns
Different Support Types in 3D
5 unknowns
5 unknowns
5 unknowns
5 unknowns
6 unknowns
Different Support Types in 3D
All 2D SUPPORTS
Some 3D SUPPORTS
Example: Ball-and-socket joint at A.
Free Body Diagram FBD 3D
Smooth journal bearing at B.
Roller support at C.
Example:
Free Body Diagram FBD 3D
Ball-and-socket joint at A.
BD and BE are cables.
Example: Free Body Diagram FBD
Free Body Diagram FBD
Solving
3D
rigid-body
equilibrium problems
Determine the reactions due to force F = 300 N, at the fixed support O.
Example:
Example:
Solution:
FBD
Rod AB subjected to 200N force. Determine the reactions at the ball-and-socket
joint A and the tension in cables BD and BE. Note that point C is acting at the
middle of the rod AB.
Example:
x
y
z
Solution:
FBD
At point A, ball and socket reaction means only reaction forces may occur along
3 axes (no moment). Cable BD under tension along y-axis only and cable BE
under tension along x-axis only. Note that point C is at the middle of the two
points A and B.
Determine the support reaction at C (ball and socket type) if the applied force at
B is F = 3.6 kN.
Example T: