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Statics
Chapter 4
Force System Resultant
Eng. Iqbal Marie
Hibbeler, Engineering Mechanics: Statics, 13e, Prentice Hall
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Equilibrium requires the body to
have No Translation and No rotation.
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4.1 Moment of a Force - Scalar Formulation
The moment of the force about some other point O is
defined as the tendency of the force to rotate the object
about point O.
Moment = Torque = Twist
Mo = (F)(d) (force x distance)
d = "perpendicular distance"
d
F O
Scalar Formulation
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Determine the Moment of the force about point A if =45
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Moments in 3D
M = 0
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Direction of Moments
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Resultant Moment in 2-D
Resultant Moment. For 2-D
problems, where all the forces lie
within the same plane, the resultant
moment MO about point O can be
determined by summing of the
moments caused by all the forces in
the system
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And O
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4.2 Cross product
If C = A X B, then C will be perpendicular to A and B.
C = A B sin
Scalar formulation
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The cross product of two vectors is
+ - +
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Cartesian Vector Formulation
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4.3 Moment of a Force - Vector Formulation
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Resultant moment of a system of forces
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Transmissibility of a force
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4.4 Principle of Moments (Varignon's Theorem)
r
F1
F2
F
O
The moment of a force = the sum of the moments due its components
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The man pulls on the rope with a force of F = 20 N.
Determine the moment that this force exerts about the
base of the pole at O.
(4, - 3, 1.5)
(0,0,10.5)
rAB= {4i , -3 j, -9k}
rAB= 10.3
FAB = 20 ( 4i -3j -9k ) 10.3
roA= {0i , 0 j, 10.5k}
I j k
0 0 10.5
7.77 -5.83 -17.48
Mo = r X F
= 61.22i +81.59j + 0k
Mo = 102 Nm
roA
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The force F={6i+8j+10k} N creates a moment about point O of Mo={-14i+8j+2k} N m.
If the force passes through a point having an x coordinate of 1 m, determine the y and z
coordinates of the point. Also, realizing that Mo=Fd, determine the perpendicular distance
d from point O to the line of action of F.
Solution:
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F r
o
Mo = rXF
a
a
ua-a
4.5 Moment of a Force about a Specified Axis
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uax uay uaz
rx ry rz
Fx Fy Fz
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y
r
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4.6 Moment of a Couple:
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The moment of a couple does not depend on the point one takes the
moment about. In other words, a moment of a couple is the same about all
points in space.
Called Free vector
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4.7 Moment of a Force on a Rigid Body
Point O is on the line of action of the force
Principle of Transmissibility:
The external effect on a rigid body remains unchanged when a force acting at a given
point on the body, is applied to another point lying on the line of action of the force.
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Point O Not on Line of Action of Force
Equivalent to a force in the same direction and a moment of magnitude = r x F
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4.8 Resultant of a Force and Couple System
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4.8. Resultant of a Force and Couple
System – 2D
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Eg. Replace the forces acting on the brace shown below with an
equivalent resultant force and couple moment at point A.
o1
R
R1
22
R
oR
yR
R
oR
xR
6.668.382
8.882tan
F
Ftan
N962)8.882()8.382(
N8.882F
N8.88245sin400N600F
FF
N8.382F
N8.38245cos400N100F
FF
y
y
y
y
y
x
x
x
RF
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The couple moment has a magnitude of 220 N.m determine
the magnitude of F of the couple forces
+
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o1
22
oyyR
oxxR
7.33350
233tan
N5.420)233()350(
N233N200N60sin500FF
N350N100N60cos500FF
RF
R EE
o o
( ccw) M M
500 sin60 4 500 cos60 0
100 0.5 200 2.5
1182.1 N m
+
Equivalent
system at E ?
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4.9 Further Reduction of a Force System
Reduced to single force at a point.
If a system of forces is either concurrent, coplanar or
parallel, it can always be reduced to a single resultant
force acting through a unique point .
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Concurrent Force Systems
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Coplanar Systems
=
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Reduction to a Wrench ( Screw)
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Parallel Force Systems
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Determine the magnitude, direction, and location on the beam of the resultant
force that is equivalent to the system of forces shown.
o1
22
oyyR
oxxR
7.33350
233tan
N5.420)233()350(
N233N200N60sin500FF
N350N100N60cos500FF
RF
R EE
o o
( ccw) M M
500 sin60 4 500 cos60 0
100 0.5 200 2.5
1182.1 N m
+
m07.5d
mN1.1182)0(350d233
mN1.11825.22005.0100
060cos500460sin500MM)ccw(oo
EER
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Determine the magnitude, direction, and location on the slab of the resultant
force that is equivalent to the system of forces shown.
mN4200 (0)500 (0)400 (6)100 (8)600 M
mN3500 - (0)500 (10)400 -(5)100 (0)600 M
N-1400 N500 -N400 -N100 N600 -
x
x
O
O
FFR
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A 0
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Replace the force system with an equivalent force system and specify a location
(0,y) for a single equivalent force to be applied.
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4.10 Reduction of a Simple Distributed Loads
wind, fluids, the weight of objects
Intensity measured as a pressure
lb/ft (psf), lb/in2 (psi) Pascal 1Pa=1 N/m2
Units
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d
x3
x1
x2
d = Xi F
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Replacing a distributed load by single resultant load:
F must pass through the centroid of the area under the curve w(x).
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Magnitude of resultant force is equal to the
total area under the loading diagram
w = w(x)
d = Xi F
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L
h
2/3 L
1/3 L
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Find the equivalent resultant force and specify the magnitude and location
of the force measured from A.
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F2 F3
F1
Replace the loading by a simple resultant force, and specify the location of
the force on the beam measured from point O.
Solution:
51 kN
d= 17.9m
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OR
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Replace the force system acting on the beam by an equivalent force and
specify its location from point A.
1.5 kN
3 kN
D = 3*1 + 1.5*1.5 + 1.5 * 4
3+1.5+1.5 = 1.625 m
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4.5 m 2m
8 m
18 kN 3kN 9kN