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Composition and Resolution of
Forces
Unit -1
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Force
• Force is an agent which tend to cause eitherchange in motion or deformation.
• It is denoted by N (Newton) = kg.m/s2
• Effect of Force
– Change the motion of body
– Retard the motion of body
– Deform the body – Bring the body to rest/equilibrium
– Change the stress in the body
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Characteristics of Force
1. Magnitude of force (N)
2. Line of action of force (direction)3. Nature of force (pull/push)
4. Point of action of force
System of Forces
1. Coplanar forces
2. Collinear forces
3. Concurrent forces
4. Coplanar forces
5. Coplanar concurrent forces
6. Non-coplanar concurrent forces
7. Non coplanar non concurrent forces
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• Principle of independence of forces
If a number of forces are simultaneously acting on a
body, then the resultant of these forces will have the
same effect as produced by all the individual forces.
• Principle of Transmissibility of forces
The effect of an external force on a rigid body
remains unchanged even if the force is moved, along its lineof action.
= + +
ine of actionP = =
=P P P
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Resultant force
• If a body is subjected to n number of forces, acting
simultaneously, the representive single force whichproduces the same effect is called the resultant force.
• The process of determining the resultant force of a givensystem of forces on a rigid body is composition of forcesor compounding of forces.
• Methods for determining the resultant force
• Analytical Method
–Parallelogram law of forces –Resolution of force
• Method of resolution
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Analytical Method
• Parallelogram law of forces
–If two forces acting simultaneously on a body, they canbe represented by adjacent sides of parallelogram in
which the resultant may be represented in magnitude
and direction by the diagonal of the parallelogram
which passes through their point of intersection
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= 2 cos
tan = sin
cos
Cases1. If the forces are in straight line( = 0), R = P + Q
2. If the forces are perpendicular( = 0), =
3. If the forces are acting opposite in straight line ( = 180), R=P - Q
4. If the forces are equal (P=Q), R = 2P cos
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Numerical
• Two forces acting at an angle of 120o of which the higher
magnitude of force was 40N and the resultant is
perpendicular to the other force, find the other force.
F2
120oR
90o
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Graphical method
• Draw a line OA and OB to scale for forces P and Q
respectively, with angle with each other.
• Construct the parallelogram OABC by erecting parallel to
OA and OB to intersect at C.
• Draw a diagonal OC, which represent the reaction force,the measure of OC, multiplied with scale gives the
magnitude of R.
O A
CB
D
P
Q
R
• The angle AOC is the direction of
reaction R.• Extent line OA and mark a point D
such that OD CD
• OD represents the projection of the
resultant R.
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Analytical Method
• Resolution of force
– The most common method adopted for resolution of forces is
into rectangular components.
– So, as per the parallelogram law of forces, Force F can be
written as
F = Fx + Fy
– In vector form, the same can be represented as
F = Fx i + Fy j
– Considering the angle between horizontal component Fx and
F is ,
Fx = F cos =
Fy = F sin = tan−
F
Fx
Fy
i
j
x
y
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Hence, in the given system of forces acting on the body in different
directions, the resultant of the forces can be determined by
resolving the forces in two directions and obtain the vector sum ofthese components.
Mathematically, it can be determined as follows:
R = F1 + F2 + F3 +. . .= (F1x i + F1y j) + (F2x i + F2y j) + (F3x i + F3y j) + . . .
Rx i+Ry j = (F1x + F2x + F3x+ . . .)i + (F1y + F2y + F3y + . . .)j
R =
and = tan−
R is the reaction force magnitude
is the direction, represented by angle of reaction w.r.t. Horizontal
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Various situations and solutions
Fx = F sin
Fy = F cos
x
y
F
x
y
F
Fx = - F cos
Fy = - F sin
x
yF
Fx = F sin (-)
Fy = - F cos (-)
x
yF
x
yF
Fx = F cos (-)
Fy = F sin (-)
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The foces F1, F2, and F3, all of which act
on point A of the bracket, are specified
in three different ways. Determine the xand y scalar components of each of the
three forces.
F 1x = 600 cos 35° = 491 N
F1y = 600 sin 35°
= 344 N
F 2x = -500 (4/5) = - 400 N
F2y = 500 (3/5) = 300 N
= tan- 1 [0.2/0.4] = 26.6 °
F 3x = 800 sin 26.6 = 358 N
F 3y = - 800 cos 26.6 = - 716 N
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R x = 100 cos 30o + 80 cos 20o = 161.8 N
Ry = 100 sin 30o
- 80 sin 20o
= 22.6 NR = 161.8 22.6 = 163.4 N
= tan-1 (22.6/161.8) = 8o
R2 = (80)2 + (100)2 + 2(80)(100) cos 50o
tan =00 50
0+00 50
R = 163.4 N = 28o wrt F2,
i.e. 8o wrt X axis
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Moment• In addition to the tendency to move a body in the direction of its
application, a force can also tend to rotate a body about an axis.• This rotational tendency is known as the moment M of the force,
also referred to as torque.
• The moment is a vector M perpendicular to the plane of the
body.• Therefore, the magnitude of the
moment is defined as
M = F d
where d is also termed as lever armdistance, perpendicular to the line offorce.
• The concept is highly used in hinges(for doors, windows etc.)
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Varignon's Theorem states that the moment of a force about
any point is equal to the sum of the moment of the
components of the force about the same point.
Mo = Q × r + r × P = R × r Mo = Q.q + (- P.p) = R.d
M = F d = r × F =F r sin
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Additional information
Hinge support: The structural component whichsupports vertical force, horizontal force, but momentis zero.
Roller support: The structural component whichsupports vertical force, but both horizontal force andmoment is zero.
Fixed support: The structural component whichsupports all vertical force, horizontal force andmoment.
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Couple• The moment produced by two equal, opposite, and non-collinear
forces is called a couple.
• These two forces cannot be combined into a single force as their
resultant in every direction is zero, their only effect is to produce a
tendency of rotation, same as moment.
• Hence, for the above situation,
Couple M = F.(a+d) – F.a = F.d (anticlock)
where d is the distance between the forces
• This indicates that the couple is independent on the point of moment.
• In vector form, M = ra ×F + rb ×(-F) = (ra-rb )×F = r × F
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Clockwise
couple
Anti-Clockwise
couple
A couple is not affected if the forces act in a different but. parallel plane
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Translation of forces
A
B
.
.
F A
B
.
.
F
FF
d= AB
.
F
M= F d
.
=
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ProblemIn the given figure, if the force of 100 N is
replaced with force P = 400 N, causing the
same couple. determine the angle,
The magnitude of the moment due to force of
100 N is
M = 100 (100) = 104 N.mm = 10 N.m (anti-clock)
Similarly, the moment due to forces P,produce a moment of
M = 400 (40 cos ) = 1.6 cos N.m (anti-clock)
Equating both, = 51.3o
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Replace the given force by equivalent
force-couple at O80 N
Solution:
Moment at O is
M = 80 (9 sin 60o)
= 624 N.m
Force = 80 N
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Resultants
• The resultant of a system of forces is the simplest force
combination which can replace the original forces without altering
the external effect on the rigid body to which the forces are applied
R = F1+F2+F3+…. = F
Rx= Fx , Ry= Fy , =
= tan−
= tan−
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Problem• Determine the resultant {If the four forces and one
couple which act on the plate shown
Rx= 66.9 N
Ry= 132.4 N
R = 148.3 N
= 63.2o
Mo
= 140 – 50 (5) + 60 cos 45o (4) – 60 sin 45o (7)
= -237 N.m = 237 N.m (anticlock)
.OMo= 237 N.m
R = 148.3 N
= 63.2o
.OMo
= 63.2o
Final Line of resultant of R is at distance d from O,
i.e. d = 237/148.3 = 1.6m
So the line of action forms tangent making an angle 63.2o with horizontal
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End of Unit 1
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