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BA/BSc II (Mathematics)
Paper: Mechanics
Topic of Presentation: Statics
Dr. Baljeet Singh
Department of Mathematics
Post Graduate Government College
Sector 11, Chandigarh-160011
Terminology
Statics:
deals with the equilibrium of bodies, that is those
that are either at rest or move with a constant velocity.
•Particle: A particle has a mass but a size that can be neglected.
When a body is idealised as a particle, the principles of
mechanics reduce to a simplified form, since the geometry
of the body will not be concerned in the analysis of the problem.

Force
A force is a vector quantity and must have magnitude, direction
and point of action.
F

P
Here, the point P is the point of action of force and and are
directions. To notify that F is a vector, it is printed as bold. Its
magnitude is denoted as |F| or simply F.
Two vectors are equal if they are equal in
magnitude and act in the same direction.

P
p
Q

Forces equal in magnitude can act in
opposite directions
R
S
Types of Force Systems
Collinear: all forces acting along the same straight line.
Coplanar: all forces acting in the same plane.
 Concurrent: the lines of action of all the forces
intersect at a common point
 Parallel: the lines of action of all the forces are
parallel
 Non-concurrent: If not concurrent or parallel
Non coplanar: the lines of actions of all the forces do not lie
in the same plane.
Resultant of forces:
The resultant of a system of forces on a particle is the single
force which has the same effect as the system of forces.
Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
• Law of cosines,
C
B
C
R 2  P 2  Q 2  2 PQ cos B
  
R  PQ
• Law of sines,
B
sin A sin B sin C


Q
R
P
Resolution of force
A given force, F can be resolved into components. There are
two major cases:
(a) When one of the two components, P is known: The second
component Q is obtained using the triangle rule. Join the tip of P to
the tip of F. The magnitude and direction of Q are determined
graphically or by trignometry.
P
Q
F=P+Q
(b) When the line of action of each component is known: The force, F can be
resolved into two components having lines of action along lines ‘a’ and ‘b’ using the
paralleogram law. From the head of F, extend a line parallel to ‘a’ until it intersects ‘b’.
Likewise, a line parallel to ‘b’ is drawn from the head of F to the point of intersection with
‘a’. The two components P and Q are then drawn such that they extend from the tail of
F to points of intersection.
a
Q
F
P
b
Forces in a plane
1. Resultant of 2 forces acting on a particle

Two forces P and Q acting on a particle A, Figure 1 (a) can be replaced by
a single force R which has the same effect on the particle, Figure 1 (c).

This force R is called the resultant of the forces P and Q and this method
for finding the resultant of the force is called the parallelogram law for the
addition of two forces.
(a)
(b)
Figure 1
(c)
Example 1: If the angle between F1 and F2 is 60o and F1 = 54N and F2 = 60N.
Find the resultant force and the angle .
Solution:

R
F1
R 2  F12  F22  2 F1 F2 cos 
R 2  60 2  54 2  2  60  54  cos120 
120o

60o
F2
R  98.77  98.8 N
sin  sin 120

R
F1
F1
sin   sin 60
R
  28.26o
R
F1
120o

60o
F2
Example 2. The two forces P and Q acts on a bolt A as shown in Figure
below (a) Determine the resultants.
(a)
(b)
Figure 2
Solution:
Example 3.
Determine graphically, the magnitude and direction of the resultant of the
two forces using (a) Paralleolegram law and (b) the triangle rule.
600 N
900 N
45o o
30
Solution: A parm. with sides equal to 900 N and 600 N is drawn to scale as shown.
The magnitude and direction of the resultant can be found by drawing to scale.
600N
45o
600 N
30o
R
15o
900 N
45o
30o
The triangle rule may also be used. Join the forces in a tip to tail fashion and
measure the magnitude and direction of the resultant.
600 N
45o
R
B
30o
135o C
900 N
900N
Using the cosine law:
R2 = 9002 + 6002 - 2 x 900 x 600 cos 1350
R
R = 1390.6 = 1391 N
Using the sine law:
B

R
600
600
sin
135
1

i
.
e
.
B

sin
sin 135 sin B
1391
 17.8
The angle of the resul tan t  30  17.8  47.8
ie. R = 139N
47.8o
600N
135o
30o 900 N
Example 4. Two structural members B and C are bolted to bracket A.
Knowing that both members are in tension and that P = 30 kN and Q =
20 kN, determine the magnitude and direction of the resultant force
exerted on the bracket.
P
25o
50o
Solution: Using Triangle rule:
75o
20 kN
30 kN
105o

25o
Q
R
R2 = 302 + 202 - 2 x 30 x 20 cos 1050 - cosine law
R = 40.13 N
Using sine rule:
4013
. N
20

Sin 105o Sin 
and
Sin
Angle R  28.8 o  25o  38
. o
i. e R  401
. N,
38
. o
1
20 sin 105o

 28.8 o
4013
.
Resultant of a system of coplanar forces
In order to determine the resultant of several coplanar forces, each
force is first resolved into its x and y components. Then, the
respective components are added using scalar algebra since they are
collinear.
The resultant force can be then formed by adding the resultants of x
and y components using the parallelogram law.
Rectangular components of a force
y
Fy = Fy j
F
j
i
Fx = Fx i
x






In many problems, it is desirable to resolve force F into
two perpendicular components in the x and y directions.
Fx and Fy are called rectangular vector components.
In two-dimensions, the cartesian unit vectors i and j are
used to designate the directions of x and y axes.
Fx = Fx i and Fy = Fy j
i.e. F = Fx i + Fy j
Fx and Fy are scalar components of F
While the scalars, Fx and Fy may be positive or negat
ive, depending on the sense ofFx
and Fy, their absolute values are respectively equal to the magnitudes of the component
forces Fx and Fy,
Scalar components of F have magnitudes:
Fx = F cos 
and Fy = F sin
F is the magnitude of force F.

Example 5: Determine the resultant of the three forces
below.
600 N
y
800 N
350 N
60o
45o
25o
x
Solution:
 F x = 350 cos 25o + 800 cos 70o - 600 cos 60o
= 317.2 + 273.6 - 300 = 290.8 N
 F y = 350 sin 25o + 800 sin 70o + 600 sin 60o
= 147.9 + 751 + 519.6 = 1419.3 N
y
i.e. F = 290.8 N i + 1419.3 N j
Resultant, F
600 N
350 N
F  290.82  1419.32  1449 N
1419.3
  tan
 78.4 0
290.8
1
F = 1449 N
800 N
78.4o
60o
45o
25o
Example 6.

A hoist trolley is subjected to the three forces shown.
Knowing that = 40o , determine (a) the magnitude of
force, P for which the resultant of the three forces is
vertical (b) the corresponding magnitude of the resultant.


2000 N
P

1000 N
Solution:
(a) The resultant being vertical means that the
horizontal component is zero.
 F x = 1000 sin 40o + P - 2000 cos 40o = 0
P = 2000 cos 40o - 1000 sin 40o =
1532.1 - 642.8 = 889.3 = 889 kN
 Fy
(b)
= - 2000 sin 40o - 1000 cos 40o =
- 1285.6 - 766 = - 2052 N =
40o
2000 N
2052 N
P
40o
1000 N
Moment of a Force




The tendency of a force to produce rotation about some
axis is called the moment of a force.
The magnitude of this tendency is equal the magnitude of
the force times the perpendicular distance between the
axis and the line of action of the force (moment arm). M
=F*d
Unit: force x distance =F*L = lb-in, k-in, lb-ft, k-ft
N-m, kN-m (SI unit)
Sign conven.: clockwise (-), counterclockwise (+)
Varignon’s Theorem
It states that the moment of a force about a point is equal
to the sum of the moments of its components about the same point.
Consider the Force, F with two components: F = F 1 + F2.
Using the distributive law of vectors
Mo = r x F1 + r x F 2 = r x (F 1 + F2) = r x F
y
F1
Moment of F about 0
F
Mo = r x F
F2
r
o
x
Moment of couple
Introduction: A couple is defined as two parallel forces
which have the same magnitude, opposite directions and are
separated by a perpendicular distance, d
-F
F
Since the resultant force of the two forces of the couple is zero,
the only effect of a couple is to produce a rotation or tendency of
a rotation in a specified direction.
-F B
r
A
F
rB rA
O
To derive the moment of the couple, consider two vectors , rA and rB
from O to points A and B lying on the line of action of F and - F.
The moment of the couple about O is:
M = rA x (F) + rB x (-F) = (rA - rB) x F
By triangle law of vector addition, rB + r = rA or r = rA - rB
So: M = r x F
This result shows that a couple moment is a free vector i.e
can act at any point, since M depends only on the position
vector directed between the forces and not position vectors
rA and rB from O to the forces.
Some remarks on couple




A couple causes rotation about an axis perpendicular to its
plane
The moment of a couple is independent of the choice of
the axis of moment (moment center)
A couple cannot be replaced with a single equivalent
resultant force
A couple may be transferred to any location in its plane
and still have the same effect
Example 7. Determine the moment of the couple shown
in Figure and the perpendicular distance between the two
forces
760 N
200 mm
A
760 N
B
100 mm
350
760
N
Solution:
200
mm
A
B
100 mm
350
760
N
FA = -760 cos(350) - 760 sin(350) =-622 i – 435.9 j N
rBA = -0.1 i + 0.2 j m
i
j
k
-0.1 0.2
0
MB =
-622 -435.9 0
|MB|=
= 168 k Nm
Mx2 + My2 + Mz2 = 168 Nm
d=M/F=168/760=0.22 m
Resolution of a force into a force and couple acting
at another point

Any force F acting on a rigid body may be moved to any
given point A (with a parallel line of action), provided that
a couple M is added. The moment M of the couple is
equal to F times the perpendicular distance between the
original line of action and the new location A.
Resultant of two parallel forces


The magnitude of the resultant R of the parallel
forces A and B equals the algebraic summation of
A and B, where R = A + B.
Location of the resultant R is obtained by the
principle of moments.
Equivalent couples


Two couples are equivalent if they produce the same
moment.
Since the moment produced by a couple is always
perpendicular to the plane containing the couple
forces, it is necessary that the forces of equal
couples lie either in the same plane or in planes that
are parallel to one another.
Resolution of a Force into a Force at B
and a couple
A Force, F acting at point A on a rigid body can be resolved
to the same force acting on another point B and in the
same direction as the original force plus a couple M
equal to r x F i.e. moment of F about B
i.e Force in (a) equal to that in (b) equal to that in ©
F = 5 kN
5 kN
A
5 kN
M = 25 kN m 5 kN
B
B
5 kN
5m
(a)
(b)
(c)
Example

A force and couple act as shown on a square plate
of side a = 625 mm. Knowing that P = 300 N, Q
= 200 N and  = 50o, replace the given force
and couple by a single force applied at a point
located (a) on line AB (b) on line AC. In each
case, determine the distance for A to the point of
application of the force.
Coplanar forces in equilibrium
Forces acting on a body are in equilibrium if they cause no change of any sort
in the motion of the body, i.e.
(a) the resultant force is zero
(b) the set of forces has no turning effect.
A particle is treated as a mass at a point so the forces acting on it must be concurrent
and cannot have any turning effect. Hence, for a particle to be in equilibrium we need
only to ensure that the resultant force in each of two directions is zero.
Two forces in equilibrium :
The resultant of the two forces is zero so they must be of equal magnitude and
act in opposite directions.
But if the forces act along distinct parallel lines thay have a turn ing effect.
So, for equilibrium, the forces must act in a straight line.
Two forces in equilibrium must be equal and opposite
and act in the same straight line.
Three forces in equilibrium
Triangle of forces :
Recall how to find the resultant of two
non-collinear forces P and Q.
If a force R is added to P and Q so that the three forces are in equilibrium,
R must cancel out the effect of the resultant of P and Q.
Hence, R is equal and opposite to this resultant and passes throug h the
intersection of P and Q.
Three forces in equilibrium must be concurrent and can be represented in
magnitude and direction by the sides of a triangle taken in order.
Lami’s theorem
An alternative to the triangle of forces is used if we know the angles between
the forces rather than any lengths.
The sine rule for ABC yields:
P
Q
R


.
sin 180    sin 180    sin 180   
But sin 180     sin   so
P
Q
R


.
sin   sin    sin   
Lami's theorem : If three forces are in equilibrium they must be concurrent and each
force is proportional to the sine of the angle between the other two forces.
Non-concurrent forces
Forces acting on a rigid body need not be concurrent. We must now take into account
that these forces are capable of producing rotation.
Consider a perfectly uniform rod pivoted at its midpoint P.
If a downward force F is applied at one end A of the rod it
is caused to turn about the pivot.
Experiments show that an additional force 2 F , applied downwards
halfway along PB, will maintain the rod in its original position.
The turning effect of a force is given by:
magnitude of force  perpendicular distance from pivot
Equilibrium under parallel forces
Consider an object in equilibrium under the action of a set of parallel forces as shown.
There are no components of force in the direction Ox, so there can be no change
of motion in that direction.
Since the object is in equilibrium
 the resultant force in the direction Oy is zero
 the resultant moment about A (or any other point) is zero.
To find the unknown forces we can
 either collect the parallel forces and take moments about one axis
 or take moments about each of two axes.
Condition for equilibrium of a particle
A particle is said to be in equilibrium if the particle:
• is at rest (if originally at rest); or
• has a constant velocity (if originally in motion).

Normally, however, the term ‘equilibrium’ of more precisely
‘static equilibrium’ is used to describe an object at rest. To
maintain equilibrium, it is necessary to satisfy Newton’s
first law of motion.

Newton’s first law of motion requires that the resultant force
acting on a particle to be equal to zero. This condition may
be expressed as :
Newton’s first law of motion requires that the resultant
force acting on a particle to be equal to zero. This
condition may be expressed as
F = 0
where  F is the vector sum of all the forces acting on a
particle.
Coplanar Force Systems
For equilibrium condition, the equation can be written
as :
F = 0
Fx i + Fy j = 0
For both of these vector equations above to be valid,
then both the x and y components must be equal to
zero. Hence,
Fx = 0 = F1x + F2x + …..
Fy = 0 = F1y + F2y + …..
The scalar equations of equilibrium shown above require
that the algebraic sum of the x and y components of all
the forces acting on the particle must be equal to zero.
Working rule for numerical exercises
Coplanar force equilibrium problems for a particle can be analyzed using the
following procedure:
1. Free-Body Diagram
2. Establish the x and y axes is any suitable orientation
3. Label all the known and unknown force magnitudes and
directions on the diagram.
4. The sense of a force having an unknown magnitude can be initially
assumed.
5. Equations Of Equilibrium
Apply the equations of equilibrium Fx = 0 and Fy = 0 .
Components are positive if they are directed along a positive axis and
negative if they are directed along the negative axis.
6. If the solution gives a negative result, this indicates that the sense of
the force is reverse of that shown on the free-body diagram.
Conditions for Rigid body equilibrium
Consider a rigid body which is at rest or moving with x y z
reference at constant velocity

F1
z

F2
x
j 
i

F3
rigid body

F4
y
Free body diagram of ith particle of the body

Fi
z
j

ri

f ij
i
External force
: gravitational, electrical, magnetic or
contact force

Fi
y
x
Internal

fi

force f i
n


j 1

f ij
Force equilibrium equation for particle i

Fi
fi  0
(Newton first law)
j i
Force equilibrium equation for the whole body


Fi

fi  0
 0
  

fi

Fi
(Newton’s 3rd law )

F
0
Moment of the forces action on the ith particle about pt. O

M jo

ri

 

Fi


fi

  


ri
Fi  ri  f i  0

Moment equilibrium equation for the
body

 

 

M jo


rj 

Fj


fj  0

rj





 rj f j  0
Fj

rj

fj  0

Mo



rj
Fj  0
Equations of equilibrium for a rigid body are



F

Mo
0
0
Equations of Equilibrium for 2D rigid
body
(1) Conditions of equilibrium

F  0

 M0  0
  Fx  0
F  0
M  0
y
0
Here:

F1
y

M2

F2

M1

F3
x
Couple moment
 Fx  algebraic sum of x components of all force on the body.
 Fy  algebraic sum of y components of all force on the body.
M
0
 algebraic sum of couple moments and moments of all
the force components about an axis ⊥ xy plane and
passing 0.
y
Example
600N
200N
A
x
B Bx
2m
3m
Ay
2m
By
100N
3 unknown Ax, Bx, By
Equations of equilibrium
 Fx  0 600 cos 45  Bx  0
  Fy  0
Ay  600 sin 45  100  200  By  0
  M B  0 100  2  Ay  7  600 sin 45  5  600 sin 45  02.  0
3 equations for 3 unknowns
General conditions for equilibrium in a
plane
In general, a body will be in equilibrium only if both the resultant
force and the resultant moment are zero.
So the general conditions necessary for an object to be in equilibrium under the
action of a set of non-parallel coplanar forces are
 the resultant force in the direction Ox is zero
 the resultant force in the direction Oy is zero
 the resultant moment about any axis is zero
In some situations it is more suitable to use an alternative set of three
independent equations:
 the resultant in the direction Ox (or Oy but not both) is zero
 the resultant moment about a particular axis is zero
 the resultant moment about a different axis is also zero
Principle of Transmissibility
The principle of transmissibility states that the condition of
equilibrium or of motion of a rigid body will remain
unchanged if a force F action at a given point of the rigid
body is replace by a force F’ of the same magnitude and the
same direction, but acting at a different point, provided that
the two forces have the same line of action.