# chapter 2 - Dr. ZM Nizam

```Forces and
equilibrium
Forces and Equilibrium
 Scalars and vectors
 Types of forces
 Resultant of forces
 Equilibrium of particles
Scalar and Vectors
 Scalar - a physical quantity that is completely described
by a
real number
e.g. time, length, mass, temperature
 Vector- is described by both magnitude (nonnegative
real
number) and direction
e.g. position of a point in space relative to
another
point, forces
- represented by bold-faced letters: u, a, W
- magnitude of vector u = |u|
- graphical representation of vectors: arrows
- direction of arrow shows the direction of
vector
- length of arrow  magnitude of vector
Vectors
Example:
rAB - position of point B relative to point A
- direction from point A to point B
- distance between A and B = |rAB|
Types of forces
Types of forces
External force
Internal force
Resultant of force in
2Dimension
• Scalar
• Added 4 m2 and 3 m2 = 7 m2
• Vector
• Added 4 km and 3 km = sum and direction
 Resultant = single vector giving the result of
the addition of the original two or more
vectors.
Resultant of forces: Graphical
 Triangle rule
- sum of vector from tail of P to
Q
P
R=P+Q
 Parallelogram rule
- the sum is independent of the
order in which the vectors are
placed head to tail
- vector addition is commutative
P+Q=Q+P=R
R
P
Q

PQ
P
Q
- vector subtraction
R=PQ
Q
 R - resultant of two forces, P and Q
Resultant of forces: Analytical
B
P
Q
C
• Trigonometric:
Law of cosine,
R 2  P 2  Q2  2PQ cos B
R=P+Q
Law of sine,
sin A sin B sinC


Q
R
A
A
Pythagorean theorem
• Resolution of vectors:
‘Resolve’ vectors into components
using the x and y axes system.
We use the “unit vectors” i and j
to designate the x and y axes

Example 1:
Figure shows an initial design sketch of part of the roof of a sports
stadium to be supported by the cables AB and AC. The forces the
cables exert on the pylon to which they are attached are
represented by the vectors FAB and FAC. The magnitude of the forces
are |FAB| = 100 kN and |FAC| = 60 kN. Determine the magnitude
and direction of the sum of the forces exerted on the pylon by the
cables (a) graphically and (b) using trigonometry.
Solution:
(a) Graphically construct the parallelogram rule with FAB and FAC
proportional to their magnitudes:
By measuring the figure, we estimate the magnitude of the
vector FAB + FAC to be 160 kN and its direction to be 19&deg; above
the horizontal.
Solution:
(b) Consider the parallelogram rule:
Since  = 180&deg;  30 &deg; = 150&deg;
Applying law of cosine to the triangle:
FAB  FAC  FAB  FAC  2 FAB FAC cos α
2
2
2
 100  60  210060cos 150
2
2
Magnitude |FAB + FAC| = 154.9 kN
To determine the angle  between FAB + FAC and the horizontal,
apply law of sines to shaded triangle:
sin 
sin α

FAB FAB  FAC
 F sin α
β  arcsin  AB
 FAB  FAC

100 sin 150 
  arcsin 
  18.8
155



Example 2:
Determine the horizontal and
vertical components of P
Solution:
Example
The ring shown in figure is subjected to two
forces, Fl and F2. If it is required that the
resultant force have a magnitude of 1 kN and be
directed vertically downward, determine the
magnitudes of Fl and F2,
provided  = 30&deg;.
Solution:
Vector addition sketch according to the parallelogram law:
Using the law of sines:
F1
1000 N

sin 30 sin 130
F1 = 635 N
F2
1000 N

sin 20 sin 130
F2 = 446 N
Example 4:
Determine the resultant of
forces that acts on bolt A.
Solution:
• Trigonometric solution - law of cosines,
R 2  P 2  Q 2  2PQ cos B
 40N  60N  240N60N cos 155
2
R  97.73N
2
Law of sines,
sinA sinB

Q
R
sinA  sinB
Q
R
 sin155
A  15.04
  20  A
  35.04
60N
97.73N
There are four concurrent
cable forces acting on the
bracket.
How do you determine the
resultant force acting on the
bracket ?
Resultant of forces (component method)
• A single force can be broken up into two
separated forces
Fy
F
Fx
• To add vectors analytically using the method
components , one should be proceed
according to the step.
• Resolve each vector into a horizontal and
vertical component
• Add the vertical components, Ry=∑Fy.
• Add the horizontal components, Rx = ∑Fx
• Combine the horizontal and vertical
components to obtain a single resultant
vector.
R  ( Rx ) 2  ( R y ) 2
 Ry 


1
  tan 
R 
 x
Exercise 1
Three concurrent forces are acting on a
bracket. Find the magnitude and angle of
the resultant force by resolving the forces.
Solution:
F1 = { 15 sin 40&deg; i + 15 cos 40&deg; j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j }
kN
= { -24 i + 10 j } kN
F3 = { 36 cos 30&deg; i – 36 sin 30&deg; j } kN
= { 31.18 i – 18 j } kN
I = x axis, j = y axis
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN

= tan-1(3.49/16.82) = 11.7&deg;
y
FR

x
Resultant force in 3 Dimension
+y
-Z
+X
-X
+Z
-y
y
6N
R
x
4N
z
3N
R1
R1  (3N ) 2  (4 N ) 2
R  (5 N ) 2  (6 N ) 2
R1  5 N
R  7.81N
R=7.81 N, coordinates (4,6,3)
Vector equation
• Or known as Cartesian vector
• Used right hand coordinate system
Example 2.13
• Express the force F shown in Figure 2.23 as
a Cartesian vector
•
Since only two coordinate direction angles are specified,
the third angle α must be determined using equation
cos2   cos2   cos2   1
cos2   cos2 60o  cos2 45o  1
cos   1  (0.5) 2  (0.707 ) 2  0.5
•
Hence, two possibilities exist, namely,
α = cos-1(0.5) = 60o or α = cos-1(0.5) = 120o
•
By inspection, it is necessary that α = 60o, since Fx must
be in the +x direction, with F = 200 N, we have
F = F cos αi + F cos βj + F cos γk
F = (200 cos 60oN)i + ( 200 cos 60o N)j + ( 200 cos 45o)k
F = {100.0i + 100.0j + 141.4k} N
•
Show that indeed the magnitude of F = 200N
F  Fx  Fy  Fz
2
2
2
F  1002  1002  141.4 2
Equilibrium of a particle
•
The term “particle” used in statics to describe a body when;
• the size and shape of the body will not significantly affect
the solution of the problem being considered.
• the mass of the body can be assumed to be concentrated
at a point.
•
A particle can be subjected only to a system of concurrent
forces and that the necessary and sufficient conditions for
equilibrium can be expressed mathematically as
R = F = 0
where F is the vector sum of all forces acting on the
particle.
•
To apply the equation of equilibrium --- account for all the
known and unknown forces (F) which act on the particle.
APPLICATIONS
For a spool of given
weight, what are
the forces in cables
AB and AC ?
APPLICATIONS (continued)
For a given cable
strength, what is the
maximum weight
that can be lifted ?
•
Procedure for Drawing a Free-Body Diagram
• Draw Outlined Shape: Imagine the particle to be isolated
or cut &quot;free&quot; from its surroundings by drawing its
outlined shape.
• Show all Forces: Indicate on this sketch all the forces that
act on the particle. These forces can be active forces,
which tend to set the particle in motion, or they can be
reactive forces which are the result of the constraints or
supports that tend to prevent motion. To account for all
these forces, it may help to trace around the particle's
boundary, carefully noting each force acting on it.
• Identify Each Force: The forces that are known should be
labeled with their proper magnitudes and directions.
Letters are used to represent the magnitudes and
directions of forces that are unknown.
EQUILIBRIUM OF PARTICLE IN 2-D
This is an example of a 2-D
or coplanar force system. If
the whole assembly is in
equilibrium, then particle A is
also in equilibrium.
To determine the tensions
in the cables for a given
weight of the engine, we
need to learn how to draw
a free body diagram and
apply equations of
equilibrium.
THE WHAT, WHY AND HOW OF A FREE
BODY DIAGRAM (FBD)
Free Body Diagrams are one of the most important things
for you to know how to draw and use.
What ? - It is a drawing that
shows all external forces acting
on the particle.
Why ? - It helps you write the
equations of equilibrium used to
solve for the unknowns (usually
forces or angles).
How ?
1. Imagine the particle to be isolated or cut free from its
surroundings.
2. Show all the forces that act on the particle.
Active forces: They want to move the particle.
Reactive forces: They tend to resist the motion.
3. Identify each force and show all known
magnitudes and directions. Show all unknown
magnitudes and / or directions as variables .
A
Note : Engine mass = 250 Kg
FBD at A
EQUATIONS OF 2-D EQUILIBRIUM
Since particle A is in equilibrium, the
net force at A is zero.
So FAB + FAD + FAC = 0
or
A
 F= 0
FBD at A
In general, for a particle in equilibrium,  F = 0 or
Fx i + Fy j = 0 = 0 i + 0 j
(A vector
equation)
Or, written in a scalar form,
Fx = 0 and  Fy = 0
These are two scalar equations of equilibrium (EofE).
They can be used to solve for up to two unknowns.
EXAMPLE
Note : Engine mass = 250 Kg
FBD at A
Write the scalar EofE:
+   Fx = TB cos 30&ordm; – TD =
0
+  Fy = TB sin 30&ordm; – 2.452 kN = 0
Solving the second equation gives: TB = 4.90 kN
From the first equation, we get: TD = 4.25 kN
• Two-Dimensional Problems
R
= Rx + Ry = Rn + Rt = 0
= Rxi + Ryj = Rnen + Rtet = 0
= Fxi + Fyj = Fnen + Ftet = 0
Satisfy only if
Rx = Rxi = Fxi = 0
Ry = Ryj = Fyj = 0
Rn = Rnen = Fnen = 0
Rt = Rtet = Ftet = 0
Example 2.14
A free-body diagram of a particle subjected to the action
of four forces is shown in Fig.2.29. Determine the
magnitudes of forces Fl and F2 so that the particle is in
equilibrium
Solution
+ Fx = F1x + F2x + F3x + F4x = 0
= F1 cos 60 + F2 cos 30 - 40 cos 56 - 10 cos 15 = 0
= 0.5 F1 + 0.866 F2 – 22.37 – 9.659 = 0
From which
F1 + 1.732 F2 = 64.06
+ Fy
(a)
= F1y + F2y + F3y + F4y = 0
= F1 sin 60 + F2 sin 30 - 40 sin 56 + 10 sin 15 = 0
= 0.8660 F1 + 0.5 F2 – 33.16 + 2.588 = 0
From which
F1 + 0.5774 F2 = 35.30
Solving Eqs (a) and (b) simultaneously yields
F1 = 20.9 kip
F2 = 24.9 kip
(b)
GROUP PROBLEM SOLVING
Given: The car is towed at
constant speed by the 600
N force and the angle  is
25&deg;.
Find:
The forces in the ropes AB
and AC.
Plan:
1. Draw a FBD for point A.
2. Apply the EofE to solve for the
forces in ropes AB and AC.
GROUP PROBLEM SOLVING (continued)
600 N
FBD at point A
A
25&deg;
FAB
30&deg;
FAC
Applying the scalar EofE at A, we get;
+  Fx = FAC cos 30&deg; – FAB cos 25&deg; = 0
+  Fy = -FAC sin 30&deg; – FAB sin 25&deg; + 600 = 0
Solving the above equations, we get;
FAB = 634 N
FAC = 664 N
EQUILIBRIUM OF PARTICLE IN 3-D
Three-Dimensional Problems
R
= F = 0
= Rx + Ry + Rz = 0
= R xi + R y j + R z k = 0
= Fxi + Fyj + Fzk = 0
satisfied only if
Rx = Rxi = Fxi = 0
Ry = Ryj = Fyj = 0
Rz = Rzk = Fzk = 0
Example 2.17
•
A 90-lb load is suspended from the hook shown in
Figure 2.28(a). The load is supported by two cables
and a spring having a stiffness k = 500 lb/ft.
Determine the force in the cables and the stretch of
the spring for equilibrium. Cable AD lies in the x-y
plane and cable AC lies in the x-z plane.
• Free-Body Diagram.
•
Equation of equilibrium
Fx = 0 FD sin 30 - (4/5)FC = 0
Fy = 0 -FD cos 30 + FB = 0
Fz = 0 (3/5) FC – 90 lb = 0
(a)
(b)
(c)
•
Solving Eq. (c) for FC, then Eq (a) for FD, and finally
Eq. (b) for FB, yields,
•
FC = 150 lb
Ans
•
FD = 240 lb
Ans
•
FB = 208 lb
Ans
•
The strech of spring is therefore
FB = ksAB
208 lb = 500 lb/ft (sAB)
sAB = 0.416 ft
Ans
Assignment No. 1:
1. Find the components of
force Q for the axis system
of A and B.
Tips:
Construct a vector parallelogram by
drawing lines parallel to axes A and B,
from the tip of Q. Then, apply the sine
law.
Assignment No. 1:
2. Three concurrent forces are acting on a bracket.
Find the magnitude and angle of the resultant force
by resolving the forces.
Assignment No. 1:
3.The members of a truss are pin-connected at joint
O. Determine the magnitudes of F1 and F2 for
equilibrium. Set  = 60o
Assignment No. 1 is due Friday, October 4, 2013 by 12 noon
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