Chapter 2
Statics of Particles
Addition of Forces
Parallelogram Rule:
The addition of two forces P and Q :
Draw two parallel lines of vectors to form a parallelogram
Draw the diagonal vector represent the resultant force
→
P
A
→
Q
→ →
→
P+ Q = R
Addition of Forces
Triangle Rule:
Alternative
method to
Parallelogram
Rule
Arrange two vectors in tip-to-tail fashion
Draw the vector from starting point to the tip of second vector
→
P
→ → →
A
→
Q
P+ Q = R
Addition of Forces
Triangle Rule:
The order of the vectors does not matter
→ → →
P
→
P+ Q = R
A
→
Q
Addition of two forces is commutative
→ →
→ →
P + Q = Q+ P
Addition of Forces
More than two forces:
Arrange the given vectors in tip-to-tail fashion
Connect the tail of first vector to the tip of last vector
→ → → →
P+ Q + S = R
→
S
→
P
A
→
Q
Addition of Forces
More than two forces:
The order of the vectors does not matter
→
→
Q
P
→
P
→
→ → → →
S
P+ Q + S = R
→
S
A
A
→
Q
→ → →
→ → →
+
=
S+ P Q
Q+ S + P
→ → →
= P+ Q + S
Rectangular Components of a Force
→ →
Fx , Fy
+
→
→
F = Fx
→
Rectangular components of F
θ : Angle between x-axis and F
measured from positive side of x-axis
→
Fy
Rectangular Components of a Force
Unit Vectors
→ →
Unit vectors
i , j




Fx  Fx i
F y  Fy j



F  Fx i  Fy j
Rectangular Components of a Force
Unit Vectors
→ →
Unit vectors
i , j




Fx  Fx i
F y  Fy j



F  Fx i  Fy j
F  F F
2
x
2
y
Fx  F cos
Fy  F sin 
Rectangular Components of a Force
Example
A Force of 800 N is applied on a bolt A.
Determine the horizontal and vertical
components of the force
Fx  F cos  800cos(1450 )   665 N


Fx   (665 ) i
Fy  F sin   800 sin(1450 )   459 N


Fy   (459 ) j



F   (655) i  (459) j
Rectangular Components of a Force
Example
A man pulls with a force of 300 N on a
rope attached to a building.
Determine the horizontal and vertical
components of the force applied by the
rope at point A
AB  8 2  6 2  10
8
8
cos  

 0.8    36.860
AB 10
  360  36.86  323.140
Fx  F cos  300 cos(323.140 )  240 N


Fy   (459 ) j
Fy  F sin   300sin(323.140 )  180 N
Addition of Forces by Summing Their
components
→
→
P
S
A
→
Q
→ → → →
P+ Q + S = R








Px i  Py j  Qx i  Qy j  S x i  S y j  Rx i  Ry j
 R x  ( Px  Q x  S x )
 R y  ( Py  Q y  S y )
 R x   Fx
 R y   Fy
Addition of Forces by Summing Their
components - Example
Problem 2.22 on page 33
Determine the resultant force applied
on the bolt
Force
7 kN
5 kN
9 kN
Magnitude
x Component
F1
5 kN
5* cos (40) = 5* sin (40) =
3.83 kN
3.21 kN
F2
7 kN
7*cos(110)=
-2.39 kN
7* sin (110)=
6.57 kN
F3
9 kN
9*cos(160)=
-8.46 kN
9* sin(160)=
3.08 kN
-7.01 kN
12.86 kN
R
y Component



R   (7.01 kN ) i  (12.86 kN ) j
Equilibrium of Particle
When the resultant of all the forces
acting on a particle is zero, the particle
is in equilibrium
Equilibrium of Particle - Example
Determine the resultant force applied on
point A
N
N
N
Force
Magnitude x Component y Component
F1
300
300
0
F2
173.2
0
-173.2
F3
200
-100
-173.2
F4
400
-200
346.4
0
0
N
200*cos(240) = -100
200*sin(240) = -173.2
R
Equilibrium of Particle – Newton’s First Law
If the resultant force acting on a particle is zero,
• the particle remains at rest (if originally at rest)
• the particle moves with a constant speed in a
straight line ( if originally in motion)
Equilibrium State
Equilibrium of Particle - Example
Load with mass of 75 kg
Determine the tensions in each ropes of
AB and AC
W  g m  ( 9.81 m / s 2 ) ( 75 kg )  736 N
N
Rx   Fx  TAB x  TAC x  0
N
N
200*cos(240) = -100
200*sin(240) = -173.2
Since the load is in equilibrium state,
The resultant force at A is zero.
 TAB cos(1300 )  TAC cos(300 )  0
Ry   Fy  TAB y  TAC y  0
 TAB sin(1300 )  TAC sin(300 )  736  0
Equilibrium of Particle – Example
(continued)
 T AB cos(1300 )  T AC cos(300 )  0
 (0.643) T AB  (0.866) T AC  0
N
 TAB sin(1300 )  TAC sin(300 )  736  0
 (0.766) TAB  (0.500) TAC  736  0
(2)
N
N
 TAB  647 N
 TAC  480 N
200*cos(240) = -100
200*sin(240) = -173.2
(1)
Equilibrium of Particle – Example-2
Two cables are tied together
at C and they are loaded as
shown
Determine the tensions in
cable AC and BC
Problem 2.44 / page 41
Equilibrium of Particle – Example-2
Draw “Free Body” Diagram

TBC

T AC
C
W  3 kN
Equilibrium of Particle – Example-2
For simplicity

F1

TBC

T AC

F2
C
W  3 kN

F3
Equilibrium of Particle – Example-2
Since the system is in equilibrium
The resultant force at C is zero.

F2

F1
1

2
C

F3  3 kN
Ry   Fy
 F1y  F2 y  F3 y  0
 F1 sin(1 )  F2 sin( 2 )  F3 sin(3 )  0
Re memberthe definitionof 
 F1 sin(180  )  F2 sin( 2 )  F3 sin(2700 )  0
Equilibrium of Particle – Example-2
Since the system is in equilibrium

2
The resultant force at C is zero.

F2

F1
1

2
C

F3  3 kN
 500
0
  43.6
 525
 300 
0
  Arc tan
  36.86
 400
 2  Arc tan
 1  180  36.86  143.130
 F1 sin(143.130 )  F2 sin(43.60 )  F3 sin(2700 )  0
Equilibrium of Particle – Example-2

F2

F1
C
2
1
 F1 sin(143.130 )  F2 sin(43.60 )  F3 sin(2700 )  0
 (0.6) F1  (0.69) F2  3000 0 (1)

F3  3 kN
Rx   Fx
 F1x  F2 x  F3x  0
 F1 cos(1 )  F2 cos( 2 )  F3 cos(3 )  0
 F1 cos(143.130 )  F2 cos(43.60 )  F3 cos(2700 )  0
 (0.8) F1  (0.724) F2  0 (2)
Equilibrium of Particle – Example-2

F2

F1
C
2
1
 (0.6) F1  (0.69) F2  3000 0
 (0.8) F1  (0.724) F2  0

F3  3 kN
 F1  2202 N
F2  2433N
(1)
(2)
Forces in Space ( 3 D )
Fx  F cos( x )
Fy  F cos( y )
Forces in Space ( 3 D )
Fz  F cos( z )
Forces in Space ( 3 D )
Fx  F cos( x )
Fy  F cos( y )
Fz  F cos( z )
•The three angle  x ,  y ,  z define the direction of the force F
•They are measured from the positive side of the axis to the force F
•They are always between 0 and 180º
Forces in Space ( 3 D )




F  Fx i  Fy j  Fz k
Forces in Space ( 3 D )

•The vector  is the unit vector along
the line of action of F

F
vector
 

F
its m agnitude



 F cos x i  F cos y j  F cos z k
 
F



 F (cos x i  cos y j  cos z k )
 
F




  cos x i  cos y j  cos z k

Forces in Space ( 3 D )




F  Fx i  Fy j  Fz k



 F cos x i  F cos y j  F cos z k



 F cos x i  cos y j  cos z k

 F




  cos x i  cos y j  cos z k


•The vector  is the unit vector along the line of action of F

Forces in Space ( 3 D )
Fx  F cos( x )
Fy  F cos( y )
Fz  F cos( z )
When thecomponentsFx , Fy and Fz of a forceF are given
F  Fx2  Fy2  Fz2
Fx
cos x 
F
cos y 
Fy
F
Fz
cos z 
F
Forces in Space ( 3 D )
Direction of the force is defined by
the location of two points,
M  x1 , y1 , z1  and N  x2 , y2 , z 2 
Forces in Space ( 3 D )

d  vectorjoining M and N



 d xi  d y j  d z k
where
d x  x 2  x1
d y  y 2  y1
d  d x2  d y2  d z2
d z  z 2  z1


F  F



1 
FF
d xi  d y j  d z k
d
 Fd x  Fd y  Fd z 
F
i
j
k
d
d
d
Fd y
Fd x
Fd z
Fx 
Fy 
Fz 
d
d
d



d

d

 1


  d xi  d y j  d z k
d



Sample Problem 2.7
A (40, 0,  30)
B (0, 80, 0)
d x  x 2  x1  0  40  40m

d
d y  y 2  y1  80  0  80m


d z  z 2  z1  0  (30)  30m

d  vectorjoining A and B




d  -( 40m) i  ( 80m) j  ( 30m) k
The tension in the guy wire is 2500 N. d  d x2  d y2  d z2  94.3m
Determine:

 d
a) components Fx, Fy, Fz of the force  
d
acting on the bolt at A,




1

 40i  80 j  30k
b) the angles x, y, z defining the
94.3




direction of the force
  0.424i  0.848 j  0.318k


Sample Problem 2.7



  0.424i  0.848j  0.318k

• Determine the components of the force.


F  F



 2500 N  0.424 i  0.848 j  0.318k 



  1060 N i  2120 N  j  795 N k
Sample Problem 2.7
• Noting that the components of the unit vector are
the direction cosines for the vector, calculate the
corresponding angles.



  cos x i  cos y j  cos z k



 0.424 i  0.848 j  0.318k

 x  115.1
 y  32.0 
 z  71.5
or
cos x 
Fx
 1060N

F
2500N
cos y 
Fy
F

2120
2500
cos z 
Fz
795

F
2500
Addition of Concurrent Forces in Space
• The resultant R of two or more vectors in space
R   Forces
Rx   F x ' s
Ry   F y ' s
Rz   F z ' s
R  R x2  R y2  R z2
R
cos x  x 
R
cos y 
Ry
R

cos z 
Rz

R
Sample Problem 2.8
m
A concrete wall is temporarily held
by the cables shown.
m
The tension is 840 N in cable AB
and 1200 N in cable AC.
m
m
Determine the magnitude and
direction of the resultant vector on
stake A
Sample Problem 2.8
Equilibrium of a Particle in Space
When the resultant of all the forces acting on a particle is
zero, the particle is in equilibrium
F
x
0
F
y
0
F
z
0
Problem 2.103 on page 60