BEM unit I.pdf

BASIC ENGINEERING MECHANICS
Unit I
Dr. M.V.Rama Rao
M.Tech,PhD,Postdoc,FIE,FICI,FACCE(I)
Professor, Department of Civil Engineering
Vasavi College of Engineering, Hyderabad

SYSTEM OF UNITS
• cgs system – centimetre , gram , second
• mks system – metre, kilogram , second
• fps system – foot , pound, second
• SI system of units – Systeme Internationale
• Newtons for force
• Metres for length
• Seconds for time
As per law, Indians are required to follow SI system of units

CARTESIAN COORDINATE SYSTEM
X
Y
X
Y
Z
PLANAR(TWO-DIMENSIONAL) SPATIAL(THREE-DIMENSIONAL)

X
Y
Z
O
RIGHT HANDED COORDINATE
SYSTEM
RIGHT-UP-FORWARD
RIGHT-UP-BACKWARD
X
Y
Z
LEFT HANDED COORDINATE
SYSTEM

X
Y
Z
RIGHT - POSITIVE
LEFT-NEGATIVE
DOWN-NEGATIVE
UP - POSITIVE
FORWARD - POSITIVE
BACKWARD-NEGATIVE

REFERENCE FRAME
• The set of three coordinate axes along with the origin of the
coordinate system is called reference frame
• All measurements in space are made with reference to the
reference frame
• Reference frame can be inertial or non-inertial(accelerating)
• An inertial reference frame has zero acceleration
• WE ADOPT INERTIAL REFERENCE FRAME IN ENGINEEERING MECHANICS
• A non-inertial reference frame moves with a constant acceleration

SCALARS AND VECTORS
• Scalars are physical quantities having only magnitude but no direction.
Example: mass , work , energy , distance etc.
Denoted as m , W , U , d etc.
• Vectors are physical quantities having both magnitude and direction
Example: weight, force , displacement , velocity , acceleration , torque
Denoted as etc.
, , ,
F d v a

VECTORS
• Vectors are represented graphically with their length
proportional to their magnitude. The direction of the vector is
specified as an arrow directed from its tail to the tip.
• The tail of the arrow represents the initial point of the vector
• The tip of the arrow represents the final point of the vector.
• The length of the arrow is proportional to the magnitude of
the vector TIP
TAIL

VECTORS
DIRECTION
MAGNITUDE
THE LENGTH OF A VECTOR IS PROPORTIONAL TO ITS MAGNITUDE
10kN
5kN
20kN

ORIENTATION OF VECTORS
F
20kN
F IS A VECTOR OF MAGNITUDE 20kN, ORIENTED ALONG
THE POSITIVE DIRECTION OF X-AXIS
20kN
P
P IS A VECTOR OF MAGNITUDE 20kN, ORIENTED ALONG
THE NEGATIVE DIRECTION OF X-AXIS
P ACTS HORIZONTALLY TO THE LEFT
F ACTS HORIZONTALLY TO THE RIGHT

Q
10kN
Q IS A VECTOR OF MAGNITUDE 10kN, ORIENTED ALONG
THE POSITIVE DIRECTION OF Y-AXIS
10kN
T
T IS A VECTOR OF MAGNITUDE 10kN, ORIENTED ALONG
THE NEGATIVE DIRECTION OF Y-AXIS
T ACTS VERTICALLY DOWN
Q ACTS VERTICALLY UP

P IS A VECTOR OF MAGNITUDE 10kN, ACTING RIGHT-UP
DRAW A HORIZONTAL LINE PASSING THROUGH THE TAIL AND
A VERTICAL LINE PASSING THROUGH THE TIP
P
10kN
PX
PY
PX acts horizontally to the right
PY acts vertically up
PX and PY are rectangular components of vector P

P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-UP
PX
PY
PX acts horizontally to the left
PY acts vertically up

P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-DOWN
P
10kN
PX
PY
PX acts horizontally to the left
PY acts vertically down

P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING RIGHT-DOWN
PX
PY
PX acts horizontally to the right
PY acts vertically down

RESOLUTION OF A FORCE INTO COMPONENTS
A PLANAR FORCE HAS TWO COMPONENTS :
ONE COMPONENT ALONG X-AXIS AND
THE OTHER COMPONENT ALONG THE Y-AXIS
X-COMPONENT CAN BE ALONG THE POSITIVE DIRECTION OF X-AXIS(TO THE RIGHT) OR
NEGATIVE DIRECTION OF X-AXIS(TO THE LEFT)
Y-COMPONENT CAN BE ALONG THE POSITIVE DIRECTION OF Y-AXIS(UPWARD) OR
NEGATIVE DIRECTION OF Y-AXIS(DOWNWARD)
THERE ARE FOUR POSSIBLE COMBINATIONS OF X AND Y COMPONENTS
- RIGHT-UP
- LEFT-UP
- LEFT-DOWN
- RIGHT-DOWN RIGHT-UP LEFT-UP LEFT-
DOWN
RIGHT-
DOWN

RESOLUTION OF A PLANAR FORCE
P
PX
PY
X

cos
sin
X X
Y X
P P
P P


= →
= 
P=100N , θX=300
0
0
100cos30 86.6
100sin30 50
X
Y
P N
P N
= = →
= = 

P
PX
PY
X

cos
sin
X X
Y X
P P
P P


= 
= 
P=100N , θX=300
0
0
100cos30 86.6
100sin30 50
X
Y
P N
P N
= = 
= = 

P
PX
PY
Y

sin
cos
X Y
Y Y
P P
P P


= 
= 
P=100N , θY=300
0
0
100sin30 50.0
100cos30 86.6
X
Y
P N
P N
= = 
= = 

P
PX
PY
Y

sin
cos
X Y
Y Y
P P
P P


= →
= 
P=100N , θ=300
0
0
100sin30 50
100cos30 86.6
X
Y
P N
P N
= = →
= = 

VECTORS
• Vectors are represented graphically with their length
proportional to their magnitude. The direction of the vector is
specified as an arrow directed from its tail to the tip.
• The tail of the arrow represents the initial point of the vector
• The tip of the arrow represents the final point of the arrow.
• The length of the arrow is proportional to the magnitude of
the vector TIP
TAIL

RESOLUTION OF A FORCE INTO COMPONENTS
A PLANAR FORCE HAS TWO COMPONENTS :
ONE COMPONENT ALONG X-AXIS AND
THE OTHER COMPONENT ALONG THE Y-AXIS
X-COMPONENT CAN BE ALONG THE POSITIVE DIRECTION OF X-AXIS(TO THE RIGHT) OR
NEGATIVE DIRECTION OF X-AXIS(TO THE LEFT)
Y-COMPONENT CAN BE ALONG THE POSITIVE DIRECTION OF Y-AXIS(UPWARD) OR
NEGATIVE DIRECTION OF X-AXIS(DOWNWARD)
THERE ARE FOUR POSSIBLE COMBINATIONS OF X AND Y COMPONENTS
- RIGHT-UP
- LEFT-UP
- LEFT-DOWN
- RIGHT-DOWN RIGHT-UP LEFT-UP RIGHT-
DOWN
LEFT-
DOWN

P
PX
PY
Y

sin
cos
X Y
Y Y
P P
P P


= 
= 
P=100N , θ=300
0
0
100cos30 86.6
100sin30 50
X
Y
P N
P N
= = 
= = 

P
PX
PY
Y

sin
cos
X Y
Y Y
P P
P P


= →
= 
P=100N , θ=300
0
0
100sin30 50
100sin30 86.6
X
Y
P N
P N
= = →
= = 

F
m
n
2 2
m n
+
X
F
Y
F
2 2
2 2
2 2
X Y
X
Y
F
F
F
F F
m n m n
m
F
m n
n
F
m n
 
=  
+
 
 
= 
=

=
+

+ 

F
4
3
2 2
4 3 5
+ =
X
F
Y
F
n
4 3
100 80 100 60
5
a
4 3 5
d
5
X Y
X Y
F F
N
F
F F N
=  = → =  =
= =  
F=100N

5 12
260 100
1
and
5 1
260 240
3 3
2 1 1
3
X Y
X Y
F
F N F
F F
N
=  =  = =
= =   
F=260N
F
5
12
X
F
Y
F
13

F
4
3
5
X
F
Y
F
n
4 3
100 80 100 60
5
a
4 3 5
d
5
X Y
X Y
F F
N
F
F F N
=  =  =  =
= =  
F=100N

RESULTANT OF A SYSTEM OF CO-PLANAR CONCURRENT FORCES
P
Q F
T R
R P Q F T
= + + + IS THE RESULTANT FORCE
RESULTANT IS THE SINGLE
FORCE THAT CAN REPLACE
THE GIVEN SYSTEM OF
FORCES ACTING ON A BODY


NUMERICAL EXAMPLE-1
COMPUTE THE RESULTANT OF THE SYSTEM OF FORCES SHOWN BELOW:
X
Y
0
30
200
P N
=
250
Q N
=
0
60
223.61
S N
=
1
2
4
3
500
T N
=
5
5
0
0
200cos30 173.2
200sin30 100.0
X
Y
P N
P N
= =
= =
0
0
250sin 60 217.506
250cos60 125.0
X
Y
Q N
Q N
= − = −
= =
1 2 5
223.61
100
5
223.61
2 200
5
X Y
X
Y
S S S
S N
S N
= =
= − = −
= −  = −
4 3 5
4 500
400
5
3 500
300
5
X Y
X
Y
T T T
T N
T N
= =

= =

= = −

NUMERICAL EXAMPLE-1
2 2
1 0
173.2 217.506 100 400 255.694
100 125 200 300 275
375.506
tan 47.083
X X X X X
Y Y Y Y Y
X Y
Y
X
R X P Q S T N
R Y P Q S T N
R R R N
R
R
 −
= = + + + = − − + =
= = + + + = + − − = −
= + =
 
= =
 
 



NUMERICAL EXAMPLE-2
The body on the incline is subjected to the vertical and horizontal forces
shown. Compute the components of each force along X and Y axes oriented
parallel and perpendicular to the incline.
F
P
1
400
200
P N
F N
=
=
X
Y
4
3

Y
X
X
F
F
Y
F



4
cos
5
3
sin
5


=
=
cos 400 0.8 320
sin 400 0.6 240
X
Y
F F N
F F N


= =  =
= − = −  = −
X
P
Y
P
sin 1200 0.6 720
cos 1200 0.8 960
X
Y
P P N
P P N


= = −  = −
= = −  = −
P

NUMERICAL EXAMPLE-3
P
3 1
260
6
P N
F N
=
=
X Y
4
3 
3
2

5
12


P

X
P
Y
P
13
F
X
X
( )
cos( ) cos cos sin sin
3 2
361 0.8 0.6
13 1
.
3
120 15
X
X
P
N
P P
P
     
= + = −
 
=  −  =
 
 
( )
sin( ) sin cos cos sin
3 2
361 0.6 0.8
13 1
.
3
340 42
Y
Y
P
N
P P
P
     
= + = +
 
= −  −
+  =
 
 

NUMERICAL EXAMPLE-3
P
3 1
260
6
P N
F N
=
=
X Y
4
3 
3
2

5
12

F

X
F
Y
F
13


13
F
X
Y
( )
cos( ) cos cos sin sin
12 5
260 0. 2
8 0.6
13 3
3
1
1
X
X
F
N
F F
F
     
= + = −
 
= − −
 −  =
 
 
( )
12 5
260 0. 4
6 0.8
13 3
2
1
2
Y
Y
F
N
P P
F
     
= + = +
 
= − −
 +  =
 
 

NUMERICAL EXAMPLE - 4
2
2
1
1
X
Y
4
3
P




X

Y
P
X
P
Y
P
( )
( )
1 2 2
0.8 0.6 893
5 5 5
c
9
os( ) cos c
98.4
446.
os sin sin
2 1
0.8 0.6 5
X
X
Y
Y
P N
N
P P P
P
P P N
P P P
P P
     
     
= + = +
 
=  +  = = 
 
 
= + = −
 
=  −  =

=

Compute P if its X-component is 893N

NUMERICAL EXAMPLE-5
The force system shown in the figure has a resultant of 200N pointing up along
the Y-axis. Compute the values of F and ϴ required to give this resultant.
0
0
0
2 2
1
500 cos 240cos30 0
cos 292.15
sin 240sin30 2
. 433.
00
sin 320
292 15
4
320
320
t .
an
29 5
3
6
1
1
7
2.
X
Y
R
N
X F
F N
R Y F
F N
F




 −
= = − + + =
 =
= = − =
 =
= + =
 
= =
 
 


X
Y

F
500N
240N

NUMERICAL EXAMPLE-6
The block shown in the figure is acted upon by its weight W=400N, a horizontal
force F=600N , and the pressure P exerted by the inclined plane. The resultant of
these forces is parallel to the incline. Compute P and R. Does the block move up
or down the incline?
F
W
4 0
600
0
W N
F N
=
=
0
30
R
0
45
P
( )
0 0
0 0
cos30 600 sin 45 ....(1)
sin30 400 cos45 ....(2)
1.366 200
600 0.866 146.42
0
0
0.866 0.7 7 600
0.5
3
.707
0.707 400
146.42
669. 2
R
N
R P
R
P
R P
R
P
P
R N
+ =
−
=
= −
=
=
= − 
= − + 
= 
− 
Block moves up the incline

PARALLELOGRAM LAW OF FORCES
Q
R
 
2 2
2 cos
R P Q PQ 
= + +
P cos
Q 

sin
Q 
sin
tan
cos
Q
P Q



=
+

NUMERICAL EXAMPLE-7
Two forces, each of magnitude P, are inclined at 600 to each other. Their
resultant is 346.41N. Compute P.
2 2
0
2 2 0
2 cos
Here, =60
2 cos60 3 346.41
346.41
200
3
R P Q PQ
P Q
R P P P P P N
P N


= + +
=
= + +  = =
= =

MOMENT OF A FORCE
Moment of a force about a given point is defined as the product of its magnitude
with the perpendicular distance of the point from the line of action of the force.
UNITS : Nm , kNm , Nmm
F
O
d
F
O
M Fd
=
X Y
F F
F
O O O
M M M
= +
MOMENT OF A FORCE ABOUT A GIVEN MOMENT
CENTRE IS THE SUM OF THE MOMENTS OF ITS
COMPONENTS ABOUT THE SAME MOMENT
CENTRE – VARIGNON ‘S THEOREM
MOMENT CENTRE
PERPENDICULAR
DISTANCE
FORCE

MOMENT OF A PLANAR FORCE
X Y
F F
F
O O O X Y
F
O y X x Y
M M M yF xF
M i F i F
= + = −
= =
CLOCKWISE POSITIVE
X
F
X
F
Y
F
y
x
( )
,
P x y
y
i
x
i
Y
X
F
Y
F
Y
F F
X
F

NUMERICAL EXAMPLE-8
A
D
C
B
IN THE FIGURE, ASSUMING CLOCKWISE
MOMENTS TO BE POSITIVE,COMPUTE
MOMENTS OF A FORCE F=450N AND
P=361N ABOUT POINTS A,B,C AND D.
F
P
X
F
Y
F
0.8 0.8 450 360
4 3 5
0.6 0.6 450 270
y
x
x
y
F
F F
F F N
F F N
= =  = =  = →
= =  = 
CLOCKWISE POSITIVE
3 3 360 270
3 4 3 360 4 270
0 5 5 270
3
1350
2
0
3 360 270
160
135
810
F
A x y
F
B x y
F
C y
F
D x y
Nm
Nm
Nm
Nm
M F F
M F F
M F
M F F
= − − = −  − =
= + =  +  =
= + =  =
= − = =
−
 −

NUMERICAL EXAMPLE-8
A
D
C
B
F
P
X
P
Y
P
2 2 361
2 3 13 13 13
3 3 361
13 13
200
300
y
x
x
y
P
P P
P
N
P
P
P
N
→


= =  = = =

= = =
CLOCKWISE POSITIVE
3 2 3 200 2 300
3 3 3 2
9
00 3 300
0 3 0 20
3
0
1
0 3 300
3 2 200
300
00
2
2 00 00
3
P
A x y
P
B x y
P
C x y
P
D x y
Nm
Nm
Nm
M P P
M P P
M P P
M P P
= − + = −  +  =
= − =  −  =
= − =  −  =
= + =
−
−
=  + 

NUMERICAL EXAMPLE-9
361
F N
=
X
F
Y
F
A
y
2
A
x m
=
A
y
i
x
i
Y
X
F
Y
F
Y
F F
X
F
X
3
2
3 2 361
a
2 361
= and
0
3 2 13 13 1
n
3 13
3 0 d 200
x
y
x
x y
y
F
F F
F N F N
F
F F
 
= =  =

→ =
=
=
In the given figure, find the y-coordinate of
point A such that the 361N force will have a
clockwise moment of 400Nm about O. Also
compute the X and Y intercepts of the action
line of the force.
0
8
2
300 2 2 0 400 2.667
3
y
x
F
F
F
O O O A x
A A
y
y m
M M M y F F
y  = =
= + = − 
− = 
3 1
0 400 00 .333
y
x
y
F
F
F
O O O y x y
M m
M M i F i i
 =
= + = + = 
0 4 0 2
0 200
y
x
F
F
F
O O O x
x y x
M m
M i F i i
M
= + = +  =
 =

NUMERICAL EXAMPLE-10
4
4
3
3


250
P N
=
Y
P
Y
P
X
Y
sin 2 2 sin cos 2 0.6 0.8 0.96
Y
P P P P P
  
= = =   =
5m
X
Y
F 2
1

COUPLE
A couple is defined as a pair of equal, parallel and oppositely
directed forces.
The perpendicular distance d between the lines of action of
forces is called the moment arm of the couple.
d
F
F
C Fd
=
Units are Nm or Nmm or kNm
A couple is a vector. It has magnitude and sense
The resultant of these forces is zero, but the
moment sum is not zero.
Thus, the only effect of the couple on a body is to
rotate the body about an axis perpendicular to
the plane of the couple.
+ SIGN-CONVENTION

PROPERTIES OF COUPLE
A unique property of the couple is that the moment sum of its forces
is constant and is independent of the choice of moment centre.
C Fd
=
F
F
A
B
D
O
( )
d x
−
x
At A 0
C F Fd
= + =
At B 0
C Fd Fd
= + =
( )
At O C F d x Fx Fd
= − + =
( )
At D C F d y Fy Fd
= + − =
y
+ SIGN-CONVENTION

Since the only effect of a couple is to produce a moment that is independent
of moment centre, the effect of the couple is unchanged if the couple is
rotated through any angle in its plane
F
F
d
F
F
d
d
F
F
F
F
C Fd
=
d
The effect of the couple is unchanged if the couple is shifted to any position
in its plane or is shifted to a parallel plane

Since the only effect of a couple is to produce a moment that is independent of
moment centre, the effect of the couple is unchanged if the couple is replaced by
another pair of forces in its plane whose product Fd and sense of rotation is
unchanged
F
F
C Fd
= ( ) ( )
2 0.5
C F d Fd
=  =
( ) ( )
0.5 2
C F d Fd
=  =
2F
2F
0.5F
0.5F
d 0.5d
2d

RESULTANT OF NON-CONCURRENT FORCES

Q
P
Q
T
T
Q
T
P
P
O
P
Q
T


P
Q
T
T
Q
T
P
P
O
Q Q
T
P
O
Q
+
P
Q
T
T
P
P
O

Q
T
P
O
C

A system of non-concurrent forces acting on a body
can be replaced by a single resultant force and a
resultant couple.
P
Q
T

R
O
C

VARIGNON’S THEOREM FOR NON-CONCURRENT FORCES
For a body being acted upon by a system of non-concurrent forces, sum of
the moments of individual forces is equal to the moment of the resultant of
the system of forces about the same moment centre plus the resultant
couple.
P
Q
T
O
P Q T R
O O O O
M M M M C
+ +  +

A vertical force P at A and another vertical force F at B which act on the bar shown in Figure
produce a resultant force of 150N down at D and a counter-clockwise couple C=300 Nm.
Find the magnitudes of forces P and F. P
A
B
F
3m
4m
D
150
R N
=
300
C Nm
=
150 ...(1)
R P F N
= + = −
150 7 300 3 0 450 ...(2)
450 150 0
450
300
3 0
R P F
B B B
M
F
P N
C M M
P F P
F N
F
N
N
= 
= 
+ = + 
−  − = +  = −
− + = −  =
+
SIGN-
CONVENTION
VARIGNON’S THEOREM

Find the values of P and F such that the four forces shown in the figure
produce an upward resultant of 300N acting at 4m from the left end of the
bar.
100N P
R
F 200N
2m 2m
3m
4m
300 100 200
...(1)
200.
R F
F
P
P
= = − + − +
 − =
+
SIGN-
CONVENTION
2 5 200
4 4 300 0 100 2 5 7 200
...(2)
R P
F
F
P
− = −  =  − + − 
− + =
N
2 2 400
2 5 20
....(1) 2
...(2)
200 P 0
0
=40
P
N
P F
F
F
−
=

=
− + =

The beam in the figure supports a load which varies uniformly from an
intensity of 60N/m at the left end to 180N/m at the right end. Compute the
magnitude and position of the resultant.
+
SIGN-
CONVENTION
12m
60 N m
180 N m
60 N m
12m
12m
120 N m
( )
1
12 60 12 120
2
1 2
12 60 6 12 12
3
1440
7
0 12
2
N
x m
R
R x
=  +   =
 
 =   +    


=


x
R

0.6 60 and
3 4 5
0.8 80
y
x
x
y
T
T T
T T N
T T N
= =  = =
= =
The three forces shown in the figure produce a horizontal resultant passing through A.
Compute P and F. 100
T N
=
X
T
Y
T
3
4
P
F
A
R
80 0 80
0 2 1 2 0
2 80
2
2 6
0
80
0 0 20
Y Y
R
A x y
R Y T F F F N
M P F T T
P P
P N
F
N
N
= 
= 
= = − = − =  =
= = −  + +  
− +  =  = −

20 6 40
4
0
0
x
R N
N
P
R
T
= →
= + = − + =
+
SIGN-
CONVENTION

P
x
P
Y
P
2
3
X
T
F
A
T Y
T
X
F
y
F
1
3
2 3
and
2 3 13 13 13
X Y
X Y
P P P P P
P P
= =  = → = 
1 3 10
316
and
1
1
0 10
0
3 316
00
30
10
X Y
X
Y
T
T N
T N
T T
T


= = 
= = =

= =
2
and
2 1 5 5 5
X Y
X Y
F F F F F
F F
= =  = → = 
The three forces shown in the figure cause a horizontal resultant to pass through A. If T=316N,
compute P and F

P
x
P
Y
P
2
3
X
T
F
A
T Y
T
X
F
y
F
1
3
R
B
R
( )
3 0 0 3 2
1
3 100 2 300 100
3
X Y
R F P T
B B B B
T T
B B X Y
M M M M
R M M T T
R N
= + +
 = + + + = − +
= −  +  =
100 100
)
2 ....(
00 1
X X
X
X X
X
X
R F P T
P
P
F
F
= + −  =
=
+
+
−
0 300 0
300...(2)
Y Y Y Y Y Y
Y Y
R F P T F P
F P
= − + + =  − + + =
− + = −
3 600
2
1 180.28
335.4
3
0.5 and 1.5 0.5 1.5 00
....(2)
00
13
2
300
5
1
Y X Y X X
X
X
X
X
X
P N
F N
F F P P F P
F P
P
P
F
F
= =  − + = −
− + = −
= = −
=

= −

= =
+
SIGN-
CONVENTION

X
Y
Z
(4,0,0)
A
4
9
(0,9,0)
B
6
(0,0,6)
C
O
(0, 5,0)
D −
5
8
(0,0, 8)
E −
7
( 7,0,0)
F −
THREE- DIMENSIONAL COORDINATE SYSTEM

X
Y
Z
(4,12, 5)
A −
4
12
(0,9,4)
D
O
5
8
(0, 3, 8)
C − −
7
( 7, 3,4)
E − −
3
5
(4,12,3)
B
3
4
9
3
4
(2, 5,0)
F −
2
THREE- DIMENSIONAL COORDINATE SYSTEM

X
Y
Z
2 2 2
( , , )
B x y z
2
x
O
2
y
2
z
1
y
1
z
1
x
SPATIAL VECTOR
1 1 1
( , , )
A x y z

COMPONENTS OF A PLANAR FORCE
1 1
( , )
A x y
2 2
( , )
B x y
X
F
Y
F
F
X Y
F F F
x y d
= =
2 1
2 1
x x x
y y y
= −
= −
x
y
d
2 2
d x y
= +

COMPONENTS OF A SPATIAL FORCE
1 1 1
( , , )
A z
x y
2 2 2
( , , )
B z
x y
X
F
Y
F
F
m
X Z
Y F
F F F
x d
z
y
F
= = =
=
1
2 1
2 1
2
x x x
y y y
z z z
= −
=
= −
−
x
y
d
m
X
Z
m
Y m
F
y
F
xF
F
z
F
F
=
=
=
X Z
Y F
F F F
= + +
2
2 2
d z
x y
= + +
FORCE
MULTIPLIER

RESULTANT OF SPATIAL FORCES
Find the resultant of force system shown in the figure in which P=280N,T=260N
and F=210 N
X
Y
Z
6
12
(0,12,0)
A
O
(6,0,4)
D
4
( 4,0, 3)
B − −
3
6
( 4,0,6)
C −
4
T
P
F

0
0
0
cos 84.04
cos 11.69
cos 79.75
x
x x
y
y y
z
z z
R
R
R
R
R
R
 
 
 
=  =
=  =
=  =
2 1 2 1 2 1
2 2 2
2 2 2
70
660
120
674
x
y
z
x y z
x x x y y y z z z
d x y z
R X N
R Y N
R Z N
R R R R N
= − = − = −
= + +
= = −
= = −
= =
= + + =



Desscription x y z d Force Multiplier X Y Z
P=280N -4 -12 6 14 20 -80 -240 120
T=260N -4 -12 -3 13 20 -80 -240 -60
F=210N 6 -12 4 14 15 90 -180 60
-70 -660 120
X
 Y
 Z


COMPONENTS OF A SPATIAL FORCE
X Y Z k
F F F F
i j
= + + X m Y m Z m
F xF F yF F zF
= = =
( )
m m m m
F xF yF zF k
F x y
i k z
j i j
= + + + +
=
A force F of magnitude 260N acts along the line A(2,3,7) to B(5,7,19).
Express F in vectorial form
2 1
2 1
2 1
2 2 2 2 2 2
5 2 3
7 3 4
19 7 12
3 4 12 13
260
20
13
m
x x x
y y y
z z z
d x y z m
F
F N m
d
= − = − =
= − = − =
= − = − =
= + + = + + =
= = =
( ) 20(3 4 12 )
m i j k
y i k
F F x z j
= + + = + +

The system of forces shown in the figure produces a vertical resultant. Given
FAD=252N, compute FAB and FAC
X
Y
Z
6
12
(0,12,0)
A
O
(6,0,4)
D
4
(0,0, 9)
B −
3
( 4,0,3)
C −
4
9
FAB
FAC
FAD
Vertical resultant means RX=0 and RZ=0
0
0
X
Z
R X
R Z
= =
= =



Desscription x y z d Force Multiplier X Y Z
FAB 0 -12 -9 15 0
FAC -4 -12 3 13
FAD=252N 6 -12 4 14 18 108 -216 72
X
 Y
 Z

15
AB
F 12
15
AB
F
−
9
15
AB
F
−
13
AC
F 4
13
AC
F
−
12
13
AC
F
−
3
13
AC
F
4
0 108 0
13
9 3 9 3
72 0 351 72 0
15 13 15 1
351
255
3
AC
A
AC
AB AC A B
B
F N
F N
X F
Z F F F
= + − = 
= − + + =  − +
=
=
+  = 



UNIT VECTOR
( )
( )
( )
2 2 2
1 1
ˆ ˆ
ˆ
1
m m m
m
m
i j k
n n i j k
n i j k
F F x y z F F F x y z F d
F F F F x y z
F F d
x y z
d
= + +  = = + + =
 
 
=  = = + +
 
 
   
= + + is a unit vector in the direction of F
F
n̂
2 2 2
1 1
ˆ (3 4 12 ) (3 4 12 )
13
3 4 12
20(3 4 12 )
F
n i j k i j
F i j k
k
= + + = + +
+
+
= +
+

DOT PRODUCT
Dot product(SCALAR PRODUCT of two vectors is defined as
a
b
. cos
a b ab 
=

a is the pre-multiplying vector
b is the post multiplying vector
1
ˆ
. cos . b
a b a a n
b

 
= 
 
 
cos
a 
1
ˆb
n

DOT PRODUCT
Dot product(SCALAR PRODUCT of two vectors is defined as
a
b . cos
a b ab 
=

1 1
ˆ ˆ
. cos . cos
a b
a b n n
a b
 
   
=  =
   
   
ˆa
n
ˆb
n
1
1

DOT PRODUCT OF TWO VECTORS
( ) ( )
1 2 3
1 2 3
1 2 3 1 2 3 1 1 2 2 3 3
. .
a a i a j a k
b bi b j b k
a b a i a j a k bi b j b k a b a b a b
= + +
= + +
= + + + + = + +
0
. . . 1 1 cos0 1
. . . . . . 1 1 cos90 0
i i j j k k
i j j i j k k j k i i k
= = =   =
= = = = = =   =
. cos cos .
b a ba ab a b
 
= = = DOT PRODUCT IS COMMUTATIVE
( )
. . .
a b c a b a c
+ = + DOT PRODUCT IS DISTRIBUTIVE
( ) ( )
. . . .
a b c a b c
= DOT PRODUCT IS ASSOCIATIVE
Example:
( ) ( )
2 3 4 4 5 6
. 2 3 4 . 4 5 6 2 4 3 5 4 6 27
a i j k b i j k
a b i j k i j k
= − + = + −
= − + + − =  −  −  = −

A boom AC is supported by a ball-and-socket joint at C and by cables BE and AD. If the force
multiplier of a force F acting from B to E is Fm=10N/m and that of a force P acting from A to
D is Pm=20N/m, find the component of each force along AC.
X
Y
Z
4
10
(4, 5,0)
B −
O
(0, 10,0)
C −
6
(0,0, 3)
D −
3
(0,3,6)
E
4
3
(8,0,0)
A
5
F
P
( )
( )
10 10( 4 8 6 )
20 20( 8 0 3 )
m m
m m
F N m F F xi yj zk i j k
P N m P P xi yj zk i j k
= = + + = − + +
= = + + = − + −
8 10 0
AC i j k
= − − +
( )
( )
2 2 2
1 1
8 10 0
8 10 0
1
8 10 0
164
A
C
C
A
k
n AC i
A
n i
j k
C
j
= = − −
+
=
+
+
− − +
( )
( )
0
1
8 10
ˆ
. .
20
8 8
1
2
6
0(
4
10 3
1
8 0
0 99.9
3
5
)
64
0
AC AC
AC
F F n
F N
j i j
i k
k
= =
= − − + − −
+
 =
−
− +
− −
( )
( )
1
10( 4 8 6
ˆ
. .
0
4 8 8 10 6 0 37.
6
1
8
8
48
1
) 10 0
0
4
1
AC AC
AC N
i j
F i j k
F n k
F =
− + − − +
+
= =
− − + − +  = −

In the system shown in figure below, the force multiplier of a force F acting from B to D is
Fm=150N/m and that of force P acting from A to E is 100N/m. Find the component of each
force along AC. What angle does each force make with AC?
X
Y
Z
8
9
(8, 3,0)
B −
O
(0, 9,0)
C −
6
(0,4, 6)
E −
(0,0,6)
D
3
6
(12,0,0)
A
F
P
4
( )
( ) 100( 12 4 6 )
3
0
150
1
5
0
1 0( 8 6 )
m m
m
m
P
F N m F xi yj zk
N m P xi
F i j
P y
k
j i
k j k
z
= = +
=
−
= − +
+ =
= +
+ +
−
+
( )
( )
2 2 2
12 9 0
1 1
ˆ 12 9 0
12 9
1
ˆ 1
0
2 9 0
15
A
C
C
A
n i j k
AC i j k
n AC i j k
AC
= − − +
= = − − +
+
= − +
+
−

X
Y
Z
8
9
(8, 3,0)
B −
O
(0, 9,0)
C −
6
(0,4, 6)
E −
(0,0,6)
D
3
6
(12,0,0)
A
F
P
4
)
150( 1
8 3 00( 12 4
6 ) 6
P i j k
F i j k =
= − + +
+ − −
( )
1
ˆ 12 9 0
15
AC
n i j k
= − − +
( )
( )
0
ˆ
. .
150
9
6
6 27 0
150( 8 3 )
690
1
12 9
1
5
5
1
AC AC
AC
F
N
i i j k
j
F n k
F
−
−
=
− +
=
+
+
− =
+
=
( )
( )
ˆ
. .
100
1
9
1
100( 12 4 1
72
4
6
4 36
)
1
2 0
0
5
5
0
1
AC AC
AC N
P i j
P i
P n
F
k k
j
= −
= =
= − +
+
− −
=
−
+

2 2 2
2 2 2
1
( 8 3 6 ) =
)
8 3 6
1
( 12 4 6 )
12
ˆ
4
1
(
9
1
ˆ ( 12 4
8 3 6 )
1
6
14
6
0
P
F
n i
i j k
i j k
n
k
i
j
j k
− + +
+ +
+
= = +
= −
− + −
+
−
+
−
+
( )
1
ˆ 12 9 0
15
AC
n i j k
= − − +
( ) ( )
( ) 0
1 1
1
1
1 0
1
( 8 3 6 )
109
63.
ˆ ˆ
cos . cos .
96 27 0
cos
9
2
8
1
1
5
8
5
5
9
10
F F AC
F
i j k i k
n j
n


− −
−
− − +
 
= =  
 
−
− + +
+
= =
( ) ( )
( )
1
0
1
1
ˆ ˆ
cos . cos .
144 36 0
c
1
( 12 4 2
5 0
6 )
14
os
21
5
1
1
5
0
1
9.
9 0
P P AC
P
i i
n n j k
j k


− −
−
− − +
=
− +
 
= =  
 
−
=
−
+

The force makes an angle of 600 with the line
Find the value of
2 3 Z
F i j F k
= + + 4 3 0
L i j k
= + +
Z
F
( ) ( )
( ) ( )
( ) ( )
2 2 2 2
2 2 2
2 2 2
0 2
2
2
1 1 1
ˆ 2 2
1 2 5
4 3 0
1 1 1
ˆ 4 3 0 4 3 0
5
4 3 0
1 4 6 0 2
ˆ ˆ
cos . 2 . 4 3 0
5 5 5 5 5
2 1
cos60 5 6
3
4
2
5
3. 16
z
F z z
z z
L
z
F L z
z z z
z Z
z
F i j F k
n F i j F k i j F k
F F F
L i j k
n L i j k i j k
L
n n i j F k i j k
F F
F
F
F
F

= + +
= = + + = + +
+ + +
= + +
= = + + = + +
+ +
+ +
= = + + + + = =
+ + +
= =  + = 
+
= 

CROSS PRODUCT
Cross product(VECTOR PRODUCT of two vectors is defined as
a
b
sin
a b ab 
 =

Angle θ is swept from the pre-multiplying vector to
the post-multiplying vector in the anti-clockwise
sense.
1
ˆb
n
ˆa
n
Fingers of right-hand curl in the direction of angle swept from a to b. The cross product
is in the direction of the thumb.
1 1
ˆ ˆ
sin sin
a b
a b n n
a b
 
   
 =   =
   
   

CROSS PRODUCT OF TWO VECTORS
( ) ( ) ( )
1 2 3 1 2 3
2 3 1 3 1 2
1 2 3 2 3 3 2 1 3 3 1 1 2 2 1
2 3 1 3 1 2
1 2 3
a a i a j a k b bi b j b k
i j k
a a a a a a
a b a a a i j k a b a b i a b a b j a b a b k
b b b b b b
b b b
= + + = + +
 = = − + = − − − + −
0
1 1 sin 0 0
1 1 sin90 1
i i j j k k
i j j k k i
j i i j k j j k i k k i
 =  =  =   =
 =  =  =   =
 = −   = −   = − 
( )
sin sin
b a ba ab a b
 
 = − = − = −  CROSS PRODUCT IS NOT COMMUTATIVE
( )
a b c a b a c
 + =  +  CROSS PRODUCT IS DISTRIBUTIVE
( ) ( )
a b c a b c
  =   CROSS PRODUCT IS ASSOCIATIVE

CROSS PRODUCT OF TWO VECTORS
2 3 4 3 5 7
3 4 2 4 2 3
2 3 4 2
5 7 3 7 3 5
3 5 7
a i j k b i j k
i j k
a b i j k i j k
= + + = + +
 = = − + = − +
4 3 2 5 6 7
3 2 4 2 4 3
4 3 2 33 18 39
6 7 5 7 5 6
5 6 7
a i j k b i j k
i j k
a b i j k i j k
= − + = + +
− −
 = − = − + = − − +

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