The document discusses basic concepts in engineering mechanics including systems of units, Cartesian coordinate systems, scalars and vectors, orientation and resolution of vectors, and the resultant of a system of coplanar concurrent forces. Key concepts covered are the four types of coordinate systems, representation of vectors, resolving a force into rectangular components, and calculating the single force equivalent to a system of multiple forces.
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Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Acetabularia Information For Class 9 .docxvaibhavrinwa19
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
1. 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
2. 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
7. 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
8. 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
9. 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
11. 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
12. ORIENTATION OF VECTORS
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
13. ORIENTATION OF VECTORS
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
14. ORIENTATION OF VECTORS
P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-UP
PX
PY
DRAW A HORIZONTAL LINE PASSING THROUGH THE TAIL AND
A VERTICAL LINE PASSING THROUGH THE TIP
PX acts horizontally to the left
PY acts vertically up
PX and PY are rectangular components of vector P
15. ORIENTATION OF VECTORS
P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-DOWN
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 left
PY acts vertically down
PX and PY are rectangular components of vector P
16. ORIENTATION OF VECTORS
P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING RIGHT-DOWN
PX
PY
DRAW A HORIZONTAL LINE PASSING THROUGH THE TAIL AND
A VERTICAL LINE PASSING THROUGH THE TIP
PX acts horizontally to the right
PY acts vertically down
PX and PY are rectangular components of vector P
17. 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
20. 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
= = →
= =
21. 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
= =
= =
22. RESOLUTION OF A PLANAR FORCE
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
= =
= =
23. RESOLUTION OF A PLANAR FORCE
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
= = →
= =
24. 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
25. 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
26. 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
28. 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
29. ORIENTATION OF VECTORS
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
30. ORIENTATION OF VECTORS
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
31. ORIENTATION OF VECTORS
P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-UP
PX
PY
DRAW A HORIZONTAL LINE PASSING THROUGH THE TAIL AND
A VERTICAL LINE PASSING THROUGH THE TIP
PX acts horizontally to the left
PY acts vertically up
PX and PY are rectangular components of vector P
32. ORIENTATION OF VECTORS
P IS A VECTOR OF MAGNITUDE 10kN, ACTING LEFT-DOWN
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 left
PY acts vertically down
PX and PY are rectangular components of vector P
33. ORIENTATION OF VECTORS
P
10kN
P IS A VECTOR OF MAGNITUDE 10kN, ACTING RIGHT-DOWN
PX
PY
DRAW A HORIZONTAL LINE PASSING THROUGH THE TAIL AND
A VERTICAL LINE PASSING THROUGH THE TIP
PX acts horizontally to the right
PY acts vertically down
PX and PY are rectangular components of vector P
34. 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
37. 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
= = →
= =
38. 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
= =
= =
39. RESOLUTION OF A PLANAR FORCE
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
= =
= =
40. RESOLUTION OF A PLANAR FORCE
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
= = →
= =
41. RESOLUTION OF A PLANAR FORCE
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
=
+
=
=
=
+
+
42. RESOLUTION OF A PLANAR FORCE
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
43. RESOLUTION OF A PLANAR FORCE
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
44. RESOLUTION OF A PLANAR FORCE
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
45. 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
46. 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
= =
= =
= = −
47. 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
−
= = + + + = − − + =
= = + + + = + − − = −
= + =
= =
48. 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
49. 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
= + = +
= − −
+ =
50. 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
= + = −
= − −
− =
( )
sin( ) sin cos cos sin
12 5
260 0. 4
6 0.8
13 3
2
1
2
Y
Y
F
N
P P
F
= + = +
= − −
+ =
51. NUMERICAL EXAMPLE - 4
2
2
1
1
X
Y
4
3
P
X
Y
P
X
P
Y
P
( )
( )
sin( ) sin cos cos sin
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
52. 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
53. 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
54. 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
=
+
55. 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
= + +
=
= + + = =
= =
56. 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
57. 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
58. 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
= − − = − − =
= + = + =
= + = =
= − = =
−
−
59. 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
= − + = − + =
= − = − =
= − = − =
= + =
−
−
= +
60. 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
= + = + =
=
62. 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
63. 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
64. PROPERTIES OF COUPLE
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
65. PROPERTIES OF COUPLE
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
68. RESULTANT OF NON-CONCURRENT FORCES
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
69. 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
+ + +
70. NUMERICAL EXAMPLE-10
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
71. NUMERICAL EXAMPLE-11
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
−
=
=
− + =
72. NUMERICAL EXAMPLE-12
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
73. NUMERICAL EXAMPLE-13
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
74. NUMERICAL EXAMPLE-14
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
75. NUMERICAL EXAMPLE-14
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
79. 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
80. 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
= +
81. 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
82. 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
NUMERICAL EXAMPLE-15
83. 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
84. 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
= + + = + +
85. 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
NUMERICAL EXAMPLE-16
FAB
FAC
FAD
Vertical resultant means RX=0 and RZ=0
0
0
X
Z
R X
R Z
= =
= =
86. 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
= + − =
= − + + = − +
=
=
+ =
87. 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
= + + = + +
+
+
= +
+
88. 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
89. 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 1
ˆ ˆ
. cos . cos
a b
a b n n
a b
= =
ˆa
n
ˆb
n
1
1
90. 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
= − + = + −
= − + + − = − − = −
91. NUMERICAL EXAMPLE-17
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 =
− + − − +
+
= =
− − + − + = −
92. NUMERICAL EXAMPLE-18
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
= − − +
= = − − +
+
= − +
+
−
93. NUMERICAL EXAMPLE-18
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
= −
= =
= − +
+
− −
=
−
+
94. NUMERICAL EXAMPLE-18
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
− −
−
− − +
=
− +
= =
−
=
−
+
95. NUMERICAL EXAMPLE-19
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
= + +
= = + + = + +
+ + +
= + +
= = + + = + +
+ +
+ +
= = + + + + = =
+ + +
= = + =
+
=
96. CROSS PRODUCT
Cross product(VECTOR PRODUCT of two vectors is defined as
a
b
sin
a b ab
=
a is the pre-multiplying vector
b is the post multiplying vector
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
= =
97. 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
98. 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
= − + = + +
− −
= − = − + = − − +