ME3351 ENGG. MECHANICS MLM.pdf

ME3351 ENGINEERING MECHANICS L T P C
3 0 0 3
COURSE OBJECTIVES:
1. To Learn the use scalar and vector analytical techniques for analyzing forces in statically determinate structures
2. To introduce the equilibrium of rigid bodies, vector methods and free body diagram
3. To study and understand the distributed forces, surface, loading on beam and intensity.
4. To learn the principles of friction, forces and to determine the apply the concepts of frictional forces at the
contact surfaces of various engineering systems.
5. To develop basic dynamics concepts – force, momentum, work and energy;
UNIT I STATICS OF PARTICLES 9
Fundamental Concepts and Principles, Systems of Units, Method of Problem Solutions, Statics of Particles - Forces in a
Plane, Resultant of Forces, Resolution of a Force into Components, Rectangular Components of a Force, Unit Vectors.
Equilibrium of a Particle- Newton’s First Law of Motion, Space and Free-Body Diagrams, Forces in Space, Equilibrium
of a Particle in Space.
UNIT II EQUILIBRIUM OF RIGID BODIES 9
Principle of Transmissibility, Equivalent Forces, Vector Product of Two Vectors, Moment of a Force about a Point,
Varignon’s Theorem, Rectangular Components of the Moment of a Force, Scalar Product of Two Vectors, Mixed Triple
Product of Three Vectors, Moment of a Force about an Axis, Couple - Moment of a Couple, Equivalent Couples, Addition
of Couples, Resolution of a Given Force into a Force -Couple system, Further Reduction of a System of Forces,
Equilibrium in Two and Three Dimensions - Reactions at Supports and Connections.
UNIT III DISTRIBUTED FORCES 9
Centroids of lines and areas – symmetrical and unsymmetrical shapes, Determination of Centroids by Integration,
Theorems of Pappus-Guldinus, Distributed Loads on Beams, Centre of Gravity of a Three-Dimensional Body, Centroid
of a Volume, Composite Bodies, Determination of Centroids of Volumes by Integration. Moments of Inertia of Areas and
Mass - Determination of the Moment of Inertia of an Area by Integration, Polar Moment of Inertia, Radius of Gyration
of an Area, Parallel-Axis Theorem, Moments of Inertia of Composite Areas, Moments of Inertia of a Mass - Moments of
Inertia of Thin Plates, Determination of the Moment of Inertia of a Three-Dimensional Body by Integration.
UNIT IV FRICTION 9
The Laws of Dry Friction, Coefficients of Friction, Angles of Friction, Wedge friction, Wheel Friction, Rolling
Resistance, Ladder friction.
UNIT V DYNAMICS OF PARTICLES 9
Kinematics - Rectilinear Motion and Curvilinear Motion of Particles. Kinetics- Newton’s Second Law of
Motion -Equations of Motions, Dynamic Equilibrium, Energy and Momentum Methods - Work of a Force,
Kinetic Energy of a Particle, Principle of Work and Energy, Principle of Impulse and Momentum, Impact of
bodies.
TOTAL: 45 PERIODS
OUTCOMES:
At the end of the course the students would be able to
1. Illustrate the vector and scalar representation of forces and moments
2. Analyze the rigid body in equilibrium
3. Evaluate the properties of distributed forces
4. Determine the friction and the effects by the laws of friction
5. Calculate dynamic forces exerted in rigid body
TEXTBOOKS:
1. Beer Ferdinand P, Russel Johnston Jr., David F Mazurek, Philip J Cornwell, Sanjeev Sanghi, Vector Mechanics for
Engineers: Statics and Dynamics, McGraw Higher Education., 12thEdition, 2019.
2. Vela Murali, “Engineering Mechanics-Statics and Dynamics”, Oxford University Press, 2018.
REFERENCES:
1. Boresi P and Schmidt J, Engineering Mechanics: Statics and Dynamics, 1/e, Cengage learning, 2008.
2. Hibbeller, R.C., Engineering Mechanics: Statics, and Engineering Mechanics: Dynamics, 13th edition, Prentice Hall,
2013.
3. Irving H. Shames, Krishna Mohana Rao G, Engineering Mechanics – Statics and Dynamics, 4thEdition, Pearson
Education Asia Pvt. Ltd., 2005.
4. Meriam J L and Kraige L G, Engineering Mechanics: Statics and Engineering Mechanics: Dynamics, 7th edition, Wiley
student edition, 2013.
5. Timoshenko S, Young D H, Rao J V and SukumarPati, Engineering Mechanics, 5thEdition, McGraw Hill Higher
Education, 2013.

ME3351 – ENGINEERING MECHANICS
PART-A QUESTIONS WITH ANSWERS (TWO MARKS)
UNIT-I Statics of Particles
1. Define Engineering Mechanics
Engineering Mechanics is defined as the branch of physical science which deals with the
behavior of a body at rest or motion under the action of forces.
2. What are the branches of Engineering Mechanics?
Rigid body Mechanics
Deformable body mechanics (also called strength of materials)
Fluid Mechanics
3. What are the branches of rigid body Mechanics?
Statics
Dynamics
4. Define statics
Statics is defined as the branch of rigid body mechanics, which deals with the behavior of a
body when it is at rest.
5. Define Dynamics
Dynamics is defined as the branch of rigid body mechanics which deals with the behavior of
a body when it is in motion.
6. Distinguish between particle and rigid body
Particle Rigid Body
Mass is idealized to be located at a
point.
Size and shape of the body are influential.
Forces are idealized to be located at a
point.
Locations of forces are important. Forces
create moments acting on the body.
Forces maintain translational
equilibrium.
Forces maintain translational and
rotational equilibrium.
7. Define Force.
Force is defined as an agent which changes or tends to change the state of rest or of uniform
motion of a body. It represents the push or pull exerted by one body on another. It is a vector
quantity.
8. What are the characteristics of a force?
Magnitude
Line of action
Direction & angle of inclination
9. State Newton's laws of motion
Newton's first law: Everybody preserves in its state of rest, or of uniform motion in a straight
line, unless it is compelled to change that state by forces impressed there on.
Newton's second law: The acceleration of a particle will be proportional to the force and will
be in the direction of the force (ie. F = ma)
Newton's third law: To every action there is an equal and opposite reaction.
10. What is collinear force system?
Force acts on a common line of action.
11. What is like parallel forces?

The parallel force which acts in the same direction are called like parallel forces.
12. What is unlike parallel forces?
The parallel force which acts in the opposite direction are called unlike parallel forces.
13. What is coplanar force system?
In coplanar force system, lines of action of all forces lie on a single plane.
14. What is Non-coplanar (or spatial) force system?
In Non-coplanar (or spatial) force system, lines of action of al forces lie on different planes
15. What is collinear force system?
In collinear force system, all the forces lie on a single line.
16. What is concurrent force system?
In concurrent force system, lines action of all forces intersects at a point.
17. What is parallel force system?
In parallel force system, lines of action of all forces are parallel to each other.
18. A vector 𝑭
⃗
⃗ starts at point (2,-1, 2) and passes through the point (-1, 3, 5). Find its unit
vector.
Given: Start point 𝑂𝐴
⃗⃗⃗⃗⃗⃗ = 2𝑖 − 1𝑗 + 2𝑘
⃗ , Final point 𝑂𝐵
⃗⃗⃗⃗⃗ = −1𝑖 + 31𝑗 + 5𝑘
⃗
𝐴𝐵
⃗⃗⃗⃗⃗⃗ = 𝑂𝐵
⃗⃗⃗⃗⃗⃗⃗ − 𝑂𝐴
⃗⃗⃗⃗⃗⃗ = −3𝑖 + 4𝑗 + 3𝑘
⃗ ,
𝐴𝐵 = √32 + 42 + 32 = √34 = 5.831
Unit vector 𝑛
∧
=
𝐴𝐵
⃗⃗⃗⃗⃗
𝐴𝐵
=
−3𝑖+4𝑗+3𝑘
⃗
5.831
= −0.514𝑖 + 0.686𝑗 + 0.686𝑘
⃗
19. State the principle of transmissibility.
The state of rest of motion of a rigid body is unaltered if a force acting in the body is
replaced by another force of the same magnitude and direction but acting anywhere on the
body along the line of action of the replaced force.
For example the force F acting on a rigid body at point A. According to the principle of
transmissibility of forces, this force has the same effect on the body as a force F applied at
point B.
20. Give the static equilibrium equations.
For particle, +
In two dimensions or plane:
∑ 𝑭𝒙 = 𝟎; ∑ 𝑭𝒚 = 𝟎
In three dimensions or plane:
∑ 𝑭𝒙 = 𝟎; ∑ 𝑭𝒚 = 𝟎; ∑ 𝑭𝒛 = 𝟎
For rigid body,
In two dimensions or plane:
∑ 𝑭𝒙 = 𝟎; ∑ 𝑭𝒚 = 𝟎 ; ∑ 𝑴𝒛 = 𝟎
In three dimensions or plane:
∑ 𝑭𝒙 = 𝟎; ∑ 𝑭𝒚 = 𝟎; ∑ 𝑭𝒛 = 𝟎 ; ∑ 𝑴𝒙 = 𝟎; ∑ 𝑴𝒚 = 𝟎; ∑ 𝑴𝒛 = 𝟎

21. Find the unit vector of a force 𝑭
⃗⃗⃗ = 𝟒𝒊 − 𝟓𝒋 + 𝟖𝒌
⃗
⃗ .
𝐹𝑜𝑟𝑐𝑒𝐹 = 4𝑖 − 5𝑗 + 8𝑘
⃗ , 𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝐹 = √42 + 52 + 82 = √105 = 10.25,
Unit vector 𝑛𝐹
∧
=
𝐹
𝐹
=
4𝑖−5𝑗+8𝑘
⃗
10.25
= 𝟎. 𝟑𝟗𝒊 − 𝟎. 𝟒𝟖𝟖𝒋 + 𝟎. 𝟕𝟖𝒌
⃗
⃗
22. Define Lami’s theorem.
It states that, "If three coplanar forces acting at a point be in equilibrium, then each force is
proportional to the sine of the angle between the other two forces”.
23. What is coplanar concurrent force system?
In coplanar concurrent force system, lines of action of all forces are meet at a single point
and lie on a single plane.
24. A rigid body is in equilibrium under the action of three coplanar forces. State the
condition of forces.
If a body is in equilibrium acted upon by three forces, then the resultant of any two forces
must be equal, opposite and collinear with the third force.
In other words, if body is in equilibrium under the action of three forces, lines of action of
three forces must be concurrent and resultant of these forces is zero.
25. What is the inclination of a force with respect to the x-axis if it is inclined at 60o
with
the y-axis and 30o
with the z-axis?
Given: θy =60o
, θz = 30o
w.k.t. 𝑐𝑜𝑠2
𝜃𝑥 + 𝑐𝑜𝑠2
𝜃𝑦 + 𝑐𝑜𝑠2
𝜃𝑧 = 1
𝑐𝑜𝑠2
𝜃𝑥 = 1 − (𝑐𝑜𝑠2
𝜃𝑦 + 𝑐𝑜𝑠2
𝜃𝑧) = 1 − (𝑐𝑜𝑠2
60𝑜
+ 𝑐𝑜𝑠2
30𝑜
)
𝑐𝑜𝑠2
𝜃𝑥 = 0, 𝜃𝑥 = 𝑐𝑜𝑠−1(0) = 𝟗𝟎𝒐
26. Find the magnitude of the resultant of the two concurrent forces of magnitude of 60kN
and 40kN with an included angle of 70o
between them.
Given: P = 60kN, Q = 40kN, θ = 70o
W.k.t. Resultant 𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝑐𝑜𝑠 𝜃
𝑅 = √(602 + 402 + 2 × 60 × 40 𝑐𝑜𝑠 70𝑂) = √6841.7 = 𝟖𝟐. 𝟕𝟏𝒌𝑵
P
𝜸 β
α
Q R
𝑷
𝐬𝐢𝐧 𝜶
=
𝑸
𝐬𝐢𝐧 𝜷
=
𝑹
𝐬𝐢𝐧 𝜸
= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
O
S
R
Q
P

27. Find the length of the line joining the origin with a point (2, 1, 2) .
Given: position vector OA = (2, 1, 2)
𝑂𝐴
⃗⃗⃗⃗⃗ = (2𝑖 + 1𝑗 + 2𝑘
⃗ ), 𝐹 = √22 + 12 + 22 = √9 = 𝟑 𝒖𝒏𝒊𝒕𝒔.
28. State polygon law of equilibrium.
It states that "If a number of forces acting simultaneously on a particle be represented in
magnitude and direction, by the sides of a polygon taken in order and particle is said to be in
equilibrium, then all these forces are form the complete polygon".
29. A force of magnitude 500N is passing through the origin and a point A (0.2, 1, 0)m.
write the vector form of the force.
Given: F = 500N, starting point OO is (0, 0, 0) m, end point A (0.2, 1, 0) m
OA = (0.2, 1, 0)𝑂𝐴
⃗⃗⃗⃗⃗ = 0.2𝑖 + 1𝑗 + 0𝑘
⃗ in m,
𝑂𝐴 = √0.22 + 12 + 02 = √1.04 = 1.02 𝑚
Unit vector 𝑛𝑂𝐴
∧
=
𝑂𝐴
⃗⃗⃗⃗⃗⃗
𝑂𝐴
=
0.2𝑖+1𝑗+0𝑘
⃗
1.02
= 𝟎. 𝟏𝟗𝟔𝒊 − 𝟎. 𝟗𝟖𝒋 + 𝟎𝒌
⃗
⃗
Force vector
𝐹𝑂𝐴
⃗⃗⃗⃗⃗⃗ = 𝐹. 𝑛𝑂𝐴
∧
= 500 × (0.196𝑖 + 0.98𝑗 + 0𝑘
⃗ )
𝑭𝑶𝑨
⃗⃗⃗⃗⃗⃗⃗ = 𝟗𝟖𝒊 + 𝟒𝟗𝟎𝒋 + 𝟎𝒌
⃗
⃗ in N.
30. A force𝑭
⃗
⃗ = 𝟖. 𝟐𝟓𝒊 + 𝟏𝟐. 𝟕𝟓𝒋 − 𝟏𝟖𝒌
⃗
⃗ N acts through the origin. What is the magnitude of
the force and the angles it makes with x-axis?
Given: 𝐹 = 8.25𝑖 + 12.75𝑗 − 18𝑘
⃗ N
𝐹 = √8.252 + 12.752 + (−18)2 = √554.625 = 23.55𝑁
𝐹 = 𝐹 (𝑐𝑜𝑠 𝜃𝑥 𝑖 + 𝑐𝑜𝑠 𝜃𝑦 𝑗 + 𝑐𝑜𝑠 𝜃𝑧 𝑘
⃗ ) N
𝑐𝑜𝑠 𝜃𝑥 𝑖 + 𝑐𝑜𝑠 𝜃𝑦 𝑗 + 𝑐𝑜𝑠 𝜃𝑧 𝑘
⃗ =
𝐹
𝐹
𝑐𝑜𝑠 𝜃𝑥 𝑖 + 𝑐𝑜𝑠 𝜃𝑦 𝑗 + 𝑐𝑜𝑠 𝜃𝑧 𝑘
⃗ = 0.35𝑖 + 0.541𝑗 − 0.764𝑘
⃗
Equating and separating coefficient of 𝑖
⃗ , 𝑗and 𝑘
⃗ , we get,
𝑐𝑜𝑠 𝜃𝑥 = 0.35, 𝑐𝑜𝑠 𝜃𝑦 = 0.541, 𝑐𝑜𝑠 𝜃𝑧 = −0.764
𝜃𝑥 = 𝑐𝑜𝑠−1
(0.35), 𝜃𝑦 = 𝑐𝑜𝑠−1
(0.541) , 𝜃𝑧 = 𝑐𝑜𝑠−1
(−0.764)
𝜽𝒙 = 𝟔𝟗. 𝟓𝟏𝒐
, 𝜽𝒚 = 𝟓𝟕. 𝟐𝟓𝒐
, 𝜽𝒛 = 𝟏𝟑𝟗. 𝟖𝟐𝒐
31. The resultant of two equal forces including 60o
between them has a magnitude of 𝟗√𝟑 𝑵.
What is the magnitude of the component forces?
Given: equal forces P =Q, θ = 60o
, Resultant R = 9√3 𝑁
w.k.t. parallelogram law of forces
𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝑐𝑜𝑠 𝜃
9√3 = √𝑃2 + 𝑃2 + 2𝑃2 𝑐𝑜𝑠 60𝑜 ⇒ 9√3 = √3𝑃2 → 𝑃 = 9 𝑁
F1
F2
F3
F4
F5

32. Three coplanar forces of magnitude P in Newton each are acting on a particle. If their
lines of action make equal angle with each other, show that the forces are in
equilibrium.
33. A space force of 120N is directed from point A (0, -4, 6) towards another point B (8, 5, -
3). What is the force vector and the corresponding unit vector?
Given: F = 120N, starting point OA is (0, -4, 6) m, end point B (8, 5, -3) m
OA = (0, -4, 6) m 𝑂𝐴
⃗⃗⃗⃗⃗ = 0𝑖 − 4𝑗 + 6𝑘
⃗ in m,
OB = (8, 5, -3) m 𝑂𝐵
⃗⃗⃗⃗⃗ = 8𝑖 + 5𝑗 − 3𝑘
⃗ in m,
𝐴𝐵
⃗⃗⃗⃗⃗ = 0𝐵
⃗⃗⃗⃗⃗ − 𝑂𝐴
⃗⃗⃗⃗⃗ = 8𝑖 + 9𝑗 − 9𝑘
⃗ in m,
𝐴𝐵 = √82 + 92 + (−9)2 = √226 = 15.03 𝑚
Unit vector 𝑛
∧
𝐴𝐵 =
𝐴𝐵
⃗⃗⃗⃗⃗
𝐴𝐵
=
8𝑖+9𝑗−9𝑘
⃗
15.03
= 𝟎. 𝟓𝟑𝟐𝒊 + 𝟎. 𝟓𝟗𝟗𝒋 − 𝟎. 𝟓𝟗𝟗𝒌
⃗
⃗ in m
Force vector
𝐹𝐴𝐵 = 𝐹. 𝑛
∧
𝐴𝐵 = 120 × (0.532𝑖 + 0.599𝑗 + 0.599𝑘
⃗ )
𝑭
⃗
⃗ 𝑨𝑩 = 𝟔𝟑. 𝟖𝟒𝒊 + 𝟕𝟏. 𝟖𝟖𝒋 − 𝟕𝟏. 𝟖𝟖𝟎𝒌
⃗
⃗ in N.
34. The sum of two concurrent forces P and Q is 300N and their resultant is 200N. The
angle between the force P and resultant is 90o
. Find the magnitude of each force.
Given: P + Q = 300 N…….. (1), Resultant R = 200 N, α =90o
,
𝛼 = 𝑡𝑎𝑛−1
(
𝑄 𝑠𝑖𝑛 𝜃
𝑃 + 𝑄 𝑐𝑜𝑠 𝜃
) 𝑡𝑎𝑛 900
= (
𝑃 + 𝑄 𝑐𝑜𝑠 𝜃
)
𝑃 + 𝑄 𝐶𝑜𝑠 𝜃 = 0, → 𝑃 = − 𝑄 𝑐𝑜𝑠 𝜃………….(3)
Substitute (3) in (2) we get, (1) → 𝑄2
− 𝑃2
= 40000 …………..(4)
Substitute (1) in (4) we get, 𝑄 − 𝑃 = 133.33……..(5)
P
α α
α
P P
𝐏
𝐬𝐢𝐧 𝛂
=
𝐏
𝐬𝐢𝐧 𝛂
=
𝐏
𝐬𝐢𝐧 𝛂
= 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭
Hence it is satisfied the Lami’s
theorem, the particle is in equilibrium.
R
Q
P
α =90o
θ
W.k.t. parallelogram law of forces,
𝑅2
= 𝑃2
+ 𝑄2
+ 2𝑃𝑄 𝑐𝑜𝑠 𝜃
⇒ 2002
= 𝑃2
+ 𝑄2
+ 2𝑃𝑄 𝑐𝑜𝑠 𝜃 … … … … . (2)

Solve (1) & (5) we get, Q= 216.67 N and P = 83.33 N
35. A force of magnitude 700N is directed along PQ where P is (0.8, 0, 1.2) m and Q is
(1.4, 1.2, 0) m. writes the vector form of the force.
Given: F = 700 N, starting point OA is (0.8, 0, 1.2) m, end point B (1.4, 1.2, 0) m
OP = (0.8, 0, 1.2) m 𝑂𝑃
⃗⃗⃗⃗⃗ = 0.8𝑖 + 0𝑗 + 1.2𝑘
⃗ in m,OQ = (1.4, 1.2, 0) m 𝑂𝑄
⃗⃗⃗⃗⃗⃗ = 1.4𝑖 + 1.2𝑗 + 0𝑘
⃗ in m,
𝑃𝑄
⃗⃗⃗⃗⃗ = 0𝑄
⃗⃗⃗⃗⃗ − 𝑂𝑃
⃗⃗⃗⃗⃗ = 0.6𝑖 + 1.2𝑗 − 1.2𝑘
⃗ in m,
𝑃𝑄 = √0.62 + 1.22 + (−1.2)2 = √3.24 = 1.8 𝑚
Unit vector 𝑛
∧
𝑃𝑄 =
𝑃𝑄
⃗⃗⃗⃗⃗
𝑃𝑄
=
0.6𝑖+1.2𝑗−1.2𝑘
⃗
1.8
= 0.333𝑖 + 0.667𝑗 − 0.667𝑘
⃗
Force vector𝐹𝑃𝑄 = 𝐹. 𝑛
∧
𝑃𝑄 = 700 × (0.333𝑖 + 0.667𝑗 − 0.667𝑘
⃗ )
𝑭
⃗
⃗ 𝑷𝑸 = 𝟐𝟑𝟑. 𝟏𝒊 + 𝟒𝟔𝟔. 𝟗𝒋 − 𝟒𝟔𝟔. 𝟗𝒌
⃗
⃗ in N.
36. A force F has the components Fx = 20N, Fy = -30N, Fz = 60N. Find the angle ‘θy’ it forms with
the co-ordinates axes y.
Given: Fx = 20N, Fy = -30N, Fz = 60N
𝐹 = 𝐹𝑥𝑖 + 𝐹𝑦𝑗 + 𝐹𝑧𝑘
⃗ = 20𝑖 − 30𝑗 + 60𝑘
⃗ inN
𝐹 = √202 + (−30)2 + 602 = √49 = 70 𝑁
𝐹 = 𝐹 (𝑐𝑜𝑠 𝜃𝑥 𝑖 + 𝑐𝑜𝑠 𝜃𝑦 𝑗 + 𝑐𝑜𝑠 𝜃𝑧 𝑘
⃗ )
20𝑖 − 30𝑗 + 60𝑘
⃗ = 70 (𝑐𝑜𝑠 𝜃𝑥 𝑖 + 𝑐𝑜𝑠 𝜃𝑦 𝑗 + 𝑐𝑜𝑠 𝜃𝑧 𝑘
⃗ )
Equating and separating coefficient of 𝑖
⃗ , 𝑗and 𝑘
⃗ , we get,
𝑐𝑜𝑠 𝜃𝑥 = 0.286, 𝑐𝑜𝑠 𝜃𝑦 = −0.428, 𝑐𝑜𝑠 𝜃𝑧 = 0.857
𝜃𝑥 = 𝑐𝑜𝑠−1
(0.286), 𝜃𝑦 = 𝑐𝑜𝑠−1
(−0.428) , 𝜃𝑧 = 𝑐𝑜𝑠−1
(0.857)
𝜽𝒙 = 𝟕𝟑. 𝟒𝒐
, 𝜽𝒚 = 𝟏𝟏𝟓. 𝟑𝟖𝒐
, 𝜽𝒛 = 𝟑𝟏𝒐
37. Using lame’s theorems calculate the forces in the member CA and CB for the system shown in
figure below.
Applying lame’s theorem
𝐹𝐶𝐴
𝑠𝑖𝑛 110𝑜
=
𝐹𝐶𝐵
=
𝑊
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐹𝐶𝐴
=
𝐹𝐶𝐵
=
1962
𝑭𝑪𝑨 =
𝟏𝟗𝟔𝟐
𝒔𝒊𝒏 𝟏𝟐𝟎𝒐
𝒔𝒊𝒏 𝟏𝟏𝟎𝒐
= 𝟐𝟏𝟐𝟖. 𝟖𝟗 𝑵
𝑭𝑪𝑩 =
𝟏𝟗𝟔𝟐
𝒔𝒊𝒏 𝟏𝟐𝟎𝒐
𝒔𝒊𝒏 𝟏𝟑𝟎𝒐
= 𝟏𝟕𝟑𝟓. 𝟒𝟗 𝑵
130o
110o
200 kg
40o
C
A B
20o
W= 1962 N
C
FA FB
120o
Free Body Diagram

38. State the difference between internal and external forces.
External forces:
The forces which represent the action of other bodies on the rigid body considered and which
are responsible for the external behavior of the rigid body are called as 'External forces'.
Internal forces:
The forces which hold together the forming the rigid body or holding the component parts
together are called as internal forces.
39. Define resultant force?
Resultant force is a single equivalent force which can replace the given force system for an
equivalence of effect
40. State the principle of resolution?
The algebraic sum of the resolved parts of a number of forces in a given direction is equal to
the resolved part of their resultant in the same direction of their resultant and in the same
direction.
41. Two wires are attached to a bolt in the foundation as shown in figure below. Determine the
pull exerted by the bolt on the foundation.
w.k.t. parallelogram law of forces
Resultant 𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝑐𝑜𝑠 𝜃 and 𝛼 = 𝑡𝑎𝑛−1
(
𝑃+𝑄 𝑐𝑜𝑠 𝜃
)
𝑅 = √66502 + 36002 + 2 × 6650 × 3600 𝑐𝑜𝑠 140𝑜 = √20504292.06 = 𝟒𝟓𝟐𝟖. 𝟏𝟕 𝑵
𝛼 = 𝑡𝑎𝑛−1
(
3600 𝑠𝑖𝑛 140𝑜
6650 + 3600 𝑐𝑜𝑠 140𝑜
) = 𝟑𝟎. 𝟕𝟑𝒐
42. Define equilibrium?
A body is said to be in a state of equilibrium, if the body is either at rest or is moving at a
constant velocity.
43. What is two force equilibrium principles?
If a body is in equilibrium acted upon by two forces they must be of collinear forces of equal
magnitude and opposite sense.
44. What is three force equilibrium principles?
If a body is in equilibrium acted upon by three forces, then the resultant of any two forces
must be equal, opposite and collinear with the third force.
45. What is four force equilibrium principles?
If a body is in equilibrium, acted upon by four forces, then the resultant of any two forces
must be equal, opposite and collinear with the resultant of the other two
46. State parallelogram law of forces?
It states that "If two forces acting simultaneously on a particle be represented in magnitude
and direction by the two adjacent sides of a parallelogram their resultant may be represented
140o
25o
15o
3600 N
6650 N
Bolt
Foundation
Q = 3600 N
P =6650 N
Free Body Diagram

magnitude and direction by the diagonal of the parallelogram which passes through their
point of intersection.
Resultant 𝑅 = √𝑃2 + 𝑄2 + 2𝑃𝑄 𝑐𝑜𝑠 𝜃 𝛼 = 𝑡𝑎𝑛−1
(
𝑃+𝑄 𝑐𝑜𝑠 𝜃
)
47. Two force P =100kN and Q = 200kN act at the origin. P is directed towards a point (-2, 3, -
5) metres and Q towards (6, -8, -4) metres. What is the resultant and corresponding unit
vector?
Given: P = 100kN, Q = 200kN
𝑭
⃗
⃗ 𝑶𝑷 = −𝟑𝟐. 𝟒𝒊 + 𝟒𝟖. 𝟕𝒋 − 𝟖𝟏. 𝟏𝒌
⃗
⃗ in kN.
OQ = (6, -8, -4) m 𝑂𝑄
⃗⃗⃗⃗⃗⃗ = 6𝑖 − 8𝑗 − 4𝑘
⃗ in m,
𝑂𝑄 = √62 + (−8)2 + (−4)2 = √116 = 10.77 𝑚
Unit vector 𝑛
∧
𝑂𝑄 =
𝑂𝑄
⃗⃗⃗⃗⃗⃗
𝑂𝑄
=
6𝑖−8𝑗−4𝑘
⃗
10.77
= 0.557𝑖 − 0.743𝑗 − 0.371𝑘
⃗ in m
Force vector: 𝐹𝑂𝑄 = 𝐹. 𝑛
∧
𝑂𝑄 = 200 × (0.557𝑖 − 0.743𝑗 − 0.371𝑘
⃗ )
𝑭
⃗
⃗ 𝑶𝑸 = 𝟏𝟏𝟏. 𝟒𝒊 − 𝟏𝟒𝟖. 𝟔𝒋 − 𝟕𝟒. 𝟐𝒌
⃗
⃗ in kN.
Resultant𝑹
⃗⃗ = 𝑭
⃗
⃗ 𝑶𝑷 + 𝑭
⃗
⃗ 𝑶𝑸 = 𝟕𝟗𝒊 − 𝟗𝟗. 𝟗𝒋 − 𝟏𝟓𝟓. 𝟑𝒌
⃗
⃗ in kN.
𝑅 = √792 + (−99.9)2 + (−155.3)2 = √30359.09 = 174.24 𝑘𝑁
Unit vector 𝑛
∧
𝑅 =
𝑅
⃗
𝑅
=
𝟕𝟗𝒊−𝟗𝟗.𝟗𝒋−𝟏𝟓𝟓.𝟑𝒌
⃗
⃗
174.24
= 𝟎. 𝟒𝟓𝟑𝒊 − 𝟎. 𝟓𝟕𝟑𝒋 − 𝟎. 𝟖𝟗𝟏𝒌
⃗
⃗
48. State triangle law of forces?
It states that "If two forces acting simultaneously on a particle represented in magnitude and
direction by the two side’s triangle, taken in order, their resultant may be represented
magnitude and direction by the third side of the triangle, taken opposite order".
49. State polygon law of forces?
It states that "If a number of forces acting simultaneously on a particle be represented in
magnitude and direction, by the sides of a polygon taken in order, then the resultant of all
Q = 200 kN
(6, -8, -4) m
O
P = 100 kN
(-2, 3, -5) m
OP = (-2, 3, -5) m 𝑂𝑃
⃗⃗⃗⃗⃗ = −2𝑖 + 3𝑗 − 5𝑘
⃗ in m,
𝑂𝑃 = √(−2)2 + 32 + (−5)2 = √38 = 6.164 𝑚
Unit vector 𝑛
∧
𝑂𝑃 =
𝑂𝑃
⃗⃗⃗⃗⃗
𝑂𝑃
=
−2𝑖+3𝑗−5𝑘
⃗
6.164
= −0.324𝑖 + 0.487𝑗 − 0.811𝑘
⃗
Force vector: 𝐹𝑂𝑃 = 𝐹. 𝑛
∧
𝑂𝑃 = 100 × (−0.324𝑖 + 0.487𝑗 − 0.811𝑘
⃗ )

these forces may be represented in magnitude and direction, by the closing side of the
polygon, taken in opposite order".
50. What are the three equations of equilibrium?
∑H=0⇒the algebraic sum of the horizontal for must be zero. ie., sum of the left hand side
forces must be equal to sum of the right hand side forces.
∑V=0⇒The algebraic sum of the vertical forces must be zero. ie. Sum of the upward forces
must be equal to sum of the downward forces
∑M=0⇒The algebraic sum of the moments about a point must be zero ie., sum of the
clockwise moments about a point must be equal to sum of the anticlockwise moments about
the same Point.
51. What is stable equilibrium?
A body is said to be in stable equilibrium, if it returns back to its original position after it is
slightly displaced from its position of rest.
52. What is unstable equilibrium?
A body is said to be in unstable equilibrium, if it does not return back to its original position
and heels farther away after slightly displaced from its position of rest.
53. What is neutral equilibrium?
A body is said to be in neutral equilibrium, if it occupies a new position (also remains at rest)
after slightly displaced from its position of rest.
54. What is free body diagram?
It is a sketch of the particle which represents it as being isolated from its surroundings. It
represents all the forces acting on it.

UNIT – II Equilibrium of Rigid bodies
1. List the different supports used to support structural components.
 Fixed support
 Roller support
 Simply support or knife edge support
 Hinged support
2. Define the term couple?
A couple is that two forces are of equal magnitude opposite sensed parallel forces, which lie
in the same plane.
3. What are the characteristics of a couple?
The algebraic sum of the forces is zero.
The algebraic sum of the moments of the forces about any point is the same and equal to the
moment of the couple itself.
4. State Varignon's theorem?
Varignon’s theorem: if a number of coplanar forces are acting simultaneously on a body, the
algebraic sum of the moments of all the forces about any point is equal to the moment of the
resultant force about the same point.
5. Define moment of a force?
The moment of a force about a point is defined as the turning effect of the force about that
point.
Moment = Force X Perpendicular distance
6. For what condition the moment of a force will be zero?
A force produces zero moment about an axis or reference point which intersects the line of
action of the force.
7. What is the difference between a moment and a couple?
The couple is a pure turning effect which may be moved anywhere in its own plane, or into a
parallel plane without change of its effect on the body, but the moment of a force must include
a description of the reference axis about which the moment is taken.
8. What is the difference between a fixed vector and a free vector?
A force which is applied at a particular location on a body is a fixed vector.
Example: A moment.
A force which can be moved anywhere in its own plane or in a parallel plane without change
in its effect on the body is called free vector.
Example: A couple.
9. What are the common types of supports used in two dimensions?
 Roller support
 Hinged support
 Fixed support
10. What are the common types of supports used in three dimensions?
 Ball support
 Ball and Socket support
 Fixed (or Welded) support
11. Define equilibrant?
The force which brings the system of forces into equilibrium is called equilibrant. It is equal
to the resultant force in magnitude collinear but opposite in nature.
12. What are the common types of loads?
 Point load (or concentrated load)
 Uniformly distributed load
 Uniformly varying load

13. What is statically determinate structure?
A structure which can be completely analysed by static conditions of equilibrium (∑H=0;
∑V =0 and ∑M=0) alone is statically' determinate structure.
14. Find the magnitude and location of the single equivalent force for a beam AB of length
8m having a point C 3m from A subjected to the following forces:
(a) An upward force of 10N at A,
(b) A downward force of 10N at C,
(c) An upward force of 40N at B
Given:
∑ 𝐹 = 10 − 10 + 40 = 40 N = R (resultant)
Take moment about a point A,
𝑀𝐴 = (10 × 0) − (10 × 3) + (40 × 8) = 290 𝑁𝑚
𝑀𝐴 = 𝑅 × 𝑟 → 40 × 𝑟 = 290
𝑟 =
290
40
= 7.25𝑚,
Single force (Resultant) R=40N, act on a point 7.25 m from point A.
15. State Varignon’s theorem.
It states that “the moment about any point of the resultant of several concurrent forces is
equal to the sum of the moments of various forces about the same point”.
16. Write the equations of equilibrium of a rigid body in two dimensions. (or) What are
the equilibrium conditions that are to be satisfied for a rigid body applied with a
system of coplanar, non-concurrent forces?
 The algebraic sum of the horizontal forces must be zero. (H=0)
 The algebraic sum of the vertical forces must be zero. (V=0)
 The algebraic sum of the moments about a point must be zero (M=0).
17. .What are the reactions at fixed support of a plane beam that are possible?
 Horizontal reaction forces,
 Vertical reaction forces, and
 Moment at support
3 m
10 N 10 N
B
C
40 N
A
8 m
3 m B
C
40 N
A
7.25 m

18. What is meant by force-couple system?
Force acting at point ‘A’ is shifted to another point ‘O’. Hence to transfer a force F acting
on a rigid body to an arbitrary point ‘O’, a couple must be added whose moment is equal to
the moment of force about O.
19. Write the condition of equilibrium of a system of parallel forces acting in a plane.
 The algebraic sum of the parallel forces must be zero. (H=0)
 The algebraic sum of the moments about a point must be zero (M=0).
20. Find the moment of the force 15N acting along the positive direction of x-axis about the
point A (2, 3) m.
Given: 𝐹 = 15𝑖 in N, acts at a point A (2, 3).
𝑂𝐴
⃗⃗⃗⃗⃗ = 2𝑖 + 3𝑗 + 0𝑘
⃗
Moment of force 𝐹 about point A𝑀
⃗⃗ 𝐴 = 𝑂𝐴
⃗⃗⃗⃗⃗ × 𝐹 = |
𝑖 𝑗 𝑘
⃗
2 3 0
15 0 0
| = −45𝑘
⃗ Nm.
21. Distinguish between space diagram and free body diagram.
Space Diagram is a graphical representation of the system. It generally shows the shape
and size of the system, the weight, the externally applied loads, the connection and the
supports of the system.
Free body diagram is a sketch of the isolated or free body which shows all the pertinent
weight forces, the externally applied loads, the reaction from its supports and the
connections acting upon it by the removed elements.
22. Two forces 𝑭𝟏
⃗⃗⃗⃗ = 𝟓𝒊 and𝑭𝟐
⃗⃗⃗⃗ = 𝟖. 𝟔𝟔𝒋 pass through a point whose co-ordinates are (2, 1).
Calculate the moment of the force about the origin.
Given: 𝐹1
⃗⃗⃗ = 5𝑖 , 𝐹2
⃗⃗⃗ = 8.66𝑗, 𝐹 = 𝐹1
⃗⃗⃗ + 𝐹2
⃗⃗⃗ = 5𝑖 + 8.66𝑗
Position vector 𝑂𝐴
⃗⃗⃗⃗⃗ = 2𝑖 + 1𝑗 + 0𝑘
⃗
Moment of force 𝐹 about origin 𝑀
⃗⃗ 𝑂 = 𝑂𝐴
⃗⃗⃗⃗⃗ × 𝐹 = |
𝑖 𝑗 𝑘
⃗
2 1 0
5 8.66 0
| = 12.32𝑘
⃗ Nm.
23. Three couples +12 Nm, -35 Nm and +100 Nm are acting in the xy, yz and xz. Write the
vector form.
Given: couple +12Nm in xy plane, -35Nm in yz plane, +100Nm
Cz =+12Nm, Cx=-35Nm, Cy =+100Nm
𝐶 = 𝐶𝑥𝑖 + 𝐶𝑦𝑗 + 𝐶𝑧𝑘
⃗ = 𝟏𝟐𝒊 − 𝟑𝟓𝒋 + 𝟏𝟎𝟎𝒌
⃗
⃗ in Nm.
O
A
𝐹
O
A
𝐹
𝐹
𝐹
O
Mo =F x d
𝐹

24. In the beam shown in figure below, find the value of ‘W’ if support reaction at C is
11N upwards.
Given:
𝐹𝑜𝑟𝑐𝑒 𝐹𝐵 = 10 𝑁, 𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑅𝐶 = 11𝑁
Condition of equilibrium of rigid bodies, ∑ 𝐹 = 0, ∑ 𝑀 = 0;
∑ 𝐹 ⟹ 𝑅𝐴 + 𝑅𝐶 − 𝐹𝐵 + 𝑊 = 0,
⟹ 𝑅𝐴 + 11 − 10 + 𝑊 = 0 ⟹ 𝑅𝐴 + 𝑊 = −1……………(1)
Take moment about a point A,
∑ 𝑀𝐴 = −(10 × 2) + (𝑅𝐶 × 4) + (𝑊 × 6) = 0
∑ 𝑀𝐴 ⇒ 6𝑊 = 20 − 44 = −24 ………………..(2)
Solve (1) and (2) we get
𝑊 = −4 𝑁, W = 4N downwards, 𝑹𝑨 = 𝟑𝑵
25. A rigid body AD is subjected to a system of parallel forces as shown in figure below.
Reduce the system to an equivalent force couple system at B.
Given:
FA = -100 N, FB =150 N, FC =300N, FD = -200 N
Single equivalent force ∑ 𝐹 = -100+150+300-200 =150 N
Take moment about a point B,
∑ 𝑀𝐵 ⟹ (100 × 1) + (300 × 2) − (200 × 3.5) = 0 𝑁𝑚
The system to an equivalent force couple system at B,
26. Show how a given force could be resolved into a force and a couple.
The process of transforming one force applied at one point, into a force and a couple at
some other point is known as resolving a force into a force and a couple.
 Force vector F cannot be simply moved to O without modifying its action on the body.
 Attaching equal and opposite force vectors at O produces no net effect on the body.
W
C
2m
A
B
2m
10N
2m
D
300N
100N 200N
C
2m
A
B
1m
150N
1.5
m
D
CB=0 Nm
C
2m
A
B
1m
150N
1.5
m
D

 The three forces may be replaced by an equivalent force vector and couple vector, i.e.,
a force-couple system.

UNIT –III Distributed Forces
1. Define Centre of Gravity.
Centre of Gravity is an imaginary point at which the entire weight of the body is assumed to act.
2. Define Centre of mass.
Centre of mass is the point where the entire mass of a body is assumed to be concentrated.
3. Define Centroid.
Centre of gravity of a plane figure is referred as centroid. Centroid is the point at which the
entire area of the figure is assumed to be concentrated.
4. Differentiate centroid and Centre of gravity
Centroid is the geometric property of geometrical figures line, area and volume. Centre of
gravity is the physical property of a body like wire, rod, disc and solids
5. State the methods of determining the centre of gravity?
 By Geometrical considerations
 Graphical method
 Integration method
 Method of moments
6. State Pappus and Guldinus theorems.
Theorem I: The area of the surface generated by revolving a plane curve about a non-
intersecting axis in the plane of the curve is equal to the product of length of the curve and
the distance travelled by the centroid G of the curve during revolution.
𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 𝐴𝑠 = 𝐿 × 2𝜋𝑦
̅
Theorem II: The volume of the solid generated by revolving a plane area about a non-
intersecting axis in its plane is equal to the product of area and length of the path travelled
by centroid G of the area during revolution.
Volume 𝑉 = 𝐴 × 2𝜋𝑦
̅
7. Find the radius of gyration of a rectangular area of MI about its base 9×104
cm4
and
cross sectional area 300cm2
.
Given: moment of inertia I = 9 x 104
cm4
, Area A = 300 cm2
Radius of gyration 𝑘 = √
𝐼
𝐴
⟹ √
9×104
300
= 𝟏𝟕. 𝟑𝟐𝟏 𝒎
8. State perpendicular axis theorem.
It states that," moment of inertia of aplane lamina about an axis perpendicular to the lamina
and passing through its centroid is equal to the sum of the moment of inertia of the lamina
about two mutually perpendicular axes passing through the centroid and in the plane of the
lamina". IZZ =IXX+ IYY
X
Y
X
Z
Y
Z
G

9. Define principal moment of inertia.
The axes about which the product moment of inertia is zero, the moment of inertia about those
axes is said to be principal moment of inertia. Principal moment of inertia is also known as
maximum or minimum moment of inertia
10. State parallel axis theorem.
Parallel axis theorem states that “ if the moment of inertia of a plane area about an axis is
equal to the sum of moment of inertia about an axis passing though its centroid IG parallel to
the given axis AB and the product of area (A) and square of the distance between the two
parallel axes (h)". IAB=IG+Ah2
11. When will the centroid and centre of mass coincides?
In homogenous plate of uniform thickness, centroid and centre of gravity will coincides.
In other words, Centroid and centre of mass coincide when the density of the material is
uniform throughout the body
12. How the mass moment of inertia of a rectangular plate is related to the area moment of
inertia of the rectangle about an axis?
Let us consider rectangular plate whose width "b", height "h", and thickness "t". ρ is density
of material of rectangular plate.
Volume of rectangular plate V = b.h.t
Mass of rectangular plate M = density x volume = ρ × b.h.t
Mass moment of inertia of rectangular plate = density × thickness × area moment of inertia
(IXX)m = ρ.t.Ixx = ρ.t.
𝑏ℎ3
12
= M
ℎ2
12
(IYY)m = ρ.t.Iyy = ρ.t.
ℎ𝑏3
12
= M
𝑏2
12
(IZZ)m = (IXX)m + (IYY)m = ρ.t.(Ixx + Iyy) = ρ.t(.
𝑏ℎ3
12
+ .
ℎ𝑏3
12
) = M(
𝑏2
12
+
ℎ2
12
)
A
h
Area (A)
G
B
h
t
b
z
z
x
x
y
y

13. Differentiate between area moment of inertia and mass moment of inertia.
Area moment of inertia Mass moment of inertia.
It is an important concept for the study of
statics of rigid bodies.
It is an important concept for the study of
dynamics of rigid bodies.
It is a property of two dimensional sections
of areas of planes.
It is a property of three dimensional
sections of solid objects.
It depends on the distribution of area about
some reference axis.
It depends on the distribution of mass about
some reference axis.
It is the measure of capacity of a section to
resist bending about reference axis.
It is the measure of the resistance offered
by a solid body to rotation about some
reference axis.
14. Define first moment of area about an axis.
The first moment MX of the area about the x-axis is defined as follows. Take small area
element of area ΔA and multiply it by its y-coordinate, i.e. its perpendicular distance from
the X-axis, and then sum over the entire area; the sum obviously goes over to an integral in
the continuous limit. Thus
𝑀𝑥 = ∑ 𝑦𝑖∆𝐴𝑖 = ∫ 𝑦𝑑𝐴
𝑖
Similarly the first moment MY of the area about the y-axis is defined by multiplying the
elemental area ΔA by its x-coordinate, i.e. its perpendicular distance from the Y-axis, and
summing or integrating it over the entire area. Thus
𝑀𝑦 = ∑ 𝑥𝑖∆𝐴𝑖 = ∫ 𝑥𝑑𝐴
𝑖
15. What is the radius of gyration of a circle of diameter d about its diameter?
The radius of gyration of circle about its diameter is k = √
𝐼
𝐴
= √
𝜋𝑑4
64
⁄
𝜋𝑑2
4
⁄
=
𝑑
4
Φd

16. What is the mass moment of inertia of thin rectangular plate having length ‘b’ and
width ‘d’ about three mutually perpendicular axis?
The mass moment of inertia of thin rectangular plate having length ‘b’ and width‘d’ about
three mutually perpendicular axis is
(IXX)m = ρ.t.Ixx = ρ.t.
𝑏ℎ3
12
= M
ℎ2
12
(IYY)m = ρ.t.Iyy = ρ.t.
ℎ𝑏3
12
= M
𝑏2
12
(IZZ)m = (IXX)m + (IYY)m = ρ.t.(Ixx + Iyy) = ρ.t(.
𝑏ℎ3
12
+ .
ℎ𝑏3
12
) = M(
𝑏2
12
+
ℎ2
12
)
17. Express product of inertia of a rectangle about its sides.
18. When will the product of inertia of a lamina become zero?
If the area of the lamina is symmetrical with respect to one or both of the axes, the product
of inertia will be zero.
19. How will you locate the principal axes of inertia?
The principal axes of object will coincides with centroid of the object, it will be locate by
using formula 𝜃 =
1
2
𝑡𝑎𝑛−1
(
2𝐼𝑋𝑌
𝐼𝑌𝑌−𝐼𝑋𝑋
)
20. From a circular plate of 300mm diameter an eccentric circular hole of 225mm was
made such that the thickness of the plate was 25 mm on one side and 50 mm on other
side. Find the location of the centroid of the remainder.
Given: Area A1 =
𝜋𝑑1
2
4
=
𝜋×3002
4
= 70685.83 𝑚𝑚2
,
Area A2 =
𝜋𝑑2
2
4
=
𝜋×2252
4
= 39760.78 𝑚𝑚2
h
b
B
B
O
A A
The product of inertia of a rectangle about its sides
IAB =
𝑏2ℎ2
4
y
𝑥
Φ 300
Φ 225
50
y
X X
225
25
Total Area A = A1- A2 =30925.05 mm2
The given area is symmetric about XX axis, i.e., 𝒚 = 0.
x1= r1 =150 mm, x2 = 25+ r2 = 137.5 mm
The given area is not symmetric about YY axis,
𝒙 =
𝑨𝟏𝒙𝟏 − 𝑨𝟐𝒙𝟐
𝑨
= 𝟏𝟔𝟔. 𝟎𝟕 𝒎𝒎

21. What is the mass moment of inertia about x, y, and z axes of a cylinder having height
‘h’ and radius ‘r’?
Mass moment of inertia about XX axis (longitudinal axis) (IXX) m =
𝑀𝑟2
2
Mass moment of inertia about XX axis (lateral axis) (IYY) m =
𝑀
12
(3𝑟2
+ ℎ2
)
Mass moment of inertia about XX axis (lateral axis) (IZZ) m =
𝑀
12
(3𝑟2
+ ℎ2
)
22. A semi-circular lamina having radius 100 mm is located in the xy plane such that its
diametric edge coincides with y-axis. Determine the x co-ordinate of its centroid.
Given: Radius of semi-circle R =100 mm.
Centroid𝑥 =
4𝑅
3𝜋
=
4×100
3𝜋
= 42.44 𝑚𝑚
23. Find the polar moment of inertia of square lamina of 100 mm side.
Given: side of square a = 100mm =0.1m
Y
Y
Z
Z
X X
Polar moment of inertia
𝐼𝑍𝑍 =
𝑎4
6
=
0.14
6
= 1.667 × 10−5
𝑚4
Y
Y
X
X
Z
Z

24. Calculate 𝒚 for the shaded area shown in figure below.
Given:
The equation of parabola𝑦 = 𝑎𝑥2
............ (1)
y = 16 m, at x = 2 m
Substitute x = 2 and y = 16 in (1) we get,
16 = 𝑎 × 22
, a = 4
(1) ⟹ 𝑦 = 4𝑥2
Co-ordinates of the centre of gravity of the area A are𝑥𝑎𝑛𝑑𝑦
̅
Distance of centre of gravity of area A from Y axis 𝑥 =
∫ 𝑥×𝑑𝐴
2
0
∫ 𝑑𝐴
2
0
=
∫ 𝑥×𝑦𝑑𝑥
2
0
∫ 𝑦𝑑𝑥
2
0
=
∫ 4𝑥3
𝑑𝑥
2
0
∫ 4𝑥2𝑑𝑥
2
0
=
3 × 24
4 × 23
= 1.5 𝑚
Distance of centre of gravity of area A from X axis 𝑦
̅ =
∫ 𝑦×𝑑𝐴
16
0
∫ 𝑑𝐴
16
0
=
∫ 𝑦×𝑥𝑑𝑦
16
0
∫ 𝑥𝑑𝑦
16
0
=
∫
𝑦
3
2
2
𝑑𝑥
16
0
∫
𝑦
1
2
2
𝑑𝑥
16
0
= (
3𝑦
5
2
5𝑦
3
2
)
0
16
= 9.6 𝑚
25. Calculate moment of inertia Ixx for plane area shown in figure below. All dimensions are in
mm.
Given:
Section 1: Triangle base b1 = 12 mm, height h1= 6 mm
Section 2: Rectangle base b2 = 3 mm, height h2= 2 mm
Moment of inertia Ixx = 𝐼𝑥𝑥1
− 𝐼𝑥𝑥2
=
𝑏1ℎ1
3
36
−
𝑏2ℎ2
3
3
=
12×63
36
−
3×23
3
= 64 𝑚𝑚4
Y
X
𝒚
y = ax2
B
2m
O
16
m
A
x
x
2
6
12
3

26. Find the mass moment of inertia of a steel pipe of length 5m having external diameter
80cm and internal diameter 60cm about its axis. Take the mass density of steel as
7900kg/m3
.
Given: internal diameter di = 60cm = 0.6 m; external diameter do = 80cm = 0.8 m;
Internal radius ri = 30cm = 0.3 m; external radius ro = 40cm = 0.8 m;
Length of cylinder h = 5 m. Density ρ = 7900kg/m3
.
Mass of hallow cylinder M =𝜌𝜋(𝑟𝑜
2
− 𝑟𝑖
2
)ℎ = 7900 × 𝜋 × (0.42
− 0.32
) × 5 = 8686.5 𝑘𝑔
Mass moment of inertia about XX axis (longitudinal axis) (IXX) m =
𝑀(𝑟𝑖
2+𝑟𝑜
2)
2
= 1085.81𝑘𝑔𝑚2
Mass moment of inertia about XX axis (lateral axis) (IYY) m =
𝑀
12
(3(𝑟𝑜
2
+ 𝑟𝑖
2) + ℎ2)
= 4162.28 𝑘𝑔𝑚2
Mass moment of inertia about XX axis (lateral axis) (IZZ) m =
𝑀
12
(3(𝑟𝑜
2
− 𝑟𝑖
2
) + ℎ2
)
= 4162.28 𝑘𝑔𝑚2
27. What is Axis of revolution?
The fixed axis about which a plane curve (may be of an arc, straight line etc.,) or a plane area
is rotated is known as axis of revolution
28. Define Axis of Symmetry?
The axis about which similar configuration exist with respect to shape, size and weight on
either side is known as axis of symmetry. It may be horizontal, vertical or inclined
29. Define moment of inertia of a body.
Moment of inertia (I) about an axis is the algebraic sum of the products of the elements of
mass and the square of the distance of the respective element of mass from the axis.
30. Define Radius of gyration
Radius of gyration of any Lamina defined as the distance from the elemental parts of the
lamina about a given axis may be given axis at which all they have to be placed, so as not to
alter the moment of inertia about the given axis.
Y
Y
Z
Z
X X

UNIT – IV Friction
1. What is angle of repose?
Angle of repose (ϕ) is the angle to which an inclined plane may be raised before an object
resting on it will move under the action of the force of gravity
2. Define friction
Friction may be defined as a force of resistance acting on a body which prevents or retards
slipping of the body relative to a second body or surface with which it is in contact
3. Define static friction
Static friction between two bodies is the tangential force which opposes the sliding of one
body relative to the other.
4. Define Dynamic friction or Kinetic Friction.
Dynamic friction is the tangential force between two bodies after motion begins
5. Define Angle of Friction.
Angle of friction is the angle between the line of action of the total reaction of one body on
other and the normal to the common tangent between the' bodies when motion is impending.
6. Define Limiting Friction.
Limiting friction 'F' is the maximum value of static friction that occurs when motion is
impending.
7. Define Co-efficient of static friction.
Coefficient of static friction is the ratio of the limiting friction force (Ff max) to the normal
reaction (N)
Coefficient of static friction 𝜇𝑠 =
𝐹𝑓𝑚𝑎𝑥
𝑁
8. Define coefficient of Dynamic friction.
Coefficient of Dynamic friction is the ratio of the friction force(Ff)to the normal reaction (N).
Coefficient of static friction 𝜇𝑘 =
𝐹𝑓
𝑁
9. Define cone of friction.
It is defined as the right circular cone with vertex at the point of contact of the two bodies (or
surface) axis in the direction of normal reaction (R) and semi vertical angle equal to angle of
friction.
10. Define Solid Friction or Dry Friction.
If between two surfaces, no lubrication (oil or grease) used, the friction that exists between
two surfaces is called solid friction.
11. What is the sliding friction?
It is the friction, experienced by a body when it slides over another body.
12. What is Rolling Friction?
It is the friction, experienced by a body when it rolls over the other.
13. State the Laws of static friction?
 The force of friction always acts in a direction opposite to that in which the body tends to
move.
 The Magnitude of the force of friction is equal to the force, which tends to move the body.
 Limiting friction bears a constant ratio to the normal reaction between the two surfaces
 The force of friction is independent of the area of contact between the two surfaces
 The force of friction depends upon the roughness of the surfaces.
14. State the laws of Dynamic friction?
 The force of friction always acts in a direction, opposite to that in which the body is
moving.

 The magnitude of the kinetic friction bears a constant ratio to the normal reaction
between the two surfaces.
 For moderate speeds, the force of friction remains constant and it decreases with the
increase of speed.
15. State the laws of solid friction?
 Laws of static friction
 The force of friction always acts in a direction opposite to that in which the body tends to
move.
 The Magnitude of the force of friction is equal to the force, which tends to move the body.
 Limiting friction bears a constant ratio to the normal reaction between the two surfaces
 The force of friction is independent of the area of contact between the two surfaces
 The force of friction depends upon the roughness of the surfaces.
 Laws of Dynamic friction
 The force of friction always acts in a direction, opposite to that in which the body is
moving.
 The magnitude of the kinetic friction bears a constant ratio to the normal reaction
between the two surfaces.
 For moderate speeds, the force of friction remains constant and it decreases with the
increase of speed.
16. A wheel of radius 50cm subjected to a load of 300N rolls on a level ground at constant
speed. If the wheel is pushed by a tractive force of 60N applied horizontally at the
centre of the wheel, find the coefficient of rolling resistance.
Given: Radius r = 50cm=0.5m, weight W=300N, tractive force P=60N
W.k.t. coefficient of rolling resistance 𝑎 =
𝑃×𝑟
𝑊
=
60×0.5
300
= 0.1𝑚
17. What is the force F required to just initiate the block shown in figure to slide if the
coefficient of friction between the surfaces in contact is 0.4?
𝐵𝑦 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠,
Σ𝐹⊥𝑡𝑜 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 = 0 ⇒ 𝑅𝑛 − 𝑊 = 0 ⇒ 𝑅𝑛 = 1000𝑁
Σ𝐹𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 = 0 ⇒ −𝜇𝑅𝑛
+ 𝐹 = 0 ⇒ 𝐹 = 𝜇𝑅𝑛
⇒ 𝐹 = 400𝑁
18. Why kinetic friction is lesser than static friction?
The kinetic friction is usually not greater than the applied force. You start moving an object
which increases the static force to prevent any motion. The smoother the object and the
smoother the surface the smaller the difference between the amount of static friction and the
amount of kinetic friction.
𝐹𝑓 = 𝜇𝑅𝑛
𝑊 = 1000 𝑁
𝑅𝑛
𝐹
1000 N

19. What is the difference between Rolling Friction and Sliding Friction?
Rolling Friction Sliding Friction
Rolling friction takes place when an object
rolls on the surface.
Sliding friction takes place when two
surfaces are rubbed against each other.
Rolling friction takes place due to the
deformation of surfaces.
Sliding friction takes place due to
interlocking between microscopic surfaces.
The coefficient of rolling friction is
dependent on the radius of the rolling object,
the depth to which the object can sink, and
the toughness of the surface.
The coefficient of sliding friction depends
on the texture of the surface and temperature
to a certain extent. It is independent of
external factors.
Coefficient of rolling friction:
𝐹𝑟 = 𝜇𝑟𝑁
Coefficient of sliding friction:
𝐹 = 𝜇𝑁
20. When do we say that the motion of a body is impending?
The motion is said to be impending if the applied forces are such that the body is just about
to slide.

UNIT-V Dynamics of Particles
1. A particle is projected into space at an angle of 30o
to the horizontal at a velocity of
40m/s. find the maximum height reached by the projectile.
Given: initial velocity u =40 m/s, projected angle α =30o
W.k.t. Maximum height hmax =
𝒖𝟐
𝟐𝒈
𝒔𝒊𝒏𝟐
𝜶 =
402
2×9.81
𝑠𝑖𝑛2
30𝑜
= 𝟐𝟎. 𝟑𝟗 𝒎
2. Distinguish between perfectly plastic impact and perfectly elastic impact.
Perfectly elastic impact Perfectly plastic impact
The relative velocities of the two objects after
impact(separation velocities) are the same as
their relativevelocities before impact (approach
velocities).
The relative velocities of the colliding
objects after impact are less than
those before impact.
The total energy of motion is not changed. Some of the total energy of motion is
lost.
Coefficient of restitution of perfectly elastic
impact is 1.
Coefficient of restitution of perfectly
elastic impact is 0.
3. Define Newton’s law (second law) of motion.
It states that," if the resultant force acting on a particle is not zero, the resultant force is
directly proportional to rate of change of velocity of particle".
4. Give the equation of work energy for a rectilinear motion.
The equation of work energy for a rectilinear motion is
𝜮𝑭𝒂𝒍𝒐𝒏𝒈 𝒕𝒉𝒆 𝒎𝒐𝒕𝒊𝒐𝒏 . 𝑺 =
𝒎
𝟐
(𝒗𝟐
− 𝒖𝟐
)
Where, ΣFalong the motion is net force acting on particle along the motion, S is displacement, v
is final velocity and u is initial velocity.
5. A car runs with an initial velocity of 30m/s and uniform acceleration of 3m/s2
. Find its
velocity after 5 seconds.
Given: u = 30 m/s, a =3 m/s2
, t = 5 s.
W.k.t. v = u +at⟹ 𝑣 = 30 + (3 × 5) = 𝟒𝟓 𝒎/𝒔.
6. The velocity of a particle is given by V =4t3
-5t2
. When does the acceleration of the
particle become zero?
Given: V =4t3
-5t2
w.k.t. acceleration a =
𝒅𝑽
𝒅𝒕
=
𝑑(4𝑡3−5𝑡2)
𝑑𝑡
= 12𝑡2
− 10𝑡 = 0
t=0; 12t -10=0⟹t = 0.833s.

7. State D’Alembert’s principle.
It states that “The system of forces acting on a body in motion is in dynamic equilibrium,
with the inertia force of the body”.
D'Alembert's principle summarized as follows:
 The vector sum of all the external forces and the inertia acting upon a rigid body is
zero.ΣF - ma = 0
 The vector sum of all the external moment and the inertia torques acting upon a
rigid body also separately zero. ΣT - Iα = 0
8. Distinguish between uniform motion and uniformly accelerated motion.
Uniform motion
A particle is said to be in uniform motion if it has equal displacements in equal intervals of
time. However small these intervals may be.
Uniform acceleration
A particle is said to be in uniform accelerated motion if it has velocity increases by an equal
amount in every equal time period.
9. A body of mass 5kg accelerates at a constant rate 2m/s2
on a smooth horizontal surface
due to external force acting at 30o
with the horizontal. Find the magnitude of the force.
Given : mass m = 5 kg, acceleration a = 2 m/s2
Condition of equilibrium: 𝚺𝐅𝐩𝐞𝐫𝐩𝐞𝐧𝐝𝐢𝐜𝐮𝐥𝐚𝐫 𝐭𝐨 𝐭𝐡𝐞 𝐦𝐨𝐭𝐢𝐨𝐧 = 𝟎; 𝚺𝐅𝐚𝐥𝐨𝐧𝐠 𝐭𝐡𝐞 𝐦𝐨𝐭𝐢𝐨𝐧 − 𝐦𝒂 = 𝟎
ΣFperpendicular to the motion = 0.5F + N − 49.05 = 0
⟹ 0.5F + N = 49.05 ...................... (1)
ΣFalong the motion = F cos 30o
= 0.866F
ΣFalong the motion − m𝑎 = 0 ⇒ 0.866F − (5 × 2) = 0
⇒ 𝐅 = 𝟏𝟏. 𝟓𝟓𝐍
Substitute F=11.55N in equ.(1) we get,N = 43.275N
F
30o
m = 5 kg
F sin 30o
F cos 30o
m.a
W=5 x 9.81
= 49.05 N
N
Free Body Diagram

10. A point P moves along a straight line according to the equation x=4t3
-2t-5, where x is in
metres, t is in seconds. Determine the velocity and acceleration when t=3seconds.
Given: Displacement x=4t3
-2t-5
W.k.t. velocity v =
𝒅𝒙
𝒅𝒕
= 𝟏𝟐𝒕𝟐
− 𝟐
Velocity v at t = 3s=106 m/s.
Acceleration a =
𝒅𝒗
𝒅𝒕
= 𝟐𝟒𝒕
Acceleration aat t = 3s =72m/s2
11. What is linear momentum?
The product of mass and velocity of a body is known as momentum of the body.
Momentum M = m xv
12. A stone is dropped from the top of a tower. It strikes the ground after four seconds.
Find the height of tower.
Given: time taken by stone to strikes the ground t = 4s, initial velocity u = 0 m/s
w.k.t. distance travelled by stone = height of tower h = ut+
𝟏
𝟐
𝒈𝒕𝟐
h = (0 × 4) + (
1
2
× 9.81 × 4)2
= 𝟕𝟖. 𝟒𝟖 𝒎.
13. The motion of particle is defined by the relation 𝒙 = 𝐭𝟑
+ 𝟏𝟔𝐭𝟐
− 𝟐𝟐, x is in metres,
and t is in seconds. Determine the acceleration of the particle at t =3seconds.
Given: Displacement 𝑥 = t3
+ 16t2
− 22
W.k.t. velocity v =
𝒅𝒙
𝒅𝒕
= 𝟑𝒕𝟐
+ 𝟑𝟐𝒕
Velocity v at t = 3s=123 m/s.
Acceleration a =
𝒅𝒗
𝒅𝒕
= 𝟔𝒕 + 𝟑𝟐
Acceleration aat t = 3s =50 m/s2
14. A car accelerates uniformly from a speed of 30kmph to a speed of 75kmph in 5seconds.
Determine the acceleration of the car and also the distance travelled during 5s.
Given: initial velocity u = 30kmph=8.33m/s, final velocity v = 20.83m/s, t = 5s.
Acceleration𝒂 =
𝒗−𝒖
𝒕
=
20.83−8.33
5
= 𝟏𝟐. 𝟓 𝒎/𝒔.
Displacement 𝒔 = 𝒖𝒕 +
𝟏
𝟐
𝑎𝑡2
= (8.33 × 5) + (0.5 × 12.5 × 52
)=197.9m/s2
.
15. A shot fired into space at an elevation of 60o
to the horizontal with a velocity of 60 m/s.
what will be its velocity 2 seconds after its start?
Given: initial velocity u =60 m/s, projected angle 𝛼 = 60𝑜
, t =2s.
Horizontal component of final velocity vx =u.cosα =60cos60o
=30 m/s.
Vertical component of final velocity vy =𝑢. 𝑠𝑖𝑛𝛼 − 𝑔. 𝑡 = (60 × 𝑠𝑖𝑛60𝑜) − (9.81 × 2) = 32.34 𝑚/𝑠
Final velocity 𝒗 = √𝒗𝒙
𝟐 + 𝒗𝒚
𝟐 = √302 + 32.342 = 𝟒𝟒. 𝟏𝟏 𝒎/𝒔.

16. The velocity of a particle moving along a straight line is given by v = 4t3
-4. If the particle
starts from the origin, find the distance travelled from t = 0 to t = 2s.
Given: v = 4t3
-4, t =0s, t = 2s.at t=0s, x=0 m/s
Displacement 𝑥 = ∫ 𝑣𝑑𝑡 = ∫(4𝑡3
− 4)𝑑𝑡
⇒ 𝑥 = 𝑡4
− 4𝑡 + 𝑐.................. (1)
Substitute t=0s, v = 0m/s in equ. (1) we get c = 0.
(1) ⇒ 𝒙 = 𝒕𝟒
− 𝟒𝒕
Substitute t=2s, in equ. (1) We get x = 8 m
17. A steel ball is thrown vertically upwards from the top of a building 25m above the ground
with an initial velocity of 18 m/s. find the maximum height reached by the ball from the
ground.
Given: initial velocity u =18m/s, height of building =25m, final velocity v =0m/s.
w.k.t. 𝑣2
− 𝑢2
= −2𝑔ℎ ⇒ 𝒉 = 𝟏𝟔. 𝟓𝟏𝒎
The maximum height reached by the ball from the ground hmax =18+16.51 =34.51m
18. Define speed.
The rate of change of displacement of a body irrespective of its direction is called speed. It’s
a scalar quantity
19. Define velocity
The rate of change of displacement of a body with respect to its surroundings in a particular
direction is called the velocity. It is a vector Quantity.
20. Define acceleration
The rate of change of velocity of a body is called acceleration.
21. Define uniform acceleration
If a body moves in such a way that its velocity changes equal in magnitude in equal intervals
of time, the body is said to be moving with a uniform acceleration.
22. Define variable acceleration.
If a body moves in such a way that its velocity changes unequal in magnitude in equal intervals
of time, the body is said to be moving with a variable acceleration.
23. Write the equations of plane motion?
 𝑣 = 𝑢 + 𝑎𝑡
 𝑣2
= 𝑢2
+ 2𝑎𝑠
 𝑠 = 𝑢𝑡 +
1
2
𝑎. 𝑡2
Where, v=Final velocity, u =Initial velocity, a=acceleration, t=time taken for displacement
s=distance travelled.
24. Write the equations of motion of a body under the force of gravity?
 𝑣 = 𝑢 + 𝑔𝑡
 𝑣2
= 𝑢2
+ 2𝑔ℎ
 ℎ = 𝑢𝑡 +
1
2
𝑔. 𝑡2
Where, v=Final velocity, u =Initial velocity, a=acceleration, t=time taken for displacement,
h=height travelled
25. A flywheel has a mass moment of inertia of 11kgm2
about the axis of rotation. It run at
a constant angular velocity of 94.25rad/s. Find the kinetic energy of the flywheel.
Given: mass moment of inertia I =11kgm2
, angular velocity ω=94.25rad/s.
The kinetic energy of the flywheel 𝐸 =
1
2
𝐼. 𝜔2
= 97713.69𝑁𝑚/𝑠 = 97.714𝑘𝑊

26. A fan rotates at constant speed of 50rpm. What is the total angular displacement it
makes in 10 min. of time?
Given: N=50rpm,t=10min.=600s
Angular velocity 𝜔 =
2𝜋𝑁
60
=
2𝜋×50
60
= 5.238𝑟𝑎𝑑/𝑠
Angular displacement 𝜃 = 𝜔𝑡 = 5.238 × 600 = 3142.86𝑟𝑎𝑑.
27. The angular rotation of an accelerated disc is given by 𝛉 =
𝟗
𝟑𝟐
𝐭𝟑
+
𝟑
𝟒
𝐭𝟐
+ 𝟔𝐭 in radians.
Find its angular acceleration when t = 2sec.
Given: 𝜃 =
9
32
𝑡3
+
3
4
𝑡2
+ 6𝑡, time t=2s
Angular acceleration 𝛼 =
𝑑2𝜃
𝑑𝑡2 =
27
16
𝑡 +
3
2
= 4.875𝑟𝑎𝑑/𝑠2
28. A body is rotating with an angular velocity of 5rad/s. after 4 seconds, the angular velocity
of the body becomes 13rad/s. determines the angular acceleration of the body.
Given: initial angular velocity ωi =5rad/s, final angular velocity ωf =13rad/s, time t= 4s
𝜔𝑓−𝜔𝑖
𝑡
=
13−5
4
= 2𝑟𝑎𝑑/𝑠2
29. A body is rotating with an initial angular velocity of 3rad/s. its angular velocity
increases to 10 rad/s in 5seconds. Determine the angular acceleration of the body.
Given: initial angular velocity ωi =3rad/s, final angular velocity ωf =10rad/s, time t= s
𝜔𝑓−𝜔𝑖
𝑡
=
10−3
5
= 1.4𝑟𝑎𝑑/𝑠2
30. Write the relation between linear acceleration and angular acceleration?
Let ‘a’ be linear acceleration in m/s2
, ‘α’ be angular acceleration in rad/s2
, and ‘r’ be radius in m.
𝑙𝑖𝑛𝑒𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝒂 = 𝒓 × 𝜶
31. A rigid body rotates about a fixed axis. Write the expression for angular velocity when
the rotation is uniformly accelerated.
1. 𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡
2. 𝜔𝑓
2
= 𝜔𝑖
2
+ 2𝛼𝜃
3. 𝜃 = 𝜔𝑖𝑡 +
1
2
𝛼. 𝑡2
Where, 𝜔𝑓=Final angular velocity in rad/s, 𝜔𝑖 =Initial angular velocity in rad/s, α= angular
acceleration in rad/s2
, t=time taken for displacements in s, θ= angular displacement in rad..
32. A flywheel 500 mm diameter is brought uniformly from rest up to a speed of 280 rpm
in 15 seconds. What is the velocity of a point on the rim one second after starting from
rest?
Given:diameter d =500mm =0.5m, initial speed N1 =0rpm(start from rest), final speed N2
=280rpm, time 15s,
W.k.t. Angular velocity 𝜔 =
2𝜋𝑁
60
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝜔𝑖 =
2𝜋𝑁
60
= 0𝑟𝑎𝑑/𝑠
Final Angular velocity 𝜔𝑓 =
2𝜋𝑁
60
=
2𝜋×280
60
= 29.33𝑟𝑎𝑑/𝑠
𝜔𝑓−𝜔𝑖
𝑡
=
29.33−0
15
= 1.956𝑟𝑎𝑑/𝑠2
Angular velocity 𝜔𝑎𝑡 𝑡=1𝑠 = = 𝜔𝑖 + 𝛼𝑡 = 0 + (1.956 × 1) = 1.956𝑟𝑎𝑑/𝑠

33. Define energy.
It is the capacity to do work
34. Define potential energy.
It is the energy possessed by a body, for doing work, by virtue of its position
35. Define kinetic energy.
It is the energy possessed by a body, for doing work, by virtue of its motion.
36. State the law of conservation of energy?
It states that, "The energy can neither be created nor destroyed, though it can be transformed
from one form into any of the form in which the energy can exist.
37. Define power.
The rate of doing work is called power.
38. What is rectilinear motion?
The motion of a particle is said to be rectilinear, if it moves along a straight line.
39. What is curvilinear motion?
The motion of a particle is said to be curvilinear, if it moves along a curved path.
40. State the principle of conservation of linear momentum.
It states that, if the resultant force acting on a particle is zero, then the linear momentum of
the particle remains constant.
i.e., Final momentum = Initial momentum.
41. States Law of conservation of angular momentum.
The sum of the moments about 'O' of the forces acting on the particle is equal to twice the rate
of change of angular momentum of the particle about ‘O’).
42. What is conservative force?
A force F is said to be conservative, when the force components are derivable from a potential
and the work done by the force F between any two points is independent of the path followed.
43. States Theorem of conservation of energy.
When a particle is acted upon by conservative forces, the sum of the particles kinetic and
potential energy remains constant during the motion.
44. Define Impulse of a force.
When a large force acts over a short period of time, that force is called an impulsive force.
45. A particle starting from rest, moves in a straight line and its acceleration is given by a = 40 -
46t2
m/s2
, where t is in sec. Determine the velocity of the particle when it has travelled 52 m.
Given: a = 40 - 46t2
, initial condition t=0s, u= 0m/s, x =0m, to find velocity at x=52m.
W.K.T. velocity 𝑣 = ∫ 𝑎𝑑𝑡 = ∫(40 − 46t2)𝑑𝑡 ⇒ 𝑣 = 40𝑡 − 15.33𝑡3
+ 𝐶
Applying initial condition t=0s, v= 0m/s we get C=0, 𝒗 = 𝟒𝟎𝒕 − 𝟏𝟓. 𝟑𝟑𝒕𝟑
Displacement 𝑥 = ∫ 𝑣𝑑𝑡 = ∫(40𝑡 − 15.33𝑡3)𝑑𝑡 ⇒ 𝑥 = 20𝑡2
− 3.83𝑡4
+ 𝐶1
Applying initial condition t=0s, x= 0m/s we get C1=0, 𝒙 = 𝟐𝟎𝒕𝟐
− 𝟑. 𝟖𝟑𝒕𝟒
When x=52m, 20𝑡2
− 3.83𝑡4
= 52, substitute t2
=y weget,-3.83y2
+20y-52=0..........(1)
Solve equ.(1) we get y=1.905, t =1.38s
46. A 4000 kg automobile is driven down a 5o
inclined at a speed of 80kmph when brakes
are applied, causing a constant total braking force of 7.5kN. Determine the distance
travelled by the automobile as it comes to a stop.
Given: m=4000 kg, initial velocity u = 80kmph = 22.22m/s, Final velocity v = 0kmph = 0 m/s,
braking force = 7500N

Condition of equilibrium: ΣFperpendicular to the motion = 0:
ΣFperpendicular to the motion ⟹ Nr − 39240cos5o
= 0
⟹ Nr = 39090.68N
W.k.t.Impulse-Momentum principle𝜮𝑭𝒂𝒍𝒐𝒏𝒈 𝒕𝒉𝒆 𝒎𝒐𝒕𝒊𝒐𝒏 . 𝒕 = 𝒎(𝒗 − 𝒖)
𝛴𝐹𝑎𝑙𝑜𝑛𝑔 𝑡ℎ𝑒 𝑚𝑜𝑡𝑖𝑜𝑛 = 39240𝑠𝑖𝑛5𝑜
− 7500 = −𝟒𝟎𝟖𝟎𝑵
−4080. 𝑡 = 4000(0 − 22.22) ⇒ 𝒕 =
𝟖𝟖𝟖𝟖𝟎
𝟒𝟎𝟖𝟎
=21.78s
39240sin5o
Fb =7500N
4000 x 9.81N
5o
Nr
39240cos5o
Fb=7500N
5o

ME3351 ENGINEERING MECHANICS
PART-B QUESTIONS (13 MARKS)
UNIT-I STATICS OF PARTICLES
1. Four coplanar forces are acting at a point. Three forces have magnitude of 20 N, 50 N, 20 N at
angles of 45°, 200° and 270° respectively with respect to +x-axis. Fourth force is unknown.
Resultant force has magnitude of 50 N and acts along x-axis at an angle of 0° with respect to
+x-axis. Determine the unknown force and its direction or angle from + x-axis.
2. A lamp of mass 1 kg is hung from the ceiling by a chain and is pulled aside by a horizontal
chord until the chain makes an angle of 60° with the ceiling. Find the tension in the chain and
chord.
3. Two spheres each of weight 500N and of radius 100 mm rest in a horizontal channel of width
of 360 mm as shown in in Fig. Q. 3. Find the reactions on the points of contact A, B and C.
Assuming all the surfaces of contact are smooth.
Fig. Q. 3.
4. A 200 kg cylinder is hung by means of two cables AB and AC which are attached to the top of
a wall. A horizontal force P perpendicular to the wall holds the cylinder in the position shown
in Fig. Q. 4. Determine the magnitude of P and the tension in each cable.
5. Two cylinders, having weight WA = 2000N and WB = 1000 N are resting on smooth inclined
planes having inclination 60° and 45ᵒ with the horizontal respectively as shown in Fig. Q. 5.
They are connected by a weightless bar AB with hinge connections. The bar AB makes 15 ᵒ
angle with the horizontal. Find the magnitude of the force P required to hold the system in
equilibrium.
C
B
y
z
1.2
A
P
12
10
8
2
x
Fig. Q. 4.

Fig. Q. 5.
6. Determine the magnitude and direction of force 𝐹 shown in Fig. Q. 6. So that the particle is in
equilibrium.
Fig. Q. 6.
7. Find the tension in the cables AB, Ac and Ad if the Crate shown in Fig. Q. 7. is weighing 9.07
kg.
Fig. Q. 7
8. The truck is to be towed using two ropes. Determine the magnitudes of forces 𝐹𝐴 and 𝐹𝐵 acting
on each rope in order to develop a resultant force of 950 N directed along the positive X-axis.

Fig. Q. 8.
9. Determine the magnitude and angle θ of F so that particle P, shown in Fig. Q. 9., is in
equilibrium.
Fig. Q. 9.
10. In the figure shown, three wires are joined at D. Two ends A and B are on the wall and other
end C is on the ground. The wire CD is vertical. A force of 60 kN is applied at D and it passes
through a point E on the ground as shown in Fig. Q. 10. Find the forces in all the three wires.
Fig. Q. 10
11. A container weighing 450 kN is suspended at P by using two cables PB and PA anchored as
shown in Fig. Q. 11. A Horizontal force F keeps the container in the current position. Find the
magnitude of force F and forces in cable PA and PB.

Fig. Q. 11.
12. Two identical rollers each of weight 5 kN rest in between an inclined wall and a vertical wall as
shown in Fig. Q. 12. Determine the reactions at the points of contact P, Q and R. Assume the
wall surfaces to be smooth.
Fig. Q. 12
13. The X, Y, Z components of a force are 54 kN, -36 kN and 36 kN respectively. Find the
magnitude of the force and angle it makes with X, Y, and Z axes. Also find the components of
the force along the line joining A (1, 2, -3) and B (-1, -2, 2).
14. Four forces of magnitude 10 kN, 20 kN, 30 kN and 50 kN are acting on a particle O. The angles
made by the forces with X-axis are 45°, 90°, 150° and 240° respectively. All the angles are
measured in anticlockwise direction. Find the magnitude and direction of equilibrant.
15. Four forces act on a bolt A as shown below. Determine the resultant of the forces on the bolt.
Fig. Q. 15.
16. Consider the 75 kg crate shown in the diagram. This crate was lying between two buildings and
it is now being lifted onto a truck, which will remove it. The crate is supported by a vertical

cable, which is joined at A to two ropes which pass over pulleys attached to the buildings at B
and C. Determined the tension in each of the ropes AB and AC.
Fig. Q. 16
17. A cylindrical roller has a weight of 10kN and it is being pulled by a force which is inclined at
30˚ with the horizontal as shown in fig. While moving it comes across an obstacle of 10cm
height. Predict the force required to cross this obstacle when the diameter of the roller is 70cm.
Fig. Q. 17.
18. Two smooth circular cylinders each of weight 1000 N and radius 15 cm are connected at their
centers by a string AB of length 40 cm and rest upon a horizontal plane, supporting above them
a third cylinder of weight 2000 N and radius 15 cm as shown in Figure. Predict the force S in
the string AB and reactions on the floor at the points of contact D and E.
Fig. Q. 18
19. Find out the resultant of the system of forces given below: (i) 20N inclined at 30° towards north
of east. (ii) 25 N towards North. (iii) 30N towards north west. (iv) 35N inclined at 40° towards
south of west.
20. Two cylinders E, F of diameter 60mm and 30mm. Weighing 160N and 40 N respectively are
placed as shown in Fig. Assuming all the contact surfaces to be smooth, find the reactions at the
contact points.

Fig. Q. 20.
21. Three links PQ, QR and RS connected as shown in Fig. Support loads W and 50 N. Find the
weight W and the force in each link if the system remains in equilibrium.
Fig. Q. 21.
22. The magnitude of the resultant of two concurrent forces including an angle of 90° between them
is √13 kN. When this included angle is changed to 60°, the magnitude of the resultant becomes
√19 kN. Find the magnitude of the two forces.
23. A force of magnitude 3.5 kN makes 30°, 50° and 100° with x, y and z axes respectively. Find
the force vector and determine its components along x, y, z axes.
24. A weight of 8 kN is suspended by three cables PA, PB and PC. The coordinates of the points
are: P (1.5, 1.5, -2), A (0, 3, 4), B (2.5, 3, 2.5), C (1, 3, 0). Determine the tensions in the cables.
25. A Body of mass 900 kg is suspended by two cables PR and PQ making an angles of 40° and 50°
respectively with ceiling. Find the tension in the cables PQ and PR.
26. The forces shown in the figure below are in equilibrium. Determine the forces F1 and F2
27. Determine the tension in cables AB & AC to hold 40 Kg load shown in fig.

28. A force P is applied at ‘O’ to the string AOB as shown in fig. If the tension in each part of
wiring is 50 N, find the direction and magnitude of force P for equilibrium conditions.
29. Two identical rollers each of weight 50N are supported by an inclined plane and a vertical
wall as shown in fig. Find the reactions at the points of supports A, B, and C.
30. A tower guy wire shown below is anchored by means of a bolt at A as shown. The tension in
the wire is 2500kN.Determine (i) the components Fx,Fy& Fz of the force acting on the bolt
(ii) the angles θx, θy, θz, defining the direction of the force.

31. . Members OA, OB and OC form a three member space truss. A weight of 10 KN is
suspended at the joint ‘O’ as shown in fig. Determine the magnitude and nature of forces in
each of the three members of the truss.
32. A crane shown in figure is required to lift a load of W=10 KN. Find the forces in the
members AB and CB.
33. In the figure shown, three wires are joined at D.

UNIT – II EQUILIBRIUM OF RIGID BODIES
1. A three-bar pendulum ABCD has three bars each 2 m in length and weighing 2.5 kN as shown
in Fig. Q. 1. It is held in equilibrium by applying a horizontal force of 3 kN at the free end.
Determine the angles 𝜃1, 𝜃2 and 𝜃3made with the vertical.
Fig. Q. 1
2. A roller of radius 30 cm weighs 2.5 Kn. It is to be pulled over a rectangular obstruction of
height of 10 cm by a horizontal forces F passing through the centre of the roller. Find the
magnitude If the force F passing through the centre of the roller. Find the magnitude if the
force F required just to turn the roller over the corner of the obstruction. Also find the
magnitude and direction of the minimum force required for the same.
3. A father and his son carry a block of mass 50 kg by using a uniform bar of length 3 m and
mass 16 kg. the son can bear only half the load carried by the father. Find the location of the
block on the bar.
4. A 20-kg ladder Fig. Q. 4. used to reach high shelves in a storeroom is supported by two
flanged wheels A and B mounted on a rail and by an unflanged wheel C resting against a rail
fixed to the wall. An 80-kg man stands on the ladder and leans to the right. The line of
action of the combined weight W of the man and ladder intersects the floor at point D.
Determine the reactions at A, B, and C.
Fig. Q. 4.
5. A system of forces acts as shown in Fig. Q. 5. Find the magnitude of A and B so that the
resultant of the forces system passes through P and Q.

Fig. Q. 5.
6. Four tugboats are used to bring an ocean liner to its pier. Each tugboat exerts a 5000N force
in the direction shown. Determine (a) the equivalent force-couple system at the foremast O,
(b) the point on the hull where a single, more powerful tugboat should push to produce the
same effect as the original four tugboats.
Fig. Q. 6.
7. A light bar AD is suspended from a cable BE and supports a 50 kg block at C as shown in
Fig. Q. 7. The ends A and D of the bar are in contact with frictionless vertical walls.
Determine the reaction at A and D.
Fig. Q. 7.
P
B
Q
A
30°
1.5
m
300 N
40°
45°
2.25 m
350 Nm
1.5
m
2.25 m

8. Find the support reactions of the truss loaded as shown in Fig. Q. 8.
Fig. Q. 8.
9. A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is held in place by
a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the
components of the reactions at A and B.
Fig. Q. 9.
10. The force P has a magnitude of 250 N and is applied at the end C of a 500-mm rod AC attached
to a bracket at A and B. Replace P with (a) an equivalent force-couple system at B, (b) an
equivalent system formed by two parallel forces applied at A and B.
Fig. Q. 10.
11. A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE show in
Fig. Q. 11. Knowing that the 5-kN force acts vertically downward, determine (a) the tension
in cables CD and CE, (b) the reaction at A.

Fig. Q. 11.
12. Replace the force system by a resultant force and couple moment at point O.
Fig. Q. 12.
13. A force (10 𝑖 + 20𝑗 − 5𝑘
⃗ ) 𝑁 acts at a point P(4, 3, 2) m. Determine the moment of the force
about the point Q(2, 3, 4) m in vector form. Also find the magnitude of the moment and its
angles with respect to x, y, z axes.
14. A plate ABCD in the shape of a parallelogram is acted upon by two couples, as shown in the
figure. Determine the angle β if the resultant couple is 1.8 Nm clockwise.
Fig. Q.14
15. Two beams AB and CD are shown in Fig. Q. 15. A and D are hinged supports. B and C are
roller supports. (i) Sketch the free body diagram of the beam AB and determine the reactions
at the supports A and B. (ii) Sketch the free body diagram of the beam CD and determine the
reactions at the supports C and D.
Fig. Q.15

16. A shaft is subjected to forces in x, y and z direction as shown in Fig. Q. 16. Replace these
forces by a resultant R at origin O and a Couple.
Fig. Q.16
17. Determine the resultant of the force system acting in plane shown in Fig. Q. 17. Locate the
distance from A where the resultant cuts the x-axis.
Fig. Q.17.
18. A system of parallel forces acting on a rigid bar ABCD is shown in Fig. Q. 18. Reduce this
system to a single force and couple at A.
Fig. Q.18.
19. Determine the magnitude, direction and position of resultant force for the system of forces
shown in Fig. Q. 19.
Fig. Q.19.

20. Determine the reactions at supports A and B of the overhanging beam shown in Fig. Q. 20.
Fig. Q. 20
21. Four forces act on a 700 3 375-mm plate as shown in Fig. Q21. (a) Find the resultant of
these forces. (b) Locate the two points where the line of action of the resultant intersects
the edge of the plate.
Fig. Q. 21.
22. A 2.4-m boom is held by a ball-and-socket joint at C and by two cables AD and AE as
shown in Fig. Q. 22. Determine the tension in each cable and the reaction at C.
Fig. Q. 22.
23. Member ABC is supported by a pin and bracket at B and by an inextensible cord attached at
A and C and passing over a frictionless pulley at D. The tension may be assumed to be the
same in portions AD and CD of the cord. For the loading shown in Fig. Q. 23 and neglecting
the size of the pulley, determine the tension in the cord and the reaction at B.
Fig. Q.23.

24. A 50-kg crate is attached to the trolley-beam system shown in Fig. Q. 24. Determine (a) the
tension in cable CD, (b) the reaction at B.
Fig. Q. 24.
25. Determine the horizontal and vertical components of reaction at the pin A and the force in the
cable BC. Neglect the thickness of the members.
Fig. Q. 25.
26. Four forces act on a 700mm X 375mm plate as shown in fig. a) Find the resultant of these
forces b) Locate the two points where the line of action of the resultant intersects the edge of
the plate.

27. Four coplanar non concurrent non parallel forces act on a square plate of side 2m as shown in
fig. Locate the resultant forces.
28. In figure below, two forces act on a circular disc as shown. If the resultant moment of all these
forces about point D on the disc is zero, determine: a) Magnitude of force P (b) Magnitude of
the resultant of two forces (c) The point on the Y-axis through which the line of action of the
resultant passes through.
29. Four forces act on a square of side 1 m as shown in fig. Reduce the force system into an
equivalent force – couple system at A.
30. A rod AB of weight 200 N is supported by a cable BD and the corner of wall and floor surface
as shown in fig. Find the reaction at A and tension in the cord.

31. . Find reactions at points A & B
32. A force (10i+20j-5k)N acts at a point P(4,3,2)m.Determine the moment of this force about the
point Q(2,3,4)m in vector form. Also find the magnitude of the moment and its angles with
respect to x.y.z axes.
33. A plate ABCD in the shape of a parallelogram is acted upon by two couples, as shown in the
figure.
34. A simply supported beam AB of 6m span is loaded as shown. A is a hinged support; B is a
roller support. Determine the reactions at A and B.

UNIT –III DISTRIBUTED FORCES
1. Determine the moment of inertia of the shaded area as shown in Fig. Q. 1. with respect to the
x-axis.
Fig. Q. 1
2. For the section in Fig. Q. 2., the moments of inertia with respect to the x and y axes have been
computed and are known to be 𝐼𝑥𝑥 = 10.38 𝑚𝑚4
, 𝐼𝑦𝑦 = 6.97 𝑚𝑚4
. Determine: (i) The
orientation of the principal axes of the section about O. (ii) The values of the principal
moments of inertia of the section about O.
Fig. Q. 2
3. A cone of base diameter 200 mm is fitted to a hemisphere of diameter 200 mm centrally. What
should be the height of cone so that the centroid of the solid combination lies at the junction
between the cone and hemisphere?
4. Locate the centroid for the area shown in Fig. Q.7.
Fig. Q. 4
y
x
3 mm
3 mm
0.5 mm
0.5 mm
4
mm
0.5 mm

5. Find the moment of inertia of the section shown in Fig. Q. 4. about the x and y centroidal
axes. All dimensions are in mm.
Fig. Q. 5.
6. An inverted T section is shown in Fig. Q. 5. Calculate the moment of inertia of the section
about XX axis parallel to the base and passing through the centroid. Also calculate radius of
gyration.
Fig. Q. 6.
7. Locate the centroid and 𝐼𝑥𝑥 and 𝐼𝑦𝑦 about the axes passing through the centroid of lamina
shown in fig. Q. 6.
Fig. Q. 7.
30
0
300
100
100
500
100

8. Find moment of inertia about 1-1 and 2-2 axes for the area shown in Fig. Q. 8.
Fig. Q. 3
9. Find 𝐼𝑥𝑥, 𝐼𝑦𝑦 through centroid of the Fig. Q. 9. With uniform thickness of 3 cm throughout.
Fig. Q. 9.
10. Derive an equation for the mass moment of inertia of cone.
11. Derive from first principles, the second moment of area of a circle about its diametrical axis.
12. For the section shown in Fig. Q. 12., locate the horizontal and vertical centroidal axis.
Fig. Q. 12

13. Find the moment of inertia of the shaded area shown in Fig. Q. 13. Shown in Fig. Q. 13. About
the vertical and horizontal centroidal axes. The width of the hole is 200 mm.
Fig. Q. 13
14. Locate the centroid for the following sections as shown in Fig. 14. (a) and (b).
Fig. Q. 14. (a) and (b)
15. Determine the moment of inertia of the shaded area shown in Fig. Q. 15. with respect to each
of the coordinate axes. Also determine the radius of gyration of the shaded area with respect
to each of the coordinate axes.
Fig. Q. 15

16. A thin steel plate which is 4 mm thick is cut and bent to form the machine part shown in Fig.
Q. 16. Knowing that the density of steel is 7850 kg m3
⁄ , determine the moments of inertia of
the machine part with respect to the coordinate axes.
Fig. Q. 14
17. Determine the mass moment of inertia of the overhung crank about the x- axis. The material
is steel having a density of 7850 kg m3
⁄ .
Fig. Q. 15
18. Determine the co-ordinates of centroid of the shaded area shown in figure.

19. A Cylinder of height of 10 cm and radius of base 4 cm is placed under sphere of radius 4 cm
such that they have a common vertical axis. If both of them are made of the same material,
locate the centre of gravity of the combined unit.
20. Find the moment of inertia of the section shown in the figure about its horizontal
centroidalaxis.
21. Calculate the mass moment of inertial of the plate shown in figure with respect to the axis
AB. Thickness of the plate is 5 mm and density of the material is 6500 Kg/m3
.
22. Derive expression form mass moment of inertia of prism along three axes.
23. For the section shown in below, locate the horizontal and vertical centroidal axes.
24. Locate the centroid of the plane area shown in figure below.

UNIT IV FRICTION
1. A pull of 250 N inclined at 25° to the horizontal plane is required just to move a body kept on
a rough horizontal plane. But the push required just to move the body is 300 N. if the push is
inclined at 25° to the horizontal, find the wieght of the body and coefficient of friction.
2. A block weighing 1000 N is kept on a rough plane inlined at 40° to the horizontal. The
coefficient of friction between the block and the plane is 0.4. Determine the least force
inclined at 15° to the plane required just to move the block up the plane.
3. Two block of mass 20 kg and 40 kg are connected by a rope passing over a frictionless pulley
as shown in Fig. Q. 3. Assuming the coefficient of friction as 0.3 for all contact surfaces. Find
the tension in the string and the acceleration of the system. Also compute the velocity of the
system after 4 seconds starting from rest.
Fig. Q. 3
4. A body weighing 196.2 N slides up a ·30° inclined plane under the action o.f an applied force
300 N parallel to be plane. The coefficient of friction is 0.2. the body move from rest.
Determine at the end of 4 seconds, the acceleration, distance traveled, velocity, kinetic energy,
work done, momentum and impulse applied on the body.
The free body diagram is drawn as shown in fig. Q. 4.
Fig. Q. 4
5. A 100 N force acts on a 300N block placed on an inclined plane as shown in Fig. Q. 5. The
coefficients of friction between the block and the plane are µ𝑠 = 0.25 .and µ𝑘 = 0.20.
Determine whether the block is in equilibrium, and find the value of the friction force.
6. A ladder of weight 390 N and 6m long is placed against a vertical wall at an angle of 30° as
shown in Fig. Q. 7. The co-efficient of friction between the ladder and the wall is 0.25 and
between ladder and floor is 0.38. Find how high a man of weight 1170 N can climb without
sliding.

Fig. Q. 6
7. Two rough planes are joined together one of them is horizontal and the other is inclined at 45°
to the horizontal. A 100 kg block is on the inclined plane and is connected to a 60 kg block
on the horizontal plane through a cable passing over a smooth pulley at the junction of the
planes. A dragging force of A is applied on 60 kg block at an angle of θ to the horizontal. Find
the magnitude of the force and the value of θ for the motion is about to start. Assume𝜇 =
0.25.
8. A support block is acted upon by two forces as shown in Fig. Q.11. Knowing that the
coefficients of friction between the block and the incline are µ𝑠 = 0.35 and µ𝑘 = 0.25,
determine the force P required (i)To start the block moving up the incline, (ii) To keep it
moving up, (iii) To prevent it from sliding down.
Fig. Q. 8.
9. Two blocks A and B. are placed on inclined planes as shown in Fig. Q.13. The block A weighs
1000 N. Determine minimum weight of the block B for maintaining the equilibrium of the
system. Assume that the blocks are connected by an inextensible string passing over a
frictionless pulley. Coefficient of friction µ𝐴 between the block A and the plane is 0.25.
Assume the same value for µ𝐵.
Fig. Q. 9.
10. Determine the smallest force P required to move the block B if(a) block A is restrained by
cable CD as shown in Fig. Q. 15. (b) Cable CD removed. Take the coefficients of frictions as
µ𝑠 = 0.30, µ𝑘 = 0.15 between all surfaces of contact.

Fig. Q. 10.
11. Determine the smallest force P required to lift the 13.34 kN load shown in Fig. Q. 16. The
coefficient of static friction between A and C and B and D is 0.3 and that between A and B is
0.4.
Fig. Q. 11.
12. Determine the least and greatest value of W in Fig. Q. 14. to keep the system of connected
bodies in equilibrium. µ for surfaces of contact between plane AC and block = 0.28 and the
between plane BC and block = 0.2.
Fig. Q. 12.
13. Determine the force P required to intend the motion of the block B shown in Fig. Q. 18, to the
left. Take µ = 0.3 for all surfaces of Contact.
Fig. Q. 13.

14. A block of weight 1600 N is in contact with a place incline 30° to horizontal. A force ‘P’
parallel to the plane and acting up the plane µ= 0.2. Find (a) The value of 'P' to just cause the
motion. (b) The value of 'P' to prevent motion. (c) The magnitude and direction of frictional
force.
15. Block (2) rests on block (1) and is attached by a horizontal rope AB to the wall as shown in
fig. What force P is necessary to cause motion of block (1) to impend? The co-efficient of
friction between the blocks is ¼ and between the floor and block (1) is 1/3. Mass of blocks
(1) and (2) are 14kg and 9 kg respectively.
16. Block A weighing 1000 N rests on a rough inclined plane whose inclination to the horizontal
is 45°. It is connected to another block B, weighing 3000 N rests on a rough horizontal plane
by a weightless rigid bar inclined at an angle of 30° to the horizontal as shown in fig. Find the
horizontal force required to be applied to the block B just to move the block A in upward
direction. Assume angle of friction as 15° at all surfaces where there is sliding.
17. A 7m long ladder rests against a vertical wall, with which it makes an angle of 45° and on a
floor. If a man whose weight is one half that of the ladder climbs it, at what distance along the
ladder will he be, when the ladder is about to slip? Take coefficient of friction between the
ladder and the wall is 1/3 and that between the ladder and the floor is ½.

18. An effort of 200 N is required just to move a certain body up an inclined plane of angle 15°,
the force is acting parallel to the plane. If the angle of inclination of the plane is made 20°,
the effort required being again parallel to the plane, is found to be 230 N. Find the weight of
the body and coefficient of friction.
19. Find the force P inclined at an angle of 32° to the inclined plane making an angle of 25 degree
with the horizontal plane to slide a block weighing 125 KN (i) up the inclined plane (ii) Down
the inclined plane, when P = 0.5.
20. Two blocks A and B of mass 50kg and 100kg respectively are connected by a string C passes
through a frictionless pulley connected with the fixed wall by another string D as shown in
figure. Find the force P required to pull the block B. Also find the tension in the string D.
Take coefficient of friction at all contact surfaces as 0.3
21. A ladder is 8m long and weighs 300N.The centre of gravity of the ladder is 3m along the
bottom end. The ladder rests against a vertical wall at B and on the horizontal floor at A as
shown in Fig. Determine the safe height up to which a man weighing 900N can climb without
making the ladder slip. The coefficient of friction between ladder and floor is 0.4 and ladder
top and wall is 0.3
22. A body of weight 500N is pulled up an inclined Plane, by a force of 350N.The inclination of
the plane is 30o
to the horizontal and the force is applied parallel to the plane. Determine the
coefficient of friction.
UNIT-V DYNAMICS OF PARTICLES
1. A train is traveling from A to D along the track shown in fig. Its initial velocity at A is zero.
The train takes 5 min to cover the distance AB, 2250 m length and 2.5 minutes to cover, the
distance BC, 3000 m in length, on reaching the station C, the brakes are applied and the train
stops 2250 m beyond, at D (i) Find the retardation on CD, (ii) the time it takes the train to get
from A to D, and (iii) its average speed for the whole distance.
2. Two blocks ‘A’ and ‘B’ of masses 𝑚𝐴 = 280 𝑘𝑔 and 𝑚𝐵 = 420 𝑘𝑔 are joined by an
inextensible cable as shown in Fig. Q. 2. Assume that the pulley is frictionless and µ=0.30
between block ‘A’ and the surface. The system is initially at rest. Determine (i) acceleration

of block A, (ii) velocity after it has moved 3.5 m and (iii) velocity after 1.5 seconds.
Fig. Q. 2.
3. The two blocks in Fig. Q. 3. start from rest. The horizontal plane and the pulley are
frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of
each block and the tension in each cord.
Fig. Q. 3
4. During a brake test, the rear-engine car is stopped from an initial speed of 100 km/h in a
distance of 50 m. If it is known that all four wheels contribute equally to the braking force,
determine the braking force F at each wheel. Assume a constant deceleration for the 1500-kg
car.
5. A block and pulley system is shown in Fig. Q. 5. The coefficient of kinetic friction between
the block and the plane is 0.25. The pulley is frictionless. Find the acceleration of the blocks
and the tension in the string when the system is just released. Also find the time required for
100 kg block to come down by 2 m.
Fig. Q. 5.
6. A particle starting from rest, moves in a straight line and its acceleration is given by

𝑎 = 50 − 36t2
m s2
⁄ where t is in sec. Determine the velocity of the particle when it has
traveled 52 m.
7. A body moves along a straight-line so that its displacement from a fixed point on the line is
given by s = 4t3
− 6t2
+ 20. Find the displacement, velocity and acceleration at the end of
3 seconds.
as shown in Fig. Q. 8. Assuming the coefficient of friction as 0.3 for all contact surfaces. Find
the tension in the string and the acceleration of the system. Also compute the velocity of the
system after 4 seconds starting from rest.
Fig. Q. 8.
9. A body weighing 196.2 N slides up a 30° inclined plane under the action of an applied force
300 N parallel to be plane. The coefficient of friction is 0.2. The body move from rest.
Determine at the end of 4 seconds, the acceleration, distance traveled, velocity, kinetic energy,
work done, momentum and impulse applied on the body. The free body diagram is drawn as
shown in Fig. Q. 19.
Fig. Q. 9.
10. A block of mass 5 kg is released from rest from a position A on a 30° incline as shown in
Fig. Q. 10. Determine the maximum compression of the spring if the spring constant is
800 𝑁 𝑚
⁄ and the coefficient of friction between block and the incline is 0.2.

Fig. Q. 10.
11. A 50 N block is released from rest on an inclined plane making an angle of 35° to the
horizontal. The block starts from A slides down a distance of 1.2 m and strikes a spring with
stiffness of 8 kN/m. The μ between block and plane is 0.25. Determine:
(i) the amount of the spring gets compressed and
(ii) Distance the block will rebound up the plane from the compressed position.
Fig. Q. 11
12. Automobile A is traveling east at the constant speed of 36 km/h. As automobile A crosses the
intersection shown, automobile B starts from rest 35 m north of the intersection and moves
south with a constant acceleration of 1.2 m/s2
. Determine the position, velocity, and
acceleration of B relative to A 5 s after A crosses the intersection.
13. The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the
effect of friction in the pulleys and between block A and the incline, determine (a) the
acceleration of each block, (b) the tension in the cable.
θ

Fig. Q. 13.
14. The position of the particle is given by the relation S=1.5t3
-9t2
-22.5t+60, where S is expressed
in meters and t in Determine (i) the time at which the velocity will be zero (ii) the position
and distance traveled by the particle at that time (iii) the acceleration of the particle at that
time and (iv) the distance traveled by the particle from t = 5s to t = 7s.
15. Two Blocks A and B of weight 100 N and 200 N respectively are initially at rest on a 30°
inclined plane as shown in figure. The distance between the blocks is 6 m. The co efficient of
friction between the block A and the plane is 0.25 and that between the block B and the plane
is 0.15. If they are released at the same time, in what time the upper block (B) reaches the
Block (A).
16. Two blocks of weight 150 N and 50 N are connected by a string and passing over a frictionless
pulley as shown in figure. Determine the acceleration of blocks A and B and the tension in
the string.

17. Two weights 80 N and 20 N are connected by a thread and move along a rough horizontal
plane under the action of a force 40 N, applied to the first weight of 80 N as shown in figure.
The coefficient of friction between the sliding surfaces of the wrights and the plane is 0.3.
Determine the acceleration of the weights and the tension in the thread using work-energy
equation.
18. A stone dropped into a well is heard to strike the water after 4 seconds. Find the depth of the
well, if the velocity of sound is 350 m/sec.
as shown in Fig. Q. 19. Assuming the coefficient of friction as 0.3 for all contact surfaces.
Find the tension in the string and the acceleration of the system. Also compute the velocity of
the system after 4 seconds starting from rest.
Fig. Q. 19
20. Block P of weight 100N and block Q of weight 50N are connected by a rope that passes over
a smooth pulley as shown in figure. Find the acceleration of the blocks and the tension in the
rope, when the system is released from rest. Neglect the mass of the pulley.

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