IFET COLLEGE OF ENGINEERING

Regulation: 2013 Academic Year: 2014-2015

IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1

IFET COLLEGE OF ENGINEERING

DEPARTMENT OF MECHANICAL ENGINEERING

SUBJECT CODE: GE6253- ENGINEERING MECHANICS

YEAR/SEM: I/II

UNIT-V- FRICTION

PREPARED BY

1. Mr. R.KAMALANATHAN (LECT/MECH)

2. Mr. K.VENKATESAN (LECH/MECH)

3. Mr. K.KAMALAKANNAN (LECT/MECH)

MODULE CO-ORDINATOR

Mr.R.VETRI (ASP/MECH)



COURSE OUTCOME:

CO5: Students are able to understand the effect of friction between plane surfaces

PO a Apply knowledge of computing, mathematics, science and engineering fundamentals

appropriate to the mechanical engineering.

PO b Identify, formulate, research literature and solve complex engineering problems

reaching substantiated conclusions using first principles of mathematics and

engineering sciences.

PO c Design solutions for complex problems and design mechanical based system,

process, component, or program to meet desired needs with appropriate

consideration for public health and safety, cultural, societal and environmental

considerations.

PO d Ability to design and conduct experiments, as well as to analyze and interpret data.

PO e Create, Select and apply appropriate techniques, resources, and modern engineering

tools, including prediction and modeling, to complex engineering activities, with an

understand of the limitations.

PO f Function effectively as an individual and as a member or leader in diverse team and

in multi-disciplinary setting.

PO g Communicate effectively on complex engineering activities with the engineering

community and with society at large, such as being able to comprehend and write

effective reports on design documentation, make effective presentation, and give and

receive clear instructions.

PO h Demonstrate understand of the societal, health, safety, legal, and cultural issues and

the consequent responsibilities relevant to engineering practices.

PO i Understand and commit to professional ethics and responsibilities and norms of

engineering practice.

PO j Understand the impact of engineering solutions in a societal context and demonstrate

knowledge of and need of sustainable development.

PO k Demonstrate knowledge and understand of management and business practice such

as risk and change management and understand their limitations.

PO l Recognize the need for ,and have the ability to engage in independent and life-long

learning



Introduction- Friction

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and

material elements sliding against each other. A resistance encountered when one body moves

relative to another body with which it is in contact. The property of body by virtue which a

force is exerted by a stationary body on the moving body to resist the motion of the moving

body is called friction.

A simple way to understand is that when a horizontal force is applied to a static body with an

intention to move the same, a frictional force equal to the applied force develops in the

opposite direction resisting the motion. As long as the body does not move, this force is

called static frictional force. Now if the applied force is increased, the frictional force in the

opposite direction increases proportionately until it reaches the limit after which if the applied

force is increased, the body starts moving. This threshold force is called static or limiting

force of friction.

The friction which exists between two surfaces that are not lubricated is called as solid

friction. The two Surfaces can be at rest or one of the surfaces is moving and other surface is

at rest. The following are laws of solid friction. The force of friction acts in opposite direction

in which the surface is having tendency to move. The force of friction is equal to force

applied to surfaces, so long as surface is at rest. When surface is on point of motion, force of

friction is highest and this maximum frictional force is called as limiting friction force.

The ratio between the limiting friction and normal reaction is a bit less when the two surfaces

are in motion. The force of friction is independent of velocity of sliding. The above laws of

solid friction are called as laws of static and dynamic friction or law of friction.



Velocity- Velocity is the rate of change of the displacement, the difference between the final

and initial position of an object. Velocity is equivalent to a specification of its speed and

direction of motion, e.g. 60 km/h to the north. Velocity is an important concept in kinematics,

the branch of classical mechanics which describes the motion of bodies. Velocity is a vector

physical quantity; both magnitude and direction.

Acceleration- Acceleration, in physics, is the rate of change of velocity of an object. An

object's acceleration is the net result of any and all forces acting on the object, as described by

Newton's Second Law. The SI unit for acceleration is the meter per second squared (m/s2).

Accelerations are vector quantities ,they have magnitude and direction. As a vector, the

calculated net force is equal to the product of the object's mass a scalar quantity and the

acceleration.

The angular velocity is defined as the rate of change of angular displacement and is a vector

quantity (more precisely, a pseudovector) which specifies the angular speed (rotational speed)

of an object and the axis about which the object is rotating. The SI unit of angular velocity is

radians per second, although it may be measured in other units such as degrees per second,

degrees per hour, etc. Angular velocity is usually represented by the symbol omega (ω, rarely

Ω). The direction of the angular velocity vector is perpendicular to the plane of rotation, in a

direction which is usually specified by the right-hand rule.



SYLLABUS

Frictional force – Laws of Coloumb friction – simple contact friction – Rolling resistance –

Belt friction. Translation and Rotation of Rigid Bodies – Velocity and acceleration – General

Plane motion.

2 mark

Frictional force – Laws of Coulomb friction

1. What is meant by friction?(nov2012)

Friction is the force resisting the relative motion of solid surfaces, fluid layers,

and material elements sliding against each other. A resistance encountered when one

body moves relative to another body with which it is in contact. The property of body

by virtue which a force is exerted by a stationary body on the moving body to resist

the motion of the moving body is called friction.

2. Define frictional force.

A resistance force encountered when one body moves relative to another body

with which it is in contact. The force exerted by the body is called the friction force

and is always acting in the direction opposite to the direction of the motion.

3. What is limiting force of friction?

A simple way to understand is that when a horizontal force is applied to a

static body with an intention to move the same, a frictional force equal to the applied

force develops in the opposite direction resisting the motion. As long as the body does



not move, this force is called static frictional force. Now if the applied force is

increased, the frictional force in the opposite direction increases proportionately until

it reaches the limit after which if the applied force is increased, the body starts

moving. This threshold force is called static or limiting force of friction.

4. Define static friction?

Static friction is friction between two or more solid objects that are not

moving relative to each other. For example, static friction can prevent an object from

sliding down a sloped surface. The coefficient of static friction, typically denoted as

μs, is usually higher than the coefficient of kinetic friction. The static friction force

must be overcome by an applied force before an object can move. The maximum

possible friction force between two surfaces before sliding begins is the product of the

coefficient of static friction and the normal force.

5. Define kinetic friction.

Kinetic friction, or dynamic friction, is defined as the form of friction between

two bodies via their surface of contact (i.e., the force acting parallel to the plane of

contact) when the two surfaces of contact are slipping against each other. Kinetic

friction is the form of friction that opposes the slipping against each other of two

surfaces in contact. Its direction is opposite the direction of relative motion of the

surfaces in contact. Note that it need not oppose the net external force between the

surfaces in contact. Kinetic friction refers to the frictional force of a moving object.

6. Define co-efficient of friction.

A coefficient of friction is a value that shows the relationship between the

force of friction between two objects and the normal force between the objects. It is a

value that is used in physics sometimes to find an objects normal force or frictional

force, when other methods aren't available. The coefficient of friction is

dimensionless, meaning it does not have any units. It is a scalar, meaning the direction

of the force does not change its magnitude.

7. Write about angle of friction.(Dec2013)



Consider a block placed on a rough floor. Now, the reaction force is because it

is equal and opposite to the weight. Now apply a horizontal force so that the block

just begins to slide, i.e., the frictional force is equal to the limiting friction. When this

condition is satisfied, the angle which the resultant (between the normal and external

force) makes with the vertical is called the angle of friction.

8. Write about cone friction.

A cone in which the resultant force exerted by one flat horizontal surface on

another must be located when both surfaces are at rest, as determined by the

coefficient of static friction.

9. Write the types of friction.

10. What is solid (dry) friction?

Dry friction resists relative lateral motion of two solid surfaces in contact. Dry

friction is subdivided into static friction ("stiction") between non-moving surfaces,

and kinetic friction between moving surfaces. The force of friction is always exerted

in a direction that opposes movement (for kinetic friction) or potential movement (for

static friction) between the two surfaces.

11. State the laws of solid friction.



The friction which exists between two surfaces that are not lubricated is called

as solid friction. The two Surfaces can be at rest or one of the surfaces is moving and

other surface is at rest. The following are laws of solid friction.

i. The force of friction acts in opposite direction in which the surface is having

tendency to move.

ii. The force of friction is equal to force applied to surfaces, so long as surface is at

rest.

iii. When surface is on point of motion, force of friction is highest and this maximum

frictional force is called as limiting friction force.

vi. The ratio between the limiting friction and normal reaction is a bit less when the

two surfaces are in motion.

vii. The force of friction is independent of velocity of sliding.

The above laws of solid friction are called as laws of static and dynamic friction or

law of friction.

12. A body of weight 130N is placed on a horizontal plane. Horizontal force is given as

53N. Find the angle of friction?

Here, W=R

F=53N

RF

=53/130

=0.408

tanӨ=

Ө=tan-1 µ= tan-1(0.408)

Ө=22.19®

13. What is the causes of friction?

Following are the main causes of friction:

i.the interlocking of irregularities in the two surfaces in contact

ii.nature of surface i.e. smooth or rough

iii.mass of the body of surfaces in contact

iv.shape of surface (e.g. flat or curved)

What is the cause of friction? What is the cause of friction?

14. What is meant by fluid friction?

Fluid friction is developed between two surfaces in the presence of fluid when

adjacent layers of fluid are moving at different velocities. Fluid friction occurs

between layers within a fluid that are moving relative to each other. This internal

resistance to flow is described as viscosity. In everyday terms viscosity of a fluid is

said to have "thickness". Thus, water is "thin", having a lower viscosity, while honey

is "thick", having a higher viscosity. The less viscous the fluid, the greater its ease of

movement.

http://en.wikipedia.org/wiki/Fluid

http://en.wikipedia.org/wiki/Viscosity



15. What is meant by dynamic friction?

If the surface starts moving on the other which is at rest, then the force

experienced by the moving surface is called dynamic friction. Kinetic (or dynamic)

friction occurs when two objects are moving relative to each other and rub together

(like a sled on the ground). The coefficient of kinetic friction is typically denoted

as μk, and is usually less than the coefficient of static friction for the same

materials. However, Richard Feynman comments that "with dry metals it is very hard

to show any difference.

16. What are the types of dynamics friction?

If the surface starts moving on the other which is at rest, then the force

experienced by the moving surface is called dynamic friction.

There are two types of dynamic friction are

1. Sliding friction,

2. Rolling friction

17. A body of weight 100N is placed on a horizontal plane. Horizontal force is given as

55N. Find the coefficient of friction?

Given data: frictional force=55,normal force=100

The coefficient of friction µ=frictional force/normal reaction

µ=F/R

µ=55/100

=0.55

18. What is meant by sliding friction?

It is the friction experienced by a body when it slides over another body. The term

sliding friction refers to the resistance created by two objects sliding against each

other. This can also be called kinetic friction. Sliding friction is intended to stop an

object from moving.

Examples of sliding friction

The surface deformation of objects

The roughness/smoothness of the surface of the objeects

The original speed of either object

19. What is meant by rolling friction?

It is the friction experienced by a body when it rolls over another body. is the

force resisting the motion when a body (such as a ball, tire, or wheel) rolls on a

surface. It is mainly caused by non-elastic effects; that is, not all the energy needed

for deformation (or movement) of the wheel, roadbed, etc

http://en.wikipedia.org/wiki/Richard_Feynman

http://www.yourdictionary.com/kinetic

http://en.wikipedia.org/wiki/Motion_(physics)

http://en.wikipedia.org/wiki/Ball

http://en.wikipedia.org/wiki/Tire

http://en.wikipedia.org/wiki/Wheel

http://en.wikipedia.org/wiki/Plasticity_(physics)



20. Where is friction involved?

Friction is involved in

wedges,

ladder,

wheels,

square threaded screws,

journal bearings,

thrust bearings, etc

21. Write about impending motion.

The state of motion of a body which is just about to move or slide is called as

impending motion. When the maximum frictional force is attained and if the applied

force exceeds the limiting friction, the body starts sliding or rolling. This state is

called as impending motion.

22. Define angle of repose.

Angle of repose is defined as the minimum angle made by an inclined plane

with the horizontal such that an object placed on the inclined surface just begins to

slide.

23. Write about equilibrium of a body when it is placed on a inclined plane.



When angle of friction is more than the angle of repose the body will need an

external force to maintain equilibrium.

The external force may be applied in the following direction

i.along the plane

ii.horizontal to the plane

iii.inclined at an angle with the inclined plane

24. What is wedge?

A wedge is a place of metal or wood in the shape of a prism whose cross section is

usually a triangle or trapezoid. It is used for lifting heavy loads and tightening fits.

It can be used to separate two objects or portions of an object, lift up an object, or

hold an object in place. It functions by converting aforce applied to its blunt end

into forces perpendicular (normal) to its inclined surfaces

25. Define rolling resistance.

It is defined as one of the friction occurs because of the deformation of the

surface under a rolling load. Rolling resistance is the force that resists the rolling of a

wheel or other circular object along a surface caused by deformations in the object

and/or surface. Generally the force of rolling resistance is less than that associated

with kinetic friction. Typical values for the coefficient of rolling resistance are

0.001. One of the most common examples of rolling resistance is the movement

of motor vehicle tires on a road, a process which generates heat and sound as by-

products.

26. What is ladder friction?

Ladder is a device for climbing or scaling on the roofs or walls. It is made by

wood, iron or rope connected by number of cross pieces called ladder. The

friction occur in a ladder is

http://en.wikipedia.org/wiki/Force

http://en.wikipedia.org/wiki/Surface_normal

http://en.wikipedia.org/wiki/Motor_vehicle

http://en.wikipedia.org/wiki/Road

http://en.wikipedia.org/wiki/Roadway_noise



27. A vertical wall of weight 300KN is subjected to lateral forces of 3KN. Will it be safe

against string on a horizontal plane on which the co efficient of friction=0.01?

Friction force

F=µ*weight of wall

=0.01*300

=3kn

Here in the frictional force is equal to the lateral force. So string will take

place and hence it is safe.

28. A ladder is placed over the wall as shown in fig. Fine out whether it is in equilibrium

position. Take µ=0.3

Taking moment about A

F2*12=250*2.5

F2=625/12=52.08N

Fx=0,

F1=F2

=52.08N=F

Fy=0

R=250

Limiting friction Fmax=µR=0.3*250=75N

Here F<Fmax,so the ladder is in equilibrium.

29. Write about friction in belt.

Belt friction is a term describing the friction forces between a belt and a surface, such

as a belt wrapped around a bollard. When one end of the belt is being pulled only part

of this force is transmitted to the other end wrapped about a surface. The friction force

increases with the amount of wrap about a surface and makes it so the tension in the

belt can be different at both ends of the belt. In practice, the theoretical tension acting

on the belt or rope calculated by the belt friction equation can be compared to the

maximum tension the belt can support.

30. Draw open belt drive.

A belt drive having parallel shafts rotating in same directions.



31. What is velocity ration of belt drive?

The ratio of the velocity of the follower to the velocity of driver is known as velocity

ratio.

Velocity ratio, N1/N2 = d1/d2

N1-speed of the driver N2-speed of the follower

D1-diameter of driver d2-diameter of follower

32. Draw cross belt drive.

A belt drives having parallel shafts rotating in opposite directions. it is the

drive where the driver pulley is compressed by belt on slack side and with less friction

on the driven pulley with opposite direction of motion.

33. Write about the length of open and cross belt drive.

The length of the belt means the total length of the belt required to connect a driver

and a follower.

Open belt

L= x

xrr

rr 221 2

21

Cross belt

L= x

xrr

rr 221 2

21

where r1-radius of the larger pulley

r2- radius of the smaller pulley

L- Length of the belt

x- Distance between two centres

34. What is the ratio of belt tension?

T1/T2 = e

Where, T1- Tension in the belt on the tight side

T2- Tension in the belt on the slack side



- Angle of contact

- Co-efficient of friction between the belt and pulley

35. Determine the minimum tension in the rope required to support a cylinder of mass

500Kg, when the rope passes once over the rod.

T1/T2 = e

T2=T1/ e

=(500*9.81)/ e 2.0

T2=2616.76N

36. Write the expression for power transmitted by belt.

The expression for power transmitted by belt

P=

kWxVTT

100021

Where, T1- Tension in the belt on the tight side


V – Velocity of the belt

37. Define centrifugal tension and initial tension in belt drive.

The tension caused in the running belt by the centrifugal force is known as

centrifugal tension.

Tc= m x v2

Where, m- Mass of belt per metre length

v- Velocity of the belt

38. What is V-belt drive and write its application?

V-Belts are friction based power or torque transmitters. The power is

transmitted from one pulley to the other by means of the friction between the belt and

pulley. The rubber used as the base material plays a very vital role in this. This is

quite similar to the friction between the Tyre and road in the automobiles that enables

the automobiles to move on the road.

a. A series of V Belts used to drive a gearbox from the motor

b. A V Belt is commonly used in the Lathe

39. What is rope drive?

Wire rope, is a type of cable which consists of several strands of metal wire

laid (twisted) into a helix. The term cable is often used interchangeably with

wire rope.

Wire ropes are used dynamically for lifting and hoisting in cranes and

elevators, and for transmission of mechanical power.

40. Write the application of rope drive.

Rope drives are wildly used where large amount of power is to be transmitted

from one pulley to another pulley.



a) It is used in spinning mill.

b) The rope drive is used in lift in multi storage building. The line shaft of each floor

is solvent by rope rope by passing directly from main engine pulley.

41. What is the ratio of rope tension?

T1/T2 = e eccos

T1- Tension in the belt on the tight side


- Angle of contact

- Co-efficient of friction between the belt and pulley

2 - Angle of groove

42. Define chain drive

Chain drive is a way of transmitting mechanical power from one place to another. It is

often used to convey power to the wheels of a vehicle, particularly bicycles and

motorcycles. It is also used in a wide variety of machines besides vehicles.

Most often, the power is conveyed by a roller chain, known as the drive chain or

transmission chain,[1] passing over a sprocket gear, with the teeth of the gear meshing

with the holes in the links of the chain.

43. Define Velocity.

Velocity- Velocity is the rate of change of the displacement, the difference

between the final and initial position of an object. Velocity is equivalent to a

specification of its speed and direction of motion, e.g. 60 km/h to the north. Velocity

is an important concept in kinematics, the branch of classical mechanics which

describes the motion of bodies. Velocity is a vector physical quantity; both magnitude

and direction.

44. Define acceleration.

Acceleration- Acceleration, in physics, is the rate of change of velocity of an

object. An object's acceleration is the net result of any and all forces acting on the

object, as described by Newton's Second Law. The SI unit for acceleration is the

meter per second squared (m/s2). Accelerations are vector quantities ,they have

magnitude and direction. As a vector, the calculated net force is equal to the product

of the object's mass a scalar quantity and the acceleration.

45. Write the mathematical expression for velocity.

Velocity Formula: Velocity is the measure of the speed of the object in a specific

direction.

It is denoted by V and is given as



Where S is the displacement and, t is the time taken.

Since displacement is expressed in meters and time taken in seconds. Velocity is

expressed in meters/second or m/s.In any problem if any of these two quantities are

given we can find the missing quantity using this formula.

46. Write short notes on speed vs velocity.

Speed only gives you a number that tells you how fast you are going. Velocity,

because it adds direction, tells you how fast you are changing your position. Because

of this difference, if your position doesn't change even if you are moving very fast,

your velocity will be zero. If you were to run in place very fast, your speed may be 6

mph, but your velocity would be 0 because you aren't going anywhere. If you were to

run backwards and forwards, always returning to your same spot, your velocity would

be 0 again because you didn't go anywhere.

47. Write the equation of initial, final velocity formula.

Generally the initial velocity is denoted by ‘u’ and the final velocity is denoted by

‘v.’In one dimensional motion, it has more importance to solve the displacement(s),

velocity (u or v) and acceleration (a).

The following equations can be used when the particle moving with a constant

acceleration.

1. v = u + at

2. s = 12(u + v)t

3. s = ut + 12at2

4. v2 = u2 + 2as

48. What is average velocity?

Average velocity- Velocity is defined as the rate of change of position with respect

to time. Sometimes it is easier, or even necessary, to work with the average velocity

of an object, that is to say the constant velocity, that would provide the same

resultant displacement as a variable velocity, v(t), over some time period Δt.

Average velocity can be calculated as:

The average velocity is always less than or equal to the average speed of an

object. This can be seen by realizing that while distance is always strictly

increasing, displacement can increase or decrease in magnitude as well as change

direction.



49. Define instantaneous velocity?

Instantaneous velocity is the velocity of an object in motion at a specific point in

time. This is determined similarly to average velocity, but we narrow the period of

time so that it approaches zero. If an object has a standard velocity over a period of

time, its average and instantaneous velocities may be the same. The formula for

instantaneous velocity is the limit as t approaches zero of the change in d over the

change in t.

50. Write the formula to find out the instantaneous velocity?

Instantaneous Velocity Formula is used to determine the instantaneous

velocity of the given body at any particular instant. It is given as:

Where x is the given function with respect to time t. The Instantaneous

Velocity is expressed in m/s. If any problem contains the function of the form f(x), the

instantaneous velocity is determined using the above formula.

51. What is angular velocity?

The angular velocity is defined as the rate of change of angular displacement

and is a vector quantity (more precisely, a pseudovector) which specifies the angular

speed (rotational speed) of an object and the axis about which the object is rotating.

The SI unit of angular velocity is radians per second, although it may be measured in

other units such as degrees per second, degrees per hour, etc. Angular velocity is

usually represented by the symbol omega (ω, rarely Ω).

The direction of the angular velocity vector is perpendicular to the plane of

rotation, in a direction which is usually specified by the right-hand rule.

52. Write the advantages and disadvantages of chain and belt drive.

(i) While belt drives on a motorcycle require less maintenance and are

smoother, they are not as efficient as chain drives. Consequently, chain drives are

more popular.

(ii) A chain drive, on the other hand, delivers power more economically than a

belt drive, and it sends more of the power that the engine produces to the wheel.

While they require more maintenance, the improved performance means they are the

more common type of drive method.

53. Define co-efficient of restitution.(Dec-2002)

It is the magnitudes of impulses corresponding to the period of restitution and

to the period of deformation is called coefficient of restitution and it is given by

http://en.wikipedia.org/wiki/Angular_displacement

http://en.wikipedia.org/wiki/Vector_%28geometry%29

http://en.wikipedia.org/wiki/Pseudovector

http://en.wikipedia.org/wiki/Angular_speed

http://en.wikipedia.org/wiki/Angular_speed

http://en.wikipedia.org/wiki/Rotational_speed

http://en.wikipedia.org/wiki/SI

http://en.wikipedia.org/wiki/Radians_per_second

http://en.wikipedia.org/wiki/Omega

http://en.wikipedia.org/wiki/Right-hand_rule



E=relative velocity of separation/relative velocity of approach

AfBf

BiAi

VVVV

I

54. Define instantaneous centre of rotation in plane motion?

In general plane motion of a rigid body at any given instant the velocities of

various particles of the slab are same as if the slab bear rotating about certain

axis perpendicular to the plane of the slab.

Instantaneous centre of rotation is a point identified with in a body where the

velocity is zero.

55. A steel ball is vertically thrown upwards from the top of the building 25 m above the

ground with an initial velocity of 18 m/sec. Find the max height reached by the ball

from the ground. (Apr/May-2003)

a. Solution:

1. V = u + at

2. 0 = 18 – 9.81 x t

3. t = 1.834

4. s = 18t – (9.81/2) t2

5. = 16.51

6. H= 25 + s

7. H = 25 + 16.51 = 41.51 m.

56. Ablock having mass of 50kg has a velocity of 15m/s horizontally on a smooth

frictionless surface. Determine the value of the horizontal force to be applied to the

block for bringing the block to rest in 5 sec.

Given data:

M=50kg,v=15m/s,t=5sec

Solution.

G=momentum

G2=G1+I1-2

0=50*15-F*5

F=150N

57. Define linear momentum.

linear momentum is a vector quantity, possessing a direction as well as a

magnitude:

http://en.wikipedia.org/wiki/Euclidean_vector



Linear momentum is also a conserved quantity, meaning that if a closed

system is not affected by external forces, its total linear momentum cannot change. In

classical mechanics, conservation of linear momentum is implied byNewton's laws;

but it also holds in special relativity (with a modified formula) and, with appropriate

definitions, a (generalized) linear momentumconservation law holds

in electrodynamics, quantum mechanics, quantum field theory, and general relativity.

58. Give the relationship with velocity and acceleration.

Although velocity is defined as the rate of change of position, it is often

common to start with an expression for an object's acceleration. As seen by the three

green tangent lines in the figure, an object's instantaneous acceleration at a point in

time is the slope of the line tangent to the curve of a (v vs. t graph at that point. In

other words, acceleration is defined as the derivative of velocity with respect to time:

From there, we can obtain an expression for velocity as the area under an acceleration

vs. time (a vs. t) graph. As above, this is done using the concept of the integral:

59. A homogeneous circular disk, 1.8m diameter, has a mass of 365 kg and is made to

revolve about an axis passing through its centre by a force of 350 N applied

tangentially to its circumference. Determine the angular acceleration of the disk.

2

22

2

2

2

sec/0654.1

sec/8.1365

7008.1365

28.1350

8.13502

8.13652

rad

rad

J

mrJ

60. What is called the Coefficient of static friction?

As the force P increases, F also increases but the bogy remains at rest and is in

equilibrium. If F reaches a limiting value friction or from when P is increased it loses its

balance and hence the body slides to right.

http://en.wikipedia.org/wiki/Closed_system

http://en.wikipedia.org/wiki/Closed_system

http://en.wikipedia.org/wiki/Momentum#Conservation

http://en.wikipedia.org/wiki/Newton%27s_laws

http://en.wikipedia.org/wiki/Special_relativity

http://en.wikipedia.org/wiki/Conservation_law_(physics)

http://en.wikipedia.org/wiki/Electrodynamics

http://en.wikipedia.org/wiki/Quantum_mechanics

http://en.wikipedia.org/wiki/Quantum_field_theory

http://en.wikipedia.org/wiki/General_relativity

http://en.wikipedia.org/wiki/Acceleration

http://en.wikipedia.org/wiki/Slope

http://en.wikipedia.org/wiki/Tangent



NFN

F

NF

NF

s

tan

tan

s =is called coefficient of static friction.

16 mark

1. A ladder of weight 1000 N and length 4 m rests as shown in fig. If a 750 N weight is

applied at a distance of 3 m from the top of ladder, it is at the point of sliding.

Determine the coefficient of friction between ladder and the floor. (Anna univ,

May/June 2010) (16)

Solution:

Let the normal reaction at the floor and the wall be Ra and Rb respectively.

Frictional force at the floor=µ Ra

Frictional force at the wall =µ RB

Where,

µ=coefficient of friction.

Resolving the forces on the ladder horizontally and vertically,

1µRA BR

21750 µ BA RR

From equation 1 & 2,

1750)µ(µ AA RR

1750µ2 AA RR

1750)1µ( 2 AR

)1µ/(1750 2AR



1)(µµ1750

2BR

Taking moment about A,

60cos48.3µ60sin48.3)60cos1(750)60cos3(1000 BB RR

BB RR µ74.1)0137.3()3751500(

BB RR µ74.10137.31875

1µµ1750

74.11µµ1750

0137.31875 22

01.602-µ5076.4µ2

0.331µ

2. From the fig calculate α so that the motion of lower block can just slide down from

the plane. The weight A and B are 30 N and 90 N respectively. The coefficient of

friction for all contact surfaces is 1/3. (Anna univ, Nov/Dec 2011)

(16)

Givendata:

NWA 30 , NWB 90

0.3331

µ

Solution:

Considering 30N block

Resolving the forces vertically,

030cos-R 0,Fv 1

130cosR1

From law of friction, 11 R41

F



cos3031

1 F ,

2cos101 F

Considering 90N block:


0cos90- ,R-R ; 0Fv 12

cos90RR 12

Substituting the value of (R1) from 1 we get

)cos90cos30(R2

3cos120R2

22 R31

F

4120cos31

2 F

cos402F

To find the angle of inclination of plane:

Resolving the forces horizontally,

,0 HF

0sin9021 FF

0)sin90cos40cos10(

sin90cos50

90/50tan

'0329 .

3. An effort of 150 N is required to just move a body up a rough plane inclined at just

12º with the horizontal, the force being parallel to the plane. If the plane were 15º

with the horizontal, the force required was 172 N. Find the weight of the body and

the coefficient of friction. (Anna univ, Nov/Dec 2007) (16)

Given data:

12;150 11 NP

15;172 22 NP

?µ?; W

Soiution:



For keeping the body in equilibrium, when the force parallel to the inclined plane,

The maximum force can be calculated by the following formula,

)sinW(µcosP

Case(i)

12;150 11 NP

)12sin12W(µcos150

-1-----0.2079WµW978.0150

Case(ii)

15;172 22 NP

)15sin15W(µcos172

-2-----0.2588WµW9659.0172

Divide equation 1 by 0.978, we get

WW 2125.0µ37.153

Divide eqn 2 by 0.9659 we get

WW 2679.0µ07.178

Solving eqn 3,4………. NW 85.445

Sub the value of W in eqn 4

445.85))(0.2679µ(445.85)(07.178

0.131µ .

4. Find the least value of P for the problem shown in fig to cause the motion to impend.

(Anna univ, Nov/Dec 2011) (16)

Given data:

WA=2.5KN,

WB=2KN

α=50°,θ=50°



solution:

considering the block(B);

normal reaction=weight of the body

RB=2KN


050cos BFp

,0 HF 050cos BBRp

Assume , coefficient of friction between block (B) and the surface, 25.0B

1025.050cos BRp

Resolving forces vertically,

050sin2 pRB

Considering the block (A)


cosWRA

KNRA 607.150cos5.2


sinWFP AA

sinWRP AAA

50sin2)607.1( AAP ( 30.0A )

KNPA 014.2

Considering the block (B):


0Fv , 0sin PRW

050sin2 pR

)250sin( pR


,0 HF 0cos APFp

0014.250cos RP

0014.2)250sin(50cos pP

KNp 202.1 .

5. A block of weight 1290 N rests on a horizontal surface and supports another block of

weight 570 N on top of it as shown in fig. Find the force P applied to the lower block

that will be necessary to cause motion to impend. The coefficient of friction between



blocks 1 and 2 is 0.25 and between block 1 and the floor is 0.40.(Anna univ,

May/June 2011)

(16)

Givendata:

W1=1290N; W2=570N

Let

Coefficient of friction between 1 and 2 is µ1=0.25

Coefficient of friction between 1 and floor is µ2=0.40

Solution:

Considering block 2

43

tan , )4/3(tan1 , 5236


0)5705236sin1(;0 TRFv

15705236sin 1 RT


05236cos;0 1 FTFH

15236cos FT , 1µ5236cos RT



20.255236cos 1 RT

Using equation 1 & 2

1

1

25.0570

7499.0RR

11 5701875.0 RR , NR 4801

11 µRF , N12048025.0

Considering block 2

The downward force of block 2 is equal to R1 will also act along the weight of the

block

Resolve the forces vertically

12 1290,0 RRFv

NR 177048012902

222 µ RF

NF 708177040.02

Resolve the forces horizontally

0HF , 70812021 FFp

NP 828 .

6. Block A weighing 1.5 kN rests on a horizontal plane and supports another block

weighing 500 N on top of it as shown in fig. The block B is attached to a vertical wall

by an inclined string, which makes an angle of 45º with the vertical. What should be

the value of force P acting at an angle of 30º to the horizontal to cause the motion of

the lower block to impend? Take µ= 0.28 for all surfaces. (Anna univ, May/June

2011) (16)



Given data:

1.5 ; 500A BW kNW N

0.28

Solution:

Considering Block(B)

Resolving the forces horizontally

0Fx ,

1 sin45 500 0R T

1sin45 500 1T R

Resolving the forces vertically

1

1

1

0

cos45 0

cos45 0

cos45 0.28 0 2

Fy

T F

T R

T R

using equations 1&2

1

1

500tan45

0.28RR

1 10.28 500R R

11.28 500R

1 390.625R N

1 1F R

1 109.375F N

Considering the block A




0Fv , 2 11500 sin30R R P

2 2F R , 2 0.28(1500 390.625 sin30 )F P


0HF

1 2sin30P F F

sin30 109.375 0.28(1890.625 sin30)P P

1.28sin30 638.75 , 998.05P N .

7. Two block A and B of mass 50 kg and 100 kg respectively are connected by a string

C which passes through a frictionless pulley connected with the fixed wall by

another string D as shown in fig. Find the force P required to pull the block B. Also

find the tension in the string D. Take coefficient of friction for all contact surfaces as

0.3. (Anna univ, Nov/Dec 2010) (16)

Given data:

3.0981100

5.49050

NkgWNkgW

B

A

Solution:

Considering Block(A)

Weight of block A is equal to the normal reaction RA

NRFRF

RNW

AA

AA

AA

15.1475.4903.03.0

5.490

Since, A does not move, the tension T also,

NFT A 15.147

Considering block(B)




NpTp

TpTRp

F

B

H

65.14715.1471

0)5.14713.0(0

0

2

2

8. A screw jack has square threads of mean diameter of 10 cm and pitch 1.25 cm.

Determine the force that must be applied to the end of 50 cm lever (i) to raise (ii) to

lower a weight of 50 kN. Find the efficiency of the jack. Is itself- locking? Assume µ=

0.20. (Anna univ, Nov/Dec 2002) (16)

Given Data:

Mean diameter,d=10cm r=5cmlead = pitch = 1.25cm

Sol:

Length of lever (a) = 50cm

-1Lead angle , =tan2Lr

-1 1.25=tan

2 5X

-1

1

=tan [0.0398]2.279tan

0.2 tan

tan (0.2)11.31

(ii) Force required to raise the weight of 50KN

1 (tan )rWP

a



1

1

50x5tan(2.27 11.34)

50P 1.189KN

tantan( )

tan2.279tan(2.279 11.31 )0.165

P

(ii) Force required to lower the weight

1

0 01

1

tan( )

50 5tan(11.30 2.279 )

500.795

rWQ

aX

Q

Q KN

IT is self locking because the friction angle ( ) is larger than the lead angle ( ) .

9. (i) A rope is wrapped three turns around a rod as shown in fig. Determine the

force

required on the free end of the rope to support a load of W = 20 kN, Take µ= 0.30 (Anna

univ, May/June 2010) (8)

Given Data :

=0.30

20W KN

Solution:

Angle of Contact



2

1

3 turns=3x2=6

:FormulaT

eT

Where,

1

2

2

required on the free end of the rope

Force on the other end which is equal to W=20KN

T=20KN

T Force

T

21

1 (0.3 x 6 )

1 5.654

1

1

1

20

20

20284.430.070069 KN

70.069 N

The force required at the free end of the rope = 70.069 N

TTe

Te

Te

T

T

T

(ii) A rope is wrapped three and a half turns around a cylinder as shown in fig.

Determine the force T1 exerted on the free end of the rope that is required to support a 1

kN weight. The coefficient of friction between the rope and the cylinder is 0.25.

(Anna univ, Dec/Jan 2003) (8)

Sol :



2

1 2

1 2

(0.25 x 7 )1

1

1 KN

= 0.25

13 x 2

2=7

1x e

244.15 KN

T

T Te

T Te

T

T

10. Determine the smallest force P required to move the block B shown in fig. If (i)

block A is restrained by cable CD, (ii) the cable CD is removed. Take µs= 0.30 and

µk= 0.25


Solution:

(150 9.81) 1471.5

0.3 x 1471.5 T = 441.45

1471.5 (150 9.81)

X N

A

X

3678.75 [(250 9.81) (150 9.81)]

1103.625

3678.75

X X

P B

(i) Block ‘A’ is restrained by cable ‘CD’



[[(250 150) 9.81] ( 150 9.81)](1103.625 441.45)1545.045

P X X X XpP N

(ii) Cable ‘CD’ is resolved

[(250 150) 9.81][(0.30 (400 9.81)]1103.625

P X Xp X XP N

11. The uniform 8 kg rod shown in fig is acted upon by 30 N force which always acts

perpendicular to the bar. If the bar has an initial clockwise angular velocity ω= 10

rad/sec when Ө =0°, determine its angular velocity at the instant Ө =90°.

(Anna univ, Nov/Dec 2002)

(16)

Sol:

2

A

2

A

A

2

2

I =3

5 0.6I =

3I =0.60

(0.6 49.5 0.3sin 30 0.6)

0.6 14.85sin 18

24.75sin 30

A

ml

X

M a X

a

ddt



2

1

2 2 22 1

22

22

22

22

22

24.75sin 30

24.75sin

(24.75sin 30)

( )( 24.75cos 300)

2

10030 24.75

2 2 2

15 24.75 502

121.8732

243.74

15.612 /sec

w

w

d

dwdtdw

Wd

dwW d

w w

wX

w

w

w

w radi

Angular Velocity at ,𝛉 = 900

is 15.612 radi / sec.

12. An automobile travels to the right at a constant speed of 72 km/hr. The diameter of

wheel is 560 mm. Determine the magnitude and direction of the following:

(i) Angular velocity of the wheel

(ii) Velocity of the point B

(iii) Velocity of the point C

(iv) Velocity of the point D (Anna univ, Nov/Dec 2002) (16)

Given data:

Constant speed=72 Km/hr,=20m/sec

Diameter of the wheel= 560mm=0.56m

Solution:



The velocity at point O must be zero,(because the wheel rolls without slipping and the point

O is in contact with the surface).

(i) Angular velocity of the wheel(W)

rw

Where , r=radius of the wheel

2D

r , mr 28.0256.0

3028.0 w , sec/4.8 radw .

Velocity at point 0, EoE VVV /0

sec/20;00 mVV E

Relative velocity of the point with respect to E,

wV Eo /

(ii) Velocity at point(c):

Velocity at point(c) is , EcEc VVV /

sec/933.18

120cos2022202

120cos..2 /2

/2

mV

wwV

VVVVV

C

C

EcEECEC

(iii)Velocity at point (B)

Velocity at point (B)= EBEB VVV /

sec/4.284.820 mVB

(iv) Velocity at point (D):

cos..2 /2

/2

EDEEDED VVVVV

30cos4.828.0202)4.828.0(20 22 DV

sec/068.22 mVD



13. Two masses m1 and m2 are tied together by a rope parallel to the inclined plane

surface, as shown in fig. Their masses are 22.5 kg and 14 kg respectively. The

coefficient of friction between m1 and the plane is 0.25, while that of mass m2 and the

plane is 0.5. Determine (i) the value of the inclination of the plane surface Ө for

which the masses will just start sliding, (ii) the tension in the rope.

(Anna univ, May/June 2001) (16)

Given data:

M1=22.5kg , w1=22.5*9.81=220.725N

m2=14kg, w2=14*9.81=137.34

solution:

let, µ1=coefficient of friction between mass m1 and the plane=0.25

µ2=co efficient of friction between mass m2 and the plane=0.5

considering Block(m1)

resolving the forces along the plane,

025.0sin 11 TRw

0)cos(25.0sin 11 Tww

1)cos(25.0sin 11 wwT

resolving the forces along the plane,

0)5.0sin( 22 RTw

2cos5.0sin 22 wwT

Equating 1&2

cos5.0sin)cos(25.0sin 2211 wwww

Dividing both sides by cosѲ,

2111 5.0tan25.0tan wwww

2121 5.025.0)(tan wwww =123.85



)137034725.220(85.123

tan

3459.0tan , )3459.0(tan1 , 08.19

The value of the inclination of the plane surface 08.19 in which the masses will just start

sliding.

Substituting 08.19 in eqn 1

08.19cos)725.220(25.008.19sin725.220 T

NT 20

14. A cast iron hoop of radius 200 mm is released from rest on a 25° incline as shown in

fig. Find the angular acceleration of the hoop and the time taken by it to move a

distance of 4 m down the slope. Take µs= 0.25. Assume that the hoop rolls without

slipping. (Anna univ, Nov/Dec 2002) (16)

Given data:

R=200mm=0.2m

θ=25

distance, d=4m,us=0.25

solution:

assume the hoop rolls is weighing at 100n.

10010.19

9.81AM

0.2 , 0.25A sr mu

W.K.T. B B BFr J

B B

B

JF

r

, where , BB

B

a

2gsin 1 1BB B B

B

KM M a

r

Where as,



gsin BB B B

B

JM M a

r

2

2gsin B BB B B B

B

K aM M M a

r

2

2gsin 1BB B B

B

KM M a

r

2

gsin

1

BB

B

B

Ma

Kr

0.25B S SK K

2

9.81sin25

0.251

0.2

Ba

4.14589

1.62 /sec2.5625Ba m

Ba =angular acceleration of hoop

15. The given fig shows a stepped pulley. The smaller radius is 150 mm and the bigger

radius is 200 mm. Two loads P and Q are connected by in extensible tant cords.

P moves with an initial velocity of 0.2 m/s and has a constant acceleration of 0.25

m/s2 both downwards. Determine

(i) Number of revolutions turned by the pulley in 4 sec.

(ii) Velocity and the distance travelled by load Q after 4 sec.

(iii) Acceleration of point B located on the rim of the pulley at t=0. Give both

magnitude and direction.




Given data:

150 0.15200 0.20Q

Rp mm mR mm m

Initial velocity, 0.2 /secWo m

20.25 /secAcceleration m

Solution:

We know that,

.p p pa r

Where, pa =angular acceleration at Block P

R=radius

3

0.25150 10

pp

p

a

r

, 21.67 /secp rad

Number of rotation turned by the pully t=4sec.

Angle turned, 21

2p oWt t

21

(0.2 4) ( 1.67 4 )2p

14.16 .p rad

No of revolution, 2

n



14.16

2.252

n

Average velocity and distance travelled by load Q after 4 sec.

T=4sec

At load Q is turned after 4 sec, w=0

Angular velocity , oW W t

W.K.T . 260

N

32 1.59 10

166.5 /sec60

rad

oW W t , 166.5 0 4

166.5

41.6 /sec4

rad .

16. A cord is wrapped on a 2 m diameter disc, which weighs 250 N. If the cord is pulled

upwards with a force of 400 N, determine the acceleration of centre of gravity of the

disc, the angular acceleration of the disc and the acceleration of the cord.


Given data:

D=2m, r=1m

Weight, w=mg=250N

Force, F=400N

Solution:



Let, T= tension in chord

Mg=weight of the cylinder

Α=angular acceleration of the cylinder in (rad/sec2)

According to Newtons law,

D’Alembert’s principle

No force acting in x-direction

maxFx , maFy

dIGmg .

dIGTR . , 2.2

2

mRTR

Solving eqn 1 & 2we get,

.2

)(2mR

Rmamg

w.k.t Ra

sub in above eqn

)(2

)(2

RamR

Rmamg

2)(

2)(

maagm

mamamg

2a

ag , ag23

81.932

32

ga , sma /54.6

Angular acceleration , sec/54.6154.6

radRa

Equation 1 becomes,



47.112

194.222

.

94.2254.6)400250(.

22

MRIGC

mmaymTmg

17. In the fig given below, the blocks A and B have masses of 45 kg and 60 kg

respectively. The drum has a moment of inertia of 16 kg/m2 about its Axis of

rotation. Find the distance through which the block A falls, before it reaches a speed

of 2 m/s.

(Anna univ, May/June 2001) (16)

Given data:

W1=45kg=(45*9.81)=441.45N

W2=60kg=588.6N

M.I=16 kg/m2=156.96N/m2

Solution:



23

606.588

33.0;9.0

145.4415.441

2

1

aT

aaa

aT

B

)206.588(3.0)4545.441(9.0163.09.016

3.09.03

21

21

aaTT

TT

72.220)9.05.4616(5.4672.22016

658.1765.403.39716

aaa

2sec/81.3

72.22085.57

rad

mss

asVV

ma

a

if

583.0)429.32(4

2

sec/429.3

)81.39.0(

22

2

18(a). A uniform 5 kg rod is shown in fig below is acted upon by 30 N force which

always acts perpendicular to the bar. If the bar has an initial clockwise angular

velocity, ω0= 10 rad/sec, when Ө =0, determine the angular velocity at the instant Ө

=90°. (Anna univ, Nov/Dec 2011) (8)

Given data:

W=5kg = (5*9.81)=49.05

Force, P=30N



ω°=10 rad/sec

θ=0

solution:

Angular velocity,

1 T

Initial angular velocity, =10 rad/sec

Since, 60

2 N

r

903.0

sec.27rad

T1027

.17T

18(b). A block weighing 36 N is resisting on a rough inclined plane having an inclination

of 30º. A force of 12 N is applied at an angle of 10º up the plane and the block is just

on the point of moving down the plane. Determine the coefficient of friction.

(8)

Given data:

Weight of block,W=36N

Inclination angle, =30

External force, p=12N

Force angle, θ=10

Co efficient of friction,µ=?

Solution:

The block is just on the point of moving down the plane



)µsin(cos)µcos(sin

W

p

)10µsin10(cos)µcos3036(sin30

12

31.176µ)-(21.1608µ)1736.09849.0(12

9.34229.176µ

29.1769.342

µ

Coefficient of friction, 0.3202µ .

19. In the engine system shown in fig, the crank AB has a constant angular velocity of

3000 rpm. For the crank positioned indicated, find (i) the angular velocity of the

connecting rod, (ii) velocity of piston.(Anna univ, Nov/Dec 2002) (16)

Given data:

Constant angular velocity,N=3000 rpm,

Solution:

Motion of crank,

Angular velocity,

60

2 N

sec/3146030002

rad



The crank rotates about point A

rV abB

sec/55.23

075.0314m

Motion of connecting rod

From triangle ABC,

9.13

sin20040sin75

From the vector diagram,

50sin1.76sin

55.23)1.7650(180sin(

/BcVVc

1.76sin50sin55.23

/

BcV

sec/58.189707.0

76.055.23

m

1.76sin)1.7650(180sin(55.23

Vc

sec/60.199707.0028.19

mVc

sec/9.92

58.182.0

2.0 /

rad

V

BC

BC

BcBC

20(a).A rigid body is undergoing general plane motion write down the relationship of the

velocities of two point A and B on it and explain.(may/june2012)(4)

Solution:

We know,



General plane motion=translation+ Rotational about the centre by using the above

principle, the absolute velocity of B can be found by the following formula

ABAB VVV /

(plane motion motion)(translation)(rotation of AB about A)

Where, BV = absolute velocity of RB,

AV = absolute velocity of A and

ABV / =Relative velocity of B with respect to A

=the length AB*angular velocity of the rod AB=rω

The formula can also written as

BAAB VVV /

Where

BAV / =Relative velocity of A with respect to B

20.(B)An automobile travels to the right at a constant speed of 72km. the diameter of the

wheel is 560mm.

Determine the magnitude and direction of the following:

1.angular velocity of the wheel

2.velocity of the point B

3. velocity of the point c

4. velocity of the point D (12)

Given data:

Diameter of the wheel=560mm

Radius, r=560/2=280mm=0.28m

Speed, AV =72km/hr

sm/203600

100072

Solution:

The velocity component at point B,C and C are shown in fig

As the linear velocity at the center of wheel A is known the point A taken as pole.

a.To find velocity at point C

as the wheel rolls without slipping. The resultant velocity at the point of contact of

wheel with the ground surface, C is zero.

0CV

b. angular velocity of the wheel

rVV AC



sec/42.7128.0200rad

c. velocity at B

rVV AB

sm/40)42.7128.0(20

d. velocity at D

as the velocity component r is tangential at D, the angle between r and AV

is also 30° by parallegram law of forces.

30cos222 TVrVV AAD

)30cos42.7128.0202()42.7128.0(20 22

sm/63.38

IFET COLLEGE OF ENGINEERING

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