Post on 15-Jan-2022
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)
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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.
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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.
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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)
Regulation: 2013 Academic Year: 2014-2015
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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.
Regulation: 2013 Academic Year: 2014-2015
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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.
Regulation: 2013 Academic Year: 2014-2015
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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
Regulation: 2013 Academic Year: 2014-2015
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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.
Regulation: 2013 Academic Year: 2014-2015
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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
Regulation: 2013 Academic Year: 2014-2015
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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.
Regulation: 2013 Academic Year: 2014-2015
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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
Regulation: 2013 Academic Year: 2014-2015
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- 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
T2- Tension in the belt on the slack 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.
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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
T2- Tension in the belt on the slack 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
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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.
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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
Regulation: 2013 Academic Year: 2014-2015
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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:
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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.
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
cos3031
1 F ,
2cos101 F
Considering 90N block:
Resolving the forces vertically,
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:
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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°
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
solution:
considering the block(B);
normal reaction=weight of the body
RB=2KN
Resolving the forces horizontally,
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)
Resolving the forces vertically,
cosWRA
KNRA 607.150cos5.2
Resolving the forces horizontally,
sinWFP AA
sinWRP AAA
50sin2)607.1( AAP ( 30.0A )
KNPA 014.2
Considering the block (B):
Resolving the forces vertically,
0Fv , 0sin PRW
050sin2 pR
)250sin( pR
Resolving the forces horizontally,
,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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Resolving the forces vertically,
0)5705236sin1(;0 TRFv
15705236sin 1 RT
Resolving the forces horizontally,
05236cos;0 1 FTFH
15236cos FT , 1µ5236cos RT
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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)
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Resolving the forces vertically,
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
0Fv , 2 11500 sin30R R P
2 2F R , 2 0.28(1500 390.625 sin30 )F P
Resolving the forces horizontally,
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)
Resolving the forces horizontally,
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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 :
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
(Anna univ, Nov/Dec 2004) (16)
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’
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
[[(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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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:
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
)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,
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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.
(Anna univ, Nov/Dec 2010) (16)
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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.
(Anna univ, Nov/Dec 2011) (16)
Given data:
D=2m, r=1m
Weight, w=mg=250N
Force, F=400N
Solution:
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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,
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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:
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
)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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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,
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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
Regulation: 2013 Academic Year: 2014-2015
IFETCE/MECH/MERGED/I YR/II SEM/GE6253/EM/UNIT-5/QB/VER1.1
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