UNESCO-NIGERIA TECHNICAL & VOCATIONAL
EDUCATION REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
MECHANICAL ENGINEERING TECHNOLOGY
MECHANICAL ENGINEERING
SCIENCE[DYNAMICS]
COURSE CODE: MEC124
YEAR I- SEMESTER 2
THEORY
Version 1: December 2008
1
MECHANICAL ENGINEERING SCIENCE [DYNAMICS]
TABLE OF CONTENT
Week 1
1.0 Linear motion………………………………………………………….7
1.1 Introduction……………………………………………………………7
1.2 Motion under uniform acceleration……………………………...........7
Week 2
2.0 Velocity Time-Graphs……………………………………………….11
Week 3
3.0 Vectors………………………………………………………………17
3.1 Additions of vectors…………………………………………………17
3.2 Subtraction of vectors…………………………………………….....17
Week 4
4.0 Circular motion……………………………………………………..21
4.1 Motion along a circular path………………………………………..21
4.1.1 Introduction……………………………………………………….21
4.2 Centripetal Force…………………………………………………….21
4.3 Centrifugal force…………………………………………………….21
4.4 Motion of an object moving in a circle with uniform speed about a
fixed point………………………………………………………………22
4.5 Acceleration in a circle (centripetal)………………………………...24
4.6 Relations between linear and angular motion ………………………25
4.7 Motion under force of gravity……………………………………….25
Week 5
5.0 Relative velocity……………………………………………………28
5.1 Relative velocity in 1 dimension……………………………………..28
5.2 Relative velocity in 2 dimensions…………………………………….29
5.3 Centrifugal force acting on a body moving along a circular path……30
5.4 Projectile motion……………………………………………………...31
5.5 Equation for the path of a projectile………………………………….32
5.6 Time of flight of a projectile on a horizontal plane…………………..34
5.7 Horizontal range of a projectile……………………………………….34
5.8 Maximum height of a projectile on a horizontal plane………………..35
2
Week 6
6.0 Mass and Weight………………………………………………………37
6.1 Mass versus weight……………………………………………………..39
6.2 How are weight and mass different…………………………………….39
6.3 Newton’s three laws of motion…………………………………………39
6.3.1 Newton's First Law of Motion………………………………………..39
6.3.2 Newton's Second Law of Motion……………………………………..39
6.3.3 Newton's Third Law of Motion……………………………………….39
6.4 Inertia…………………………………………………………………..40
6.4.1 Moment of inertia……………………………………………………..40
Week 7
7.0 Momentum……………………………………………………………..42
7.1 Impulse and Momentum………………………………………………..42
7.1.1 Impulse………………………………………………………………..42
7.2 Momentum………………………………………………………………42
7.2.1 The Law of Conservation of Momentum……………………………..42
7.3 Radius of Gyration………………………………………………………43
Week 8
8.0 Work Energy and Power……………………………………………….45
8.1 Work……………………………………………………………………47
8.2 Work done by a torque…………………………………………………50
8.3 Energy………………………………………………………………….52
8.3.1 Potential energy……………………………………………………….52
8.3.2 Kinetic energy…………………………………………………………53
8.3.3 Conservation of energy………………………………………………..56
8.3.4 Kinetic energy of a rotating body……………………………………..57
8.3.5 Efficiency……………………………………………………………..58
3
Week 9
8.4 Tractive effort, tractive resistance, power available & Power
required………………………………………………………………………59
8.5 Rolling resistance…………………………………………………………59
8.6 Gradient resistance………………………………………………………..60
8.7 Air (or wind) resistance……………………………………………………62
8.8 Power available…………………………………………………………….62
8.9 Power required……………………………………………………………..62
Week 10
8.10 Power……………………………………………………………………..
8.11 Transmission of motion and power………………………………………
8.12 Belt drive………………………………………………………………….
8.13 Belt ………………………………………………………………………
8.14 Power Transmission ……………………………………………………..
8.15 Flat belts………………………………………………………………….
8.16 Round belts………………………………………………………………
8.17 Vee belts…………………………………………………………………
8.18 Timing Belts…………………………………………………………….
8.19 Pulley……………………………………………………………………
8.20 Rope and pulley systems………………………………………………..
8.21 Types of systems………………………………………………………..
8.22 Single Pulley Systems…………………………………………………..
4
Week 11
9.0 Chains and Gears………………………………………………………
9.1 Chain drive…………………………………………………………….
9.2 Common Types of Chains……………………………………………..
9.3 Gear drive………………………………………………………………
9.4 How Gears Work……………………………………………………….
9.5 Spur Gears………………………………………………………………
9.6 Worm Gears…………………………………………………………….
Week 12
10.0 Machines………………………………………………………………..
10.1 Machines……………………………………………………………….
10.2 Mechanical Advantage (M.A)………………………………………….
10.3 Velocity ratio (V.R.)……………………………………………………
10.4 Mechanical Efficiency (M.E.)………………………………………….
10.5 Screw …………………………………………………………………..
10.6 The lever………………………………………………………………..
10.7 First-class levers………………………………………………………..
10.8 Second-class levers……………………………………………………..
10.9 Third-class levers……………………………………………………….
Week 13
11.0 Wheel and axle………………………………………………………….
11.1 The screw jack ………………………………………………………….
11.2
Wheel and Axle……………………………………………………….
11.3 Windlass…………………………………………………………………
11.4 Simple belt drives, transmission belts…………………………………..
11.5 Inclined plane, ramp, screw, thread, wedge…………………………….
5
11.6 Calculating mechanical advantage…………………………………..
11.6.1 Ideal mechanical advantage……………………………………….
11.6.2 Actual mechanical advantage……………………………………..
Week 14
12.0 Simple Harmonic Motion……………………………………………
12.1 Simple harmonic motion …………………………………………….
12.2 Frequency…………………………………………………………….
12.3 Amplitude…………………………………………………………….
Week 15
12.4 Acceleration at any instant in terms of displacement…………………
12.5 Velocity at any instant in terms of displacement……………………..
12.6 Variable forces producing S.H.M…………………………………….
6
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 1 1.0 Linear Motion
1.1 Introduction
Dynamics is that branch of engineering mechanics, which deals with the forces and their effects,
while acting upon the bodies in motion.
Linear motion is simply motion in straight line.
Angular motion is rotational motion which takes place about the geometric axis of the body.
Speed of a body may be defined as its rate of change of distance with time with respect to its
surroundings. The speed of a body is irrespective of its direction and is thus a scalar quantity.
The unit of speed is m/s.
Displacement is distance in a specified direction.
Velocity of a body may be defined as its rate of change of distance with time, with respect to its
surroundings, in a particular direction. As the velocity is always expressed in a particular
direction, therefore it is a vector quantity. The unit of velocity is m/s.
Acceleration of a body may be defined as the rate of change of its velocity with time. It is said to
be positive, when the velocity of a body increases with time, and negative when the velocity
decreases with time. The negative acceleration is also called retardation. In general, the term
acceleration is used to denote the rate at which the velocity is changing. It may be uniform or
variable. The unit of acceleration is m/s2.
1.2 MOTION UNDER UNIFORM ACCELERATION
Fig. 1.1
Consider plane motion of a particle starting from O and moving along OX with a uniform
acceleration as shown in Fig. 1. let P be its position after t seconds.
7
Let
u = Initial velocity,
v = Final velocity,
t = Time (in seconds) taken by the particle to change its velocity from u to v,
a = Uniform positive acceleration, and
s = Distance travelled in t seconds.
Since in t seconds, the velocity of the particle has increased steadily from (u) to (v) at the rate of
a, therefore, total increase in velocity
v = u + at
................................. (1)
and average velocity
 u + v
=

 2 
We know that the distance travelled by the particle,
s = average velocity x time
 u + v
s=
 ×t
 2 
Subtituting the value of v from equation (1)
 u + u + at 
2
s=
 × t = ut + 21 at ................(2)


2
From equation (1), (i.e. v = u+ a.t) we find that
v−u
a
Substituting this value of t in the distance formula above equation.
t=
2
2
 u + v  v − u v − u
s=
×
=
 

 2   a 
2a
Or
2a s = v2 –u2
∴v 2 = u 2 + 2as...................(3)
8
Example 1
A car moving with a velocity of 54 km h-1 accelerates uniformly at the rate of 2 m s-2. Calculate
the distance travelled from the place where acceleration began to that where the velocity reaches
72 km h-1 , and the time taken to cover this distance.
Solution:
54 km h-1 = 15 m s-1, 72 km h-1 = 20 m s-1, acceleration a = 2 m s-2.
(i)
Using
v 2 = u 2 + 2as,
∴ 202 = 152 + 2 × 2 × s
202 − 152
∴s =
= 43 43 m.
2×2
(ii)
v = u + at
∴ 20 = 15 + 2t
20 − 15
∴t =
= 2.5 s
2
Using
Example 2
On turning a corner, a motorist rushing at 20m/s, finds a child on the road 50 m ahead. He
instantly stops the engine and applies brakes, so as to stop the car within 10m of the child.
Calculate (i) retardation, and (ii) time required to stop the car.
Solution:
(i) Retardation
Let
a = acceleration of the motorist.
We know that
v 2 = u 2 + 2as
0 = (20) + 2 × a × 40 = 400 + 80a
2
∴ a = − 400
80
= −5 m / sec 2
(The minus sign shows that the acceleration is negative i.e. retardation)
(ii) The time required to stop the car
Let t = time required to stop the car
We know that final velocity of the car(v),
9
0 = u + a.t = 20 − 5 × t
20
∴t=
= 4s
5
(Qa = −5 m / s 2 )
Exercise 1.
A body starts with a velocity of 0.3m/s and moves in a straight line with a constant acceleration.
If its velocity at the end of 5 seconds is 5.5 m/s, find (i) the uniform acceleration and (ii) distance
travelled in 10 seconds. (Ans. 0.5 m/s; 55m)
Exercise 2.
A car starts from rest and accelerates uniformly to a speed of 72 km.p.h. over a distance of 500
m. Find acceleration of the car and time taken to attain this speed.
If a further acceleration rises the speed to 90 km.p.h. in 10 seconds, find the new acceleration
and the further distance moved. (Ans. 0.4m/s2 ; 50 s ; 0.5 m/s2 ; 225 m; 62.5m)
10
MECHANICAL ENGINEERING SCIENCE
WEEK 2 2.0
DYNAMICS
Velocity time Graphs
When the velocity of a moving train is plotted against the time, a ‘velocity-time (v-t) graph’ is
obtained. Useful information can be deduced from this graph, as we shall see shortly. If the
velocity is uniform, the velocity-time graph is a straight line parallel to the time axis, as shown
by line (1) in Fig.2. If the train increases in velocity steadily from rest, the velocity-time graph is
a straight line, line (2), inclined to the time axis. If the velocity change is not steady, the velocitytime graph is curved. In Fig. 2, for example, the velocity-time graph OAB represents the velocity
of a train starting from rest which reaches a maximum velocity at A, and then comes to rest at the
time corresponding to B.
v
A
(2)
E
∆v
G
∆t
.
(1)
D
C
(3)
O
t
XY
B
Fig. 2.1 Velocity (v)-time (t) curves
Acceleration is the ‘rate of change of velocity’, that is, the change of velocity per second. We can
see that the acceleration of the train at any instant is given by the gradient to the velocity time
graph at that instant, as at E. At the peak point A of the curve OAB the gradient is zero, that is,
11
the acceleration is then zero. At any point, such as G, between A and B the gradient to the curve
is negative because the graph slows downwards. Here the train has deceleration or decrease in
velocity with time. Like velocity, acceleration is a vector.
The gradient to the curve at any point such as E is given by:
velocity change ∆v
=
time
∆t
Where ∆v represents a small change in v in a small time ∆t . In the limit, the ratio ∆v ∆t
becomes dv dt , using calculus notation.
1. When a body is moving with a uniform velocity
Y
A
B
v
X
t
O
C
Y
Fig. 2.2
(a) Uniform velocity
Consider the motion of a body, which is represented by the graph ABCD as shown
in figure (a) above.
We know that the distance traversed by the body,
S = velocity X time
Thus, the area of the figure OABC (i.e., velocity x time) represents the distance
traversed by the body, to some scale.
2. When the body is moving with a variable velocity
Y
B
B
α
A
v
D
u
X
α. t
u
X
We know that the distance
traversed by a body,
C
Fig. 2.3
Y
12
s = u.t + 12 at 2
From the geometry of the figure (b) above, we know that Area of the figure
OABC = Area OADC +ABD
But area of the figure OADC = u x t
And area of the figure ABD = 12 × t.at = 12 at 2
∴ total area OABC = u. t + 12 at 2
Thus, we see that the area of the figure OABC represents the distance traversed by the
body to some scale. From the figure it is also seen
αt
=α
tan α =
t
Thus, tan α represents the acceleration of the body.
Example
Given below is a strobe picture of a ball rolling across a table. Strobe pictures reveal the position
of the object at regular intervals of time, in this case, once each 0.1 seconds.
Notice that the ball covers an equal distance between flashes. Let's assume this distance equals
20 cm and display the ball's behavior on a graph plotting its x-position versus time.
The slope of the position versus time graph shown above would equal 20 cm divided by 0.1 sec
or 200 cm/sec. The following graph displays this exact same information in a new format, a
velocity versus time graph.
13
This graph very clearly communicates that the ball's velocity never changes since the slope of the
line equals zero. Note that during the interval of time being graphed, the ball maintained a
constant velocity of 200 cm/sec. We can also infer that it is moving in a positive direction since
the graph is in quadrant I where velocities are positive.
To determine how far the ball travels on this type of graph we must calculate the area bounded
by the "curve" and the x- or time axis.
As you can see, the area between 0.1 and 0.3 seconds confirms that the ball experienced a
displacement of 40 cm while moving in a positive direction.
14
Given below are three orientations of velocity-time graphs for one-dimensional uniform
velocity. On each graph, the height of the graph represents the object's velocity and the area
bounded by the graph and the x- or time axis represents the object's displacement, or change in
position.
v vs t - since its slope equals zero there is no acceleration, or change in
velocity. The object is traveling at a constant, steady rate.
It is moving in a positive direction since the graph is in quadrant I where
the y-axis (aka, velocity value) is positive.
We know the object was traveling in a positive direction since its
rectangular area is located in a positive quadrant.
v vs t - since its slope equals zero there is no acceleration, or change in
velocity.
This object is NOT moving since its velocity equals zero. The object is in
a state of rest and obviously has no displacement.
v vs t - since its slope equals zero there is no acceleration, or change in
velocity. The object is traveling at a constant, steady rate.
It is moving in a negative direction since the graph is in quadrant IV
where the y-axis (aka, velocity value) is negative.
We know the object was traveling in a negative direction since its
rectangular area is located in a negative quadrant.
15
Exercise
A car moving with a velocity of 10 m/s accelerates uniformly at 1 ms-2 until it reaches a velocity
of 15 ms-1. Calculate (i) the time taken , (ii) the distance travelled during acceleration, (iii) the
velocity reached 100 m from the place where the acceleration began. (Ans. (i) 5s (ii)62.5m (iii)
17m/s
16
DYNAMICS
MECHANICAL ENGINEERING SCIENCE
WEEK 3 3.0 Vectors
3.1 Addition of vectors
Suppose a ship is travelling due east at 30 km h-1 and a boy runs across the deck in a north-west
direction at 6 km h-1, Fig.(i) . We can find the velocity and direction of the boy relative to the sea
by adding the two velocities. Since velocity is a vector quantity, we draw a line OA to represent
30 km h-1 in magnitude and direction, and then, from the end of A, draw a line AC to represent
6km h-1 in magnitude and direction, Fig.(ii). The sum, or resultant, of the velocities is now
represented by the line OC in magnitude and direction, because a distance moved in one second
by the ship (represented by OA) together with a distance moved in one second by the boy
(represented by AC) is equivalent to a movement of the boy from O to C relative to the sea.
C
6 km h-1
Vector sum
6
30 km h-1
A
O
30
(i)
(ii)
FIG. 3 Addition of vectors
3.2 Subtraction of vectors
We now consider the subtraction of vectors. If a car A travelling at 50 km h-1 is moving in the
same direction as another car B travelling at 60 km h-1, the relative velocity of B to A=
difference in velocities = 60-50 = 10 km h-1. If however, the cars are travelling in opposite
directions, the relative velocity of B to A = 60-(-50)=110 km h-1, since the velocity of A is
opposite (negative) compared to B.
Suppose that a car X is travelling with a velocity v along a road 30˚ east of north, and a car Y is
travelling with a velocity u along a road due east, Fig.7(i). since ‘velocity’ has direction as well
17
magnitude, that is, ‘velocity’ is a vector quantity, we cannot subtract u and v numerically to find
the relative velocity. We must adopt a method which takes into account the direction as well as
the magnitude of the velocities, that is a vector subtraction is required.
r r r
r
The velocity of X relative to Y = difference in velocities = v − u = v + (− u ). suppose OA
represents the velocity, v, of X in magnitude and direction, Fig.7(ii). Since Y is travelling due
east, a velocity AB numerically equal to u but in the due west direction represents the vector(-u).
the vector sum of OA and AB is OB from above, which therefore represents in magnitude and
direction the velocity of X minus that of Y. By drawing an accurate diagram of the two
velocities, OB can be found.
v
30˚
B
-u
A
A
20 ms-1
X
Y
u
Relative
velocity
r r
=v −u
r
v
90˚
B
O
v
v
(i)
Example 1
(ii)
C
(iii)
FIG. 4 Subtraction of velocities
Example2
18
A car is moving round a circular track with a constant speed v of 20 ms-1, Fig.8(iii). At different
times the car is at A, B and C respectively. Find the velocity change (a) from A to C, (b) from A
to B.
Solution:
r
r
(a) Velocity change from A to C = vC − v A = (+20) − (−20) = 40 m s −1 in the direction of C .
r
r
r
r
(b) Velocity change from A to B = v B − v A = v B + (− v A ).
P
θ
VB
Q
R
-VA
FIG. 5
r
r
In Fig. 9, PQ represents the vector v B or 20 m s-1 and QR represents − v A or 20 m s-1.
So
r
r
PR = v B − v A = 20 2 − 20 2 = 28 m s −1 (approx.), and its direction θ relative to vB is
45˚.
19
Exercise
R = 11.2 km
R = 50 km
20
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 4 4.0 Circular Motion
4.1 Motion long a circular path
4.1.1 Introduction
We see that whenever a person, on a bicycle a motor cycle or a scooter, drives round a curve or a
circular track, he has to lean inward in order to maintain a perfect equilibrium. As a matter of
fact, the angle with which the man leans with the vertical is more, if he is running at a faster
speed than that when he is running at a lower speed. This type of motion is called the motion
along a circular path.
4.2 Centripetal Force
A body moving in a circle or along a circular path, with a constant velocity, suffers a continuous
change in its direction at a very point of its motion; though the magnitude of its speed remains
the same. Since the velocity involves both magnitude as well as direction, and the velocity of the
body is continuously changing due to change in direction; therefore according to Newton's First
law of motion, an external force must act continuously upon the body, to produce a change in the
direction of the moving body.
Strictly speaking, the body, due to inertia, tends to move along the tangent at every point of its
motion, with the constant velocity. Therefore, some force must act at right angles to the direction
of motion at every point, which should change the direction of motion of the body; leaving the
speed uniform. Thus the force, which acts along the center of the circle along which the body
moves, is known as centripetal force.
4.3 Centrifugal force
According to the Newton's Third Law of motion, the force, which acts opposite to the centripetal
force, is known as centrifugal force. It may be noted that the centrifugal force always acts away
from the center of the path, or in other words, the centrifugal force always tends to throw the
body away from the center of circular path.
Angular velocity is the rate of change of displacement of a body, and is expressed in r.p.m.
(revolutions per minute) or in radians per second. It is usually denoted by ω (omega).
21
Angular acceleration is the rate change of angular velocity and is expressed in radians per
second per second (rad/sec2) and is usually, denoted by α . It may be constant or variable.
Angular displacement is the total angle, through which a body is rotated, and is usually denoted
by θ . Mathematically, if a body is rotating with a uniform angular velocity ( ω ) then in t
seconds, the angular displacement, θ = ω .t
Torque is the turning moment of a force on the body on which it acts. The torque is equal to the
product of the force and the perpendicular distance from any point O to the line of action of the
force.
Mathematically, torque,
T = F ×l
where,
F = Force acting on the body, and
l = Perpendicular distance between the point O and line of action of the
force (known as arm or leverage)
The units of torque depend upon the units of force and leverage. If the force is in N and leverage
in mm, then the unit of torque will be N-mm. similarly, if the force is in kN and leverage in m,
then the unit of torque will be kN-m.
4.4 Motion of an object moving in a circle with uniform speed about a fixed
point
There are numerous cases of objects moving in a curve about some fixed point. The earth and the
moon revolve continuously around the sun, for example, and the rim of the balance-wheel of a
watch moves to-and fro in a circular path about the fixed axis of the wheel. Now we shall study
the motion of an object moving in a circle with a uniform speed around a fixed point O as center,
22
FIG. 6 Motion in a Circle
If the object moves from A to B so that the radius OA moves through an angle θ ,its angular
velocity, ω , about O is defined as the change of the angle per second. Thus if t is the time taken
by the object to move from A to B,
ω=
θ
t
........................................................... (1)
Angular velocity is usually expressed in ‘radian per second’ (rad s-1). From (1),
θ = ω t ........................................................... (2)
which is analogous to the formula ‘distance = uniform velocity x time’ for motion in a straight
line.
It will be noted that the time T to describe the circle once, known as the period of motion, is
given by
T=
2π
ω
, .......................................................... (3)
since 2 π radians is the angle in 1 revolution.
If s is the length of the arc AB, then s r = θ , by definition of an angle in radians.
∴ s = rθ .
Dividing by t, the time taken to move from A to B,
s
θ
∴ =r .
t
t
23
But s t = the speed, v , of the rotating object, and θ t is the angular velocity.
∴ v = rω .......................................................... (4)
Example 3
A model car moves round a circular track 0.3 m at 2 revolutions per second. What is (a) the
angular velocity ω , (b) the period T, (c) the speed v of the car? Find also (d) the angular
velocity of the car if it moves with a uniform speed of 2 m s-1 in a circle of radius 0.4 m?
(a) for 1 revolution, angle turned θ = 2π rad (360°). So
ω = 2 × 2π = 4π rad s-1.
(b) period T = time for 1 rev =
2π
ω
=
2π
= 0.5 s. (Or, T = 1 s/2 rev = 0.5 s.)
4π
(c) velocity v = rω = 0.3 × 4π = 12
. π = 38
. m s−1 .
(d) from v = rω ,
ω=
v 2 m s−1
=
= 5 rad s-1.
r
0.4 m
4.5 Acceleration in a circle (centripetal)
When a stone is attached to a string and whirled round at constant speed in a circle, one can feel
the force (pull) in the string needed to keep the stone moving in its circular path. Although the
stone is moving with a constant speed, the presence of the force implies that the stone has an
acceleration.
The force on the stone acts towards the center of the circle. We call it a centripetal force. The
direction of the acceleration is in the same direction as the force, that is towards the center. We
now show that if v is the uniform speed in the circle and r is the radius of the circle,
Acceleration towards center =
v2
............................................. (1)
r
Acceleration towards center =
r 2ω 2
= rω 2 ................................. (2)
r
Or since v = rω ,
24
4.6 Relations between linear and angular motion
Following are the relations between the linear motion and the angular motion of a body:
S/NO
PARTICULARS
LINEAR MOTION
ANGULAR MOTION
1
Initial velocity
U
ωo
2
Final velocity
V
ω
3
Constant acceleration
A
α
4
Total distance traversed
S
θ
5
Formula for final velocity
v = u + at
ω = ωo + α t
6
Formula for distance traversed
s = ut + 21 at 2
θ = ω o t + 21 αt 2
7
Formula for final velocity
v 2 = u 2 + 2as
ω 2 = ω o 2 + 2αθ
8
Differential formula for velocity
9
Differential formula for acceleration
v=
ds
dt
ω=
dθ
dt
a=
dv
dt
α=
dω
dt
4.7 Motion under force of gravity
It is a particular case of motion, under a constant acceleration of (g) where its value is taken as
9.8 m/sec2. if there is a free fall under gravity, the expressions for velocity and distance travelled
in terms of initial velocity, time and gravity acceleration will be:
1.
v = u + gt
2.
s = ut +
3.
v 2 = u 2 + 2 gs
1 2
gt
2
25
Example 4
A stone was thrown vertically upwards, from the ground, with a velocity of 49 m/sec. after 2
seconds, another stone was thrown vertically upwards from the same place. If both the stones
strike the ground at the same time, find the velocity, with which the second stone was thrown.
Take
g = 9.8 meters/sec2.
Solution:
0 = u-g.t = 49-9.8t First of all, consider the upward motion of the first stone.
Given: u = 49 m/sec; v = 0; g = - 9.8 m/sec2
Let
t = Time taken by the stone to reach maximum height.
We know that final velocity of the stone
∴
t = 49/9.8 = 5sec
It means that the stone will take 5 sec to reach the maximum height and 5 sec to come back to
the ground.
∴ Total time of flight
= 5+5 = 10 sec
Now consider the motion of second stone. Therefore, time taken by the second stone for going
upwards and coming back to the earth
= 10-2 = 8 sec
and time taken by the second stone to reach maximum height
= 8/2 = 4 sec
Now consider the upward motion of the second stone.
Given:
v = 0; t = 4 sec
We know that final velocity of the stone(v),
0 = u – g.t = u – 9.8 x 4 = u – 39.2
∴ u = 39.2 m/sec
26
Exercise1
A motor car of mass 1000 kg is travelling round a circular track with a velocity of 15 m/s. if the
radius of the track is 400 m, find the horizontal thrust exerted on the wheels. (Ans. 562.5N)
Exercise 2
A skater describes a circle of 10 m radius with a velocity of 4m/s. At what angle, with the
vertical he must lean inwards? (Ans. 9°16′ )
27
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 5 5.0 Relative Velocity
It has been established since long that every motion is relative as an absolute motion is
impossible to conceive. Strictly speaking our motion is always with reference to the Earth, which
is supposed to be fixed or at rest. But we know our Earth is also not at rest. It has some relative
velocity with respect to the celestial bodies such as sun, moon etc. these celestial bodies, in turn,
have some relative velocity with respect to the stars of the universe.
It will be interesting to know that when we say that a train is moving at 50 kilometers per hour,
we simply mean that the speed of the train, which appears to an observer on the earth, is 50
kilometers per hour. Thus the relative velocity of A, with respect to B, is the velocity with which
A appears to move to an observer sitting on B, neglecting the motion of B relative to the Earth.
5.1 Relative velocity in 1 dimension
Most people find relative velocity to be a relatively difficult concept. In one dimension, however,
it's reasonably straight-forward. Let's say you're walking along a road, heading west at 8 km/hr.
A train track runs parallel to the road and a train is passing by, traveling at 40 km/hr west. There
is also a car driving by on the road, going 30 km/hr east. How fast is the train traveling relative to
you? How fast is the car traveling relative to you? And how fast is the train traveling relative to
the car?
One way to look at it is this: in an hour, the train will be 40 km west of where you are now, but
you will be 8 km west, so the train will be 32 km further west than you in an hour. Relative to
you, then, the train has a velocity of 32 km/hr west. Similarly, relative to the train, you have a
velocity of 32 km/hr east.
Using a subscript y for you, t for the train, and g for the ground, we can say this: the velocity of
you relative to the ground = = 8 km/hr west the velocity of the train relative to the ground =
= 40 km/hr west
Note that if you flip the order of the subscripts, to get the velocity of the ground relative to you,
for example, you get an equal and opposite vector. You can write this equal and opposite vector
by flipping the sign, or by reversing the direction, like this: the velocity of the ground relative to
you =
= -8 km/hr west = 8 km/hr east
The velocity of the train relative to you, , can be found by adding vectors appropriately. Note
the order of the subscripts in this equation:
28
Rearranging this gives:
A similar argument can be used to show that the velocity of the car relative to you is 38 km/hr
east. the velocity of you relative to the ground = = 8 km/hr west = -8 km/hr east the velocity
of the car relative to the ground = = 30 km/hr east
The velocity of the train relative to the car is 70 km/hr west, and the velocity of the car relative to
the train is 70 km/hr east.
5.2 Relative velocity in 2 dimensions
In two dimensions, the relative velocity equations look identical to the way they look in one
dimension. The main difference is that it's harder to add and subtract the vectors, because you
have to use components. Let's change the 1D example to 2D. The train still moves at 40 km/hr
west, but the car turns on to a road going 40° south of east, and travels at 30 km/hr. What is the
velocity of the car relative to the train now?
The relative velocity equation for this situation looks like this:
The corresponding vector diagram looks like this:
Fig.5.1
Because this is a 2-D situation, we have to write this as two separate equations, one for the xcomponents (east-west) and one for the y-components (north-south):
29
Now we have to figure out what the x and y components are for these vectors. The train doesn't
have a y-component, because it is traveling west. So:
The car has both an x and y component:
Plugging these in to the x and y equations gives:
Combining these two components into the vector gives a magnitude of:
at an angle given by the inverse tangent of 19.3 / 63.0, which is 17 degrees. So, the velocity of
the car relative to the train is 66 km/hr, 17 degrees south of east.
5.3 Centrifugal force acting on a body moving along a circular path
Consider a body moving along a circular path with a constant velocity.
Let
m = Mass of the body in kg
R = Radius of the circular path, and
v = Constant angular velocity of the body.
We know that the centrifugal acceleration of the body,
a=
v2
r
and centrifugal force,
PC = Mass x Centrifugal acceleration
v 2 m.v 2
= m×
=
r
r
…..( when v is given)
= m.ω 2 .r
……(when ω is given)
30
Example 5
A railway engine of mass 60 tonnes, is moving in an arc of radius 200 meters with a speed of
36 km.p.h. Find the force exerted on the rails towards the centre of the circle.
Solution.
Given. m = 60 t; r = 200 m; v = 36 km.p.h.= 10 m/s
We know that the force exerted on the rails,
m.v 2 60(10)
=
= 30kN
r
200
2
Pc =
5.4 Projectile motion
Projectile motion is the motion of an object who's path is affected by the force of gravity. We are
all affected by gravity, but it profoundly alters the motion of objects that are thrown or shot
upward. The arching of a thrown ball is caused by gravity, as well as its falling motion in
general.
Gravity is a force that acts on objects, it makes objects accelerate "downward". While we do not
need to know about forces to analyze projectile motion we do need to know a very important
detail: gravity causes objects to accelerate downward at roughly 9.8 m/s² 9.8 m/s² is the
generally accepted amount of acceleration that happens, in some areas of Earth it is more
or less but we will use 9.8 for our calculations.
Consider a particle projected upwards from a point O at an angle α, with the horizontal, with an
initial velocity u m/sec as shown in Fig.4.
Now resolving this velocity into its vertical and horizontal components,
V = u sin α
and
H = u cos α
We see that the vertical component (u sinα ) is subjected to retardation due to gravity. The
particle will reach maximum height, when the vertical component becomes zero. After this the
particle will come down, due to gravity.
31
Y
u
u sin α
X
A
u cos α
FIG.5.2 Projectile on a horizontal plane
The horizontal component (u cos α) will remain constant, since there is no acceleration or
retardation (neglecting air resistance). The combined effect of the horizontal and the vertical
components will be to move the particle, along some path in the air and then to fall the particle
on the ground at some point A, other than the point of projection O as shown in Fig.4.
5.5 Equation for the path of a projectile
u
P
y
α
O
A
FIG.5.3 Path of Projectile
FIG.5.4 Path of Projectile
32
Consider a particle projected from a point O at a certain angle with the horizontal. As already
discussed, the particle will move along a certain path OPA, in the air, and will fall down at A, as
shown in Fig.5.
Let
u = velocity of projection, and
α = Angle of projection with the horizontal.
Consider any point P as the position of particle, after t seconds with x and y as co-ordinates as
shown in Fig. 5. We know that horizontal components of the velocity projection
Horizontal component = u cos α
And vertical component = u sin α
∴
y = u sin αt −
and
x = u cos αt
or
t =
1 2
gt
2
x
u cos α
Substituting the value of t equation (i),
 x  1  x 
y = u sin α 
 − g

 u cos α  2  u cos α 
gx 2
= x tan α − 2
2u cos 2 α
2
Since this is the equation of a parabola, therefore path of a projectile (or the equation of
trajectory) is also a parabola.
Note. It is an important equation, which helps us in obtaining the following relations of a
projectile:
1. Time of flight, 2. Horizontal range and 3. Maximum height of a projectile.
33
5.6 Time of flight of a projectile on a horizontal plane
It is the time, for which the projectile has remained in the air. Looking at the equation
y = u sin αt −
1 2
g .t
2
We know that when the particle is at A, y is zero. substituting this value of y in the above
equation,
1 2
g.t
2
1
or u sin αt = g .t 2
2
1
u sin α = gt 2
2
2u sin α
∴t=
g
0 = u sin αt −
5.7 Horizontal range of a projectile
We have already discussed, that the horizontal distance between the point of projection and the
point, where the projectile returns to the earth is called horizontal range of a projectile. We have
also discussed that the horizontal velocity of a projectile
= ucosα
And time of flight, t =
2u sin α
g
∴ Horizontal range,
R = Horizontal velocity x Time of flight
= u cosα ×
R=
Note
2u sin α 2u 2 sin α cosα
=
g
g
u 2 sin 2α
....................(Qsin 2α = 2 sin α cosα )
g
for a given velocity of projectile, the range will be maximum when sin2α = 1. Therefore
34
2α = 90˚ or α
or
Rmax =
= 45˚
u 2 sin 90° u 2
= ..................(Q sin 90° = 1)
g
g
5.8 Maximum height of a projectile on a horizontal plane
We have already discussed that the vertical component of the initial velocity of a projectile
= u sin α ……………………………….(i)
And vertical component of final velocity
= 0………………………………………(ii)
∴ Average velocity of (i ) and (ii ),
u sin α + 0 u sin α
=
=
..............................................(iii )
2
2
Let H be the maximum height reached y the particle and t be the time taken by the particle to
reach maximum height i.e., to attain zero velocity from (usinα). We have also discussed that time
taken by the projectile to reach maximum height,
=
u sin α
g
∴ Maximum height of the projectile,
H = Average vertical velocity × time
=
u sin α u sin α u 2 sin 2 α
×
=
2
g
2g
Example 6
If a particle is projected inside a horizontal tunnel which is 5 meters high with a velocity of
60m/sec, find the angle of projection and the greatest possible range.
Solution. Given : H = 5 m; u = 60 m/sec
Angle of Projection
Let
α = Angle of Projection.
We know that height of tunnel (H)
u 2 sin 2 α (60 ) sin 2 α
=
= 183.7 sin 2 α
2g
2 × 9.8
2
5=
35
∴ sin α = 0.1650 or α = 9°30′
Greatest possible range
We know that greatest possible range,
u 2 sin 2α (60 ) sin (2 × 9°30′)
(60) sin 19° = 3600 × 0.3256 m = 119.6m
R=
=
m=
g
9.8
9.8
9.8
2
2
Exercise1
A ball is thrown upwards with a speed of 10 m/s making an angle 30° with horizontal and
returning to ground on same horizontal level. Find (i) time of flight and (ii) and time to reach the
maximum height . (Ans. (i) t = 1s; (ii) 0.5s )
Exercise 2
A ball is projected upwards with a velocity of 60 m/s at an angle 60° to the vertical. Find the
velocity of the projectile after 1 second. (Ans. v= 55.68m/s )
36
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 6
6.1 Mass and Weight
The mass of an object is a fundamental property of the object; a numerical measure of its
inertia; a fundamental measure of the amount of matter in the object. Definitions of mass
often seem circular because it is such a fundamental quantity that it is hard to define in
terms of something else. All mechanical quantities can be defined in terms of mass, length, and
time. The usual symbol for mass is m and its SI unit is the kilogram. While the mass is normally
considered to be an unchanging property of an object, at speeds approaching the speed of light
one must consider the increase in the relativistic mass.
The weight of an object is the force of gravity on the object and may be defined as the
mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the
Newton. Density is mass/volume.
For an object in free fall, so that gravity is the only force acting on it, then the expression
for weight follows from Newton's second law.
37
38
6.1 Mass versus weight
6.2 How are weight and mass different?
To understand the differences we need to compare a few points:
1) Mass is a measurement of the amount of matter something contains, while Weight is the
measurement of the pull of gravity on an object.
2) Mass is measured by using a balance comparing a known amount of matter to an unknown
amount of matter. Weight is measured on a scale.
3) The Mass of an object doesn't change when an object's location changes. Weight, on the other
hand does change with location.
6.3 Newton’s three laws of motion
Let us begin our explanation of how Newton changed our understanding of the Universe by
enumerating his Three Laws of Motion.
6.3.1 Newton's First Law of Motion:
Everybody continues in its state of rest or uniform motion in a straight line, unless it is acted on
by a resultant force.
6.3.2 Newton's Second Law of Motion:
The change of momentum per unit time is proportional to the impressed force, and takes place in
the direction of the straight line along which the force acts.
6.3.4 Newton's Third Law of Motion:
For every action there is an equal and opposite reaction.
39
6.4 Inertia
Inertia is the resistance an object has to a change in its state of motion. The principle of inertia is
one of the fundamental principles of classical physics which are used to describe the motion of
matter and how it is affected by applied forces.
6.4.1 Moment of inertia
Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m2,
Former British units slug ft2), is the rotational analog of mass. That is, it is the inertia of a rigid
rotating body with respect to its rotation. The moment of inertia plays much the same role in
rotational dynamics as mass does in basic dynamics, determining the relationship between
angular momentum and angular velocity, torque and angular acceleration, and several other
quantities. While a simple scalar treatment of the moment of inertia suffices for many situations,
a more advanced tensor treatment allows the analysis of such complicated systems as spinning
tops and gyroscope motion.
The symbols I and sometimes J are usually used to refer to the moment of inertia.
Moment of inertia is the rotational analogue to mass. Review the definitions as explained in your
text book.
The following table contains moments of inertia for various common bodies. The 'M' in each
case is the total mass of the object.
slender
rod:
axis
throug
h
center
axis
throug
h end
rectangul
ar plane:
axis
throug
h
center
axis
along
edge
sphere
thinwalled
hollow
solid
cylinder
hollow
solid
40
thinwalled
hollow
41
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 7
7.0 Impulse and Momentum
7.1 Impulse
Impulse is defined as the product of force and the time for which it is applied.
Impulse = Force x Time = Ft
Example 1
A vehicle has a force of 400N applied to it for 20 seconds. Calculate the impulse?
Solution
IMPULSE = Ft = 400 x 20 = 8000 N s
7.2 Momentum
Momentum is defined as the product of mass and velocity.
Momentum = Mass x velocity = m u kg m/s
Example 2
A vehicle of mass 5000 kg changes velocity from 2 m/s to 6 m/s. calculates the change in
momentum.
Solution
Initial momentum = mu1 = 5000 x 2 = 10000 kg m/s
Final momentum = mu2 = 5000 x 8 = 40000 kg m/s
Change in momentum = 40000 – 10000 = 30000 kg m/s
7.2.1 The Law of Conservation of Momentum
The total momentum before a collision is equal to the total momentum after the collision.
Consider two bodies of mass m1 and m2 moving at velocities u1 and u2 in the same direction.
After collision the velocities change to v1 and v2 respectively.
42
10
The initial momentum = m1u1+m2u2
Figure 11
7.3 Radius of Gyration
If the entire mass of a given body assumed to be concentrated at certain point, at a distance k
from the given axis, such that
M .k 2 = I
∴k =
I
m
…..(where I is the mass moment of inertia of the body)
43
The distance k is called radius of gyration. Thus the radius of gyration of a body may be defined
as the distance from the axis of reference where the whole mass (or area) of a body is assumed to
be concentrated.
Example 3
A circular wheel of mass 50 kg and radius of 200mm is rotating at 300 r.p.m. Find its kinetic
energy.
Solution. Given M = 50 kg; r = 200 mm = 0.2 m; N = 300 r.p.m.
We know that mass moment of inertia of the circular wheel,
I = 0.5 M .r 2 = 0.5 × 50 (0.2 ) = 1 kg − m 2
2
and angular velocity of the wheel,
ω = 300 r.p.m. = 5 r.p.s.=10π rad/s
∴ Kinetic energy of the wheel,
I .ω 2 1(10 π )
E=
=
= 493.5 N − m
2
2
Exercise1
A wheel has a string of length 4m wrapped round its shaft. The string is pulled with a constant
force of 150N. It is observed that when the string leaves the axle, the wheel is rotating at 3
revolutions in a second. Find the moment of inertia of the wheel. (Ans. 3.38 kg.m2)
Exercise 2
2
44
MECHANICAL ENGINEERING SCIENCE
WEEK 8
8.0 Work Energy and Power
45
DYNAMICS
46
Motion
P
S
Fig. 1.1
8.1 WORK
Let a force P, Fig.1.1, act on a body through a distance s, then:
Work done = Force x Distance moved in the direction of force
= Ps
If the force is in newtons and the distance in metres then the units of work are newton-metres. A unit of
Work equal to one newton-metre is defined in SI system as the joule (J). 1J = 1N X 1M
The joule is also the unit of energy and heat and since it is a rather a small quantity it is more often
47
Convenient to use the following multiples:
1kilojoule (KJ) = 103J
1megajoule (MJ) = 106J
1gigajoule (GJ) = 109J
If the force is inclined to the direction of the motion of the body, then the amount of work done is the product of the
component of the force in the direction of motion and the distance moved by the body. Thus, with reference to
fig.1.2
P
θo
MOTION
S
Fig.1.2
P
PSINθ
θo
Direction of motion
PCOSθ
Component of force P in the direction of motion = Pcosθ
Hence, Work done in moving through distance S = Pcosθ x S
It should be noted that the component PSINθ acts at right angles to the direction of motion, since there is no
movement in that direction, this component does no work
48
.
Exercise
Example 2
A man exerts a force of 200 N lifting a cylinder head through a distance of 1.7 m what is the work done?
Solution 2
Work done = force x distance moved in the direction of the force
= 200 N x 1.7 m
= 340 Nm
In the strict engineering sense, no work is done if the man puts the cylinder head down again in the same place as the
total displacement of the head is zero. Similarly, if a force is exerted on an object without producing motion, no
work is done although a considerable amount of energy may have
49
8.2.1
Work done by a torque
Torque is the name given to a force which tends to produce rotary motion. When the applied torque is great enough
to overcome the frictional and other resistances and motion results, work is done. Hence
Work done = Torque x Distance moved
= F x r x 2π
Where r is the radius of the circle.
When a constant torque is applied tending to maintain rotation then
Work done =F x r x 2π x N ………………………………. 2.1
Where N is the number of revolutions.
Example3
A fitter applies a force of 100 N on each arm of a die-stock whilst turning it through 17 revolutions. If the arms are
0.15 m long, how much work does he do?
Solution
Torque T = Force x Distance at right-angles to fulcrum
= 100 x 2 x 0.15
= 30 Nm
Now Work done = Torque x Distance moved
= 30 x 0.3 x 3.142 x 17
= 480.7 Nm
Work is also done when a constant torque turns through an angle
s
θ
ω
A
O
P
Fig.2.1
Consider an arm OA Fig. 2.1 of radius r metres fixed to a shaft at O. let a constant force P newtons be applied
at A at right angles to the radius OA. Then by the argument given above,
Torque T applied to shaft = Force x Radius
T = P X r newton metres
Now suppose the force causes point A to move through distance S metres measured round the circumference of the
circle traced by A. Then:
Work done = force x distance moved in the direction of the force
= P s joules
If θ (measured in radians) is the angle turned through by arm OA, then :
S=rθ
Such that, work done = P x r θ = pr x θ
50
Then,
work done in joules = T X θ
When a constant torque is applied, the formulae for torque and distance are integrated and
Work done = 2π x N x T …………………….. 2.2
Now let us proceed to the case of the power transmitted by a constant torque. If the time taken by arm OA to turn
through the angle θ radians is t seconds, then
Power transmitted by torque T = T x θ / t watts
But θ/t = ω the angular velocity of the shaft in radians / second
Hence,
Power transmitted by torque = T ω watts
If the shaft or arm OA is rotating at a constant speed of N rev/min, then since
ω = 2 π N / 60 rad/s
Then
Power transmitted by torque = 2π x N x T / 60 watts …………….. 2.3
Fig.2.2 TORQUE PRODUCED BY A
ROTARY MOTION
51
8.3 Energy
Energy is defined as the capacity for doing work. Energy can exist in various forms such as
mechanical energy, heat energy, chemical energy, and electrical energy. In all forms, the unit of
energy is the Joule.
There are two important kinds of mechanical energy, namely potential energy and kinetic energy
8.3.1 Potential energy
Potential energy ( P.E ) is the energy a body possesses due to it’s position in a gravitational
field, i.e due to it’s height above the ground ( or any conveniently reference level ).
If a body of mass m kilogrammes is raised vertically through a height h metres as shown in
Fig.4.1, the force required will be the gravitational force acting on the body, i.e., it’s weight W =
mg newtons. Hence the work done in lifting the body will be mgh joules. This amount of work
will be ‘ stored up ‘ in the body as potential energy by virtue of it’s position relative to the
ground. In other words , if the body is allowed to fall freely to ground level again, it will be
capable of doing mgh joules of work. Thus:
Potential energy (P.E) = mgh ……………………….. 5.1
Body of mass m
W = mg
Height raised above ground
=h
An object can store energy as the result of its position. For example, the heavy ball of a
demolition machine is storing energy when it is held at an elevated position. This stored energy
of position is referred to as potential energy. Similarly, a drawn bow is able to store energy as the
result of its position. When assuming its usual position (i.e., when not drawn), there is no energy
stored in the bow. Yet when its position is altered from its usual equilibrium position, the bow is
able to store energy by virtue of its position. This stored energy of position is referred to as
potential energy. Potential energy is the stored energy of position possessed by an object.
52
Example 11
A car of mass 900 kg is raised by a garage lift to a height of 2 m above the ground. Determined
the potential energy possessed by the car due to it’s position with respect to the ground.
Solution
P.E possessed by car = mgh
= 900 x 9.81 x 2
= 17658 Nm = 17658 J = 17.658 kJ
P.E possessed by car =17.658 kJ
8.3.2 Kinetic energy
Kinetic Energy Concept
Kinetic energy is energy of motion. The kinetic energy of an object is the energy it possesses
because of its motion. The kinetic energy* of a point mass m is given by
Kinetic energy is an expression of the fact that a moving object can do work on anything it hits;
it quantifies the amount of work the object could do as a result of its motion. The total
mechanical energy of an object is the sum of its kinetic energy and potential energy.
For an object of finite size, this kinetic energy is called the translational kinetic energy of the
mass to distinguish it from any rotational kinetic energy it might possess - the total kinetic
energy of a mass can be expressed as the sum of the translational kinetic energy of its center of
mass plus the kinetic energy of rotation about its center of mass.
*This assumes that the speed is much less than the speed of light. If the speed is comparable with
c then the relativistic kinetic energy expression must be used
53
Energy as the capacity for doing work is a convertible currency. To give something kinetic
energy you must do work on it. This development uses the concept of work as well as Newton's
second law and the motion equations. It is a special case of the work-energy principle, a
powerful general principle of nature.
54
From the Newton’s first law of motion, in order to set a body in motion a force must be applied
to the body to overcome it’s inertia, and this force acting through the displacement of the body
does work. This amount of work done will be ‘stored up ‘in the body as kinetic energy.
Acceleration = a
P
m
m
Initial velocity
Fig.5.2
s
u=0
Final velocity = v
Let a constant unresisted force P act on a body of mass m, initially at rest, and displace it in a
straight line through a distance s, as shown in Fig. 4.2. Then:
Work done by force = Ps
If a is the acceleration produced, then:
Accelerating force, P= ma
so that,
Work done = mas ……………. 5.2
If the velocity reached in a distance s is v, then
2
2
2
−u
v
v
s=
0r s =
(since u=0)
2a
2a
Then:
Work done = ma x
The expression
1
mv2
2
v
2
2a
=
1
mv2 ……………………5.3
2
is the kinetic energy possessed by the body when moving with a
velocity v,
Hence,
Kinetic energy ( K.E ) =
1
mv2
2
……………………………. 5.4
Example 11
A motor vehicle of mass 2 tonnes is travelling at 50.4 km/h. What is it’s kinetic energy? (1
tonne = 1000 kg)
Solution
Mass of vehicle = m = 2 x 1000 kg = 2000 kg
50.4 x 5
Velocity of vehicle = v =
= 14 m/s
18
1
Kinetic energy ( K.E ) = mv2
2
1
= x 2000 x 142
2
= 196000 J = 196 kJ
Example 12
A vehicle of mass 1600 kg increases it’s speed uniformly from 36 km/h to 72 km/h by the
action of an average accelerating force of 2.4 kN . By how much will it’s kinetic energy have
increased during the acceleration ? Show that this increase in kinetic energy is equal to the work
done by the accelerating force.
55
Solution
Initial velocity u = 36 km/h = 10 m/s
Final velocity v = 72 km/h = 20 m/s
1
1
Increase in K.E =
m( v2 – u2 ) = x 1600 x (202 – 102) = 240000 J = 240 Kj
2
2
Now, since accelerating force, P = 2.4 KN = 2400 N
P
2400
Acceleration, a =
=
= 1.5 m/s2
m
1600
2
2
2
2
−10
−u
v
20
= 100 m
Distance travelled, s =
=
2a
2 x 1.5
Work done = Pxs
Work done = 2400 x 100 = 240000 J = 240 kJ = Increase in K.E
8.3.3 Conservation of energy
The principle of conservation of energy states that: ‘Energy can neither be created nor
destroyed’.
Energy can be converted from one form to another, but it is found that a loss of energy in one
form is always accompanied by an equivalent increase in another form. In all such conversions
the total amount of energy remains constant.
Let us consider again the body of mass m raised to the height h above the ground (Fig.4.1). We
have seen that the potential energy possessed by the body due to it’s height above the ground is
mgh joules. If the body is then allowed to fall freely from that height until it is just about to strike
the ground, all it’s available energy will be given up. If no external work is done on or by the
body during it’s time of fall then, by the principle of the conservation of energy , the body will
gain kinetic energy equal in amount to the original potential energy. Hence,
K.E on reaching the ground = Original P.E.
= mgh
1
But
Kinetic energy (K.E) = mv2
2
1
Therefore
mv2
=
mgh
2
Or
v = ( 2 gh ) …………………………… 5.5
Where v is the velocity of the body at the point of impact with the ground.
Now suppose the body falls through a vertical distance x, then:
= K.E. gained
P.E. still possessed by body = mg (h-x ) ……………………………… 5.6
Example 13
I n a drop-forging operation, the top die an it’s holder, with a total mass of 40 kg.fall freely for
3m before contacting the metal resting on the bottom die. Calculate;
(a) The velocity and kinetic energy of the top die at impact;
56
(b) The force exerted on the metal (assuming it to be constant) if the top die is brought to rest
in 20 mm after impact.
Solution
(a)
v=
=
(K.E) =
( 2 gh )
( 2 x 9.81 x3 ) = 7.67 m/s
1
mv2
2
1
x 40 x 7.672
2
= 1177 J = 1.177kJ
=
Check:
K.E gained = P.E lost
= mgh = 40 x 9.81 x 3 = 1177 J
(b) Constant force exerted on metal = P = ?
Distance moved by top die after impact = s = 0.02 m
Now
P.E. lost = Work done on metal
i.e
mgh = Ps
mgh
1177
=
= 58850 N
Therefore:
P=
s
0.02
= 58.850kN
8.3.4 Kinetic energy of a rotating body
V
m
r
O
ω
o
Fig.5.4
Suppose a very small body of mass m rotates with a tangential velocity v about centre O in a
circular path of radius r, (Fig.5.4). Then
1
K.E. of body = mv2 ……………………………. 5.4
2
Let the angular velocity of the body be ω rad/s then,
v=ωr
so that,
2
2
2 2
v = (ω r) = ω r ………………………………5.5
Substituting the value of v2 from equation 4.7 in to equation 4.6a, we get:
1
K.E. of rotating body =
m ω2 r2 …………………………… 5.6
2
57
8.3.5 EFFICIENCY
In any mechanical device, whether it is an engine, a pump, a motor or a dynamo, a considerable amount of the work
put in to the device is lost in overcoming friction, etc so that the useful work done by the device is always much less
than the actual work put in to the device.
The fraction :
Useful work done
Actual work put in
or
Work output
Power output
or
……………… 4.1
Work input
Power input
Is called the efficiency of the device, and is expressed as a ‘ per unit ‘ value or as a percentage. The percentage value
( % ) is obtained simply by multiplying the ‘ per unit ‘ value by 100.
Example 8
In a lifting machine a load of 1.8kN is lifted by a force (or effort) of 150kN, which moves 750 mm for every 25 mm
moved by the load. What is the efficiency of the machine under these conditions?.
Solution
For every 25 mm lift of load,
Useful work done by the machine = Load x Vertical distance moved
= 1.8 x 103 x 25 x10-3
= 45 J
Actual work put into machine = Force (or effort) x Distance moved
= 150 x 750x 10-3 = 112.5 J
Useful work done
45 J
=
= 0.4 per unit = 0.4 x 100
Actual work put in 112.5 J
Efficiency =
= 40 %
Example 9
A pump is required to raise 27000 litres of water per hour through a vertical height of 15 m. If the pump has an
efficiency of 75 %, determine the power required to drive it. (1 litre of water has a mass of 1 kg.)
Solution
Mass of 27000 litres of water = 27000kg
Weight of water to be raised through 15 m/second =
27000 x 9.81
= 73.58 N
60 x 60
Useful work to be done/second = 73.58 x 15 = 1104 J
i.e Power output = 1104 W (since 1W = 1 J/s)
= 1.104 kW
Power input =
Power output
100
= 1.104 x
= 1.472 k W
Efficiency
75
Power required to drive pump = 1.472 Kw
Example 10
The power developed in the cylinders of a motor car engine is 30 kW whilst the power delivered at the crankshaft is
found to be 24 kW. What is the mechanical efficiency of the engine?
Solution
Power output = Power available at the crankshaft = 24 Kw
Power input = Power developed in the engine cylinders = 30 kW
Power output
x 100
Power input
24 (kW )
=
x 100 = 80 %
30 (kW )
Efficiency (%) =
Mechanical efficiency = 80 %
58
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 9
8.4 Tractive effort, tractive resistance, power
Available & Power required
When a motor vehicle moves along a road at constant speed the only driving force or tractive
effort (Te) needed to be applied to the road wheels is that required to maintain the motion
against the various opposing forces which would otherwise tend to decelerate the vehicle and
bring it to rest. The tractive effort at constant speed is therefore equal to the sum of all the forces
opposing the motion of the vehicle. These forces are all included in the term tractive
resistance ( Tr ) and can be divided in to three groups, namely:
(1)
Rolling resistance ( Rr ).
(2)
Gradient resistance ( Rg ).
(3)
Air (or wind) resistance (Ra ).
Hence, for constant speed,
Tractive effort required = Sum of all the resistances
i.e.,
Te = Rr + Rg + Ra ……………………………………… 7.0
For a vehicle moving along a level road at constant speed, eq.(6.0) becomes:
Te = Rr + Ra ……………………………………… 7.1
NOTE If the vehicle is accelerated along the road, the tractive effort must exceed all the
resistances and provide the accelerating force.
8.5 Rolling Resistance
The rolling resistance ( Rr ) of a vehicle is due mainly to wheel bearing friction and to the
deformation of the tyres or road surface. It’s value defends upon the nature of the road surface,
type and size of tyres used, and upon the normal load on the tyres. It’s value is also found to
increase by too low an air pressure in the tyres.
The rolling resistance is generally assumed to be independent of the vehicle speed and is often
expressed in N/tone of total mass of vehicle.
Example 14
A vehicle of mass 2 tonnes is being propelled along a level road by a constant tractive effort of
1.25 kN. If the rolling resistance amounts to 100 N/tone of total mass, determine the acceleration
of the vehicle. Neglect air resistance.
Solution
Mass of vehicle, m = 2 t = 2000 kg
59
Tractive effort, Te = 1.25 kN = 1250N
Rolling resistance, Rr = 100 N/t = 100 N/t x 2t = 200 N
Acceleration = a
c.g
c
Rr = 100N/t
Te = 1250 N
W = mg
Fig. 7.0
Now, since vehicle is accelerated on the level and air resistance is neglected, then:
Accelerating force, P = Te – Rr
= 1250- 200 = 1050 N
But from:
Accelerating force, P = ma
Therefore :
Acceleration, a =
P
m
=
1050
2000
= 0.525 m/s2
Exercise
A car travelling on a level road rolls to rest from a speed of 72km/h in a distance of 600m. If the
mass of the car is 900kg, find the frictional resistance (which is assumed constant) during the
retardation. Assuming the same constant resistance, calculate the tractive effort required when
the car starts from rest and accelerates uniformly to 72km/h in 15s on the level.
Answer
Frictional resistance = 300N, Tractive effort = 1.5kN
8.6 Gradient resistance
When a vehicle is being propelled up an incline the tractive effort must also neutralize the effect
of the component of the weight of the vehicle down the slope. This is called the gradient
resistance (Rg) and depends entirely on the steepness of the slope and upon the weight of the
vehicle. (Remember that the weight of the vehicle is the gravitational force exerted by the earth
on it’s mass)
Consider a vehicle of mass m resting on an incline (Fig.6.2) whose gradient is 1 in G. Let θ be
the angle of the slope. Then:
G
C
1
cg
θ
mgSinθ
A
B
mgCosθ
W =mg
Fig.7.1
60
Component of weight W down the slope = Wsinθ = mgSinθ
BC
1
But,
Sinθ =
=
AC
G
Hence,
mg
Gradient resistance, Rg = mgsinθ =
……………………………… 7.2
G
NOTE When a gradient is specified as 1 in G, G actually refers to length AB, so that:
1
BC
=
= tan θ
G
AB
But since in most practical gradients θ is small, tanθ is approximately equal to sinθ.
Example15
A motor lorry has a mass of 4 Mg and is driven up a hill of gradient 1 in 20 at a constant speed of
36 km/h. If the rolling resistance amounts to 75 N/tonne, calculate the tractive effort exerted by
the vehicle. If the engine is switched off, how far will the vehicle travel before coming to rest ?
Solution
Constant speed of 36km/h
c.g
Te (=Rg + Rr )
Rr=75N/t
20
Rg=1.962kN
1
W=mg=39.24kN
mg
4 Χ 1000 Χ 9.81
=
= 1962 N
G
20
Rolling resistance, Rr = 75 x 4 = 300 N
To maintain a constant speed of 36 km/h up the hill (neglecting air resistance ),
Tractive effort, Te = Rg +Rr
= 1962 + 300 = 2262 N
Now , if the engine is switched off,
Retarding force, P = Tractive resistance = 2262 N
But from,
P = ma
P 2262
Retardation, a = =
= 0.5655 m/s2
m 4000
Now ,
Initial velocity, u = 36 km/h = 10 m/s
Final velocity, v = 0
Acceleration, a = -0.5655 m/s2
Distance travelled, s = ?
2
2
V – u = 2 as
0 – 102 =2 x (-0.5655) x s
Distance travelled, s = 88.42 m
Gradient resistance, Rg =
61
Exercise
A motor vehicle stands on the top of a hill of gradient 1 in 15 and the hand-brake fails. If the
total mass of the vehicle is 1.2 t and the total resistance to motion amounts to 200 N, calculate:
(a) The acceleration of the vehicle down the slope;
(b) The distance it will travel in 25 s;
(c) The speed of the vehicle after 25 s.
Answer: (a) Acceleration down the slope = 0.4875 m/s2, b) Distance travelled in 25 s = 152.4 m
( c) Speed after 25 s = 12.19 m/s
8.7 Air (or wind) resistance
Air (or wind) resistance (Ra) depends upon many factors, such as the shape and the frontal
area of the vehicle as well as it’s speed relative to the air. The air resistance, in newtons, is
generally given by the expression:
Ra = kAV2 ……………………………… 7.3
Where k is a constant depending on the shape of the vehicle, A the projected frontal area (m2),
and V the speed of the vehicle relative to the air (km/h).
8.8 Power available
The power available at the driving wheels at any given instant is equal to the power developed at
the crankshaft of the engine during that instant after, of course, making allowance for the
transmission frictional losses.
Hence, if
b.p. = brake power of the engine, in W
E = transmission efficiency
Then,
Power available at road wheels = b.p x E watts …………………………… 7.4
Also, if
Te = tractive effort, in N
v = vehicle speed, in m/s
Power available at road wheels = Te x v watts …………………………… 7.5
From eqs.(6.4) and (6.5). the power developed at the crankshaft of the engine at any given road
speed can be determined by the following expression:
Te x v
b.p. of engine =
watts ……………………………… 7.6
E
8.9 Power required
The power required at a given speed is that necessary to overcome the tractive resistance acting
on the vehicle at that speed.
Hence, if
Te = tractive effort, in N
v = vehicle speed, in m/s
Then.
Power required to overcome the resistances = Tr x v watts …………………….. 7.7
62
Example 16
A car of total mass 1.5 Mg is driven up an incline of 1 in 50 at a constant speed of 72 km/h while
the engine is developing 16 kW. Calculate the power required to overcome the road and wind
resistances. Neglect frictional losses of the transmission.
Solution:
Speed of car at 72km/h, v= 20 m/s
Engine power developed, b.p = 16kW =16000 W
Neglecting frictional losses,
b.p = Te x v
b. p 16000
Tractive effort, Te =
=
= 800 N
v
20
mg 15000 x 9.81
Gradient resistance, Rg =
=
= 294.3 N
G
50
At constant speed,
Te = Rg +Rr + Ra
800 = 294.3 + ( Rr + Ra )
Rr + Ra = 800 – 294.3 = 505.7 N
Now ,
Power required to overcome the road and wind resistances = (Rr + Ra) x v
= 505.7 x 20
= 10114 W = 10.114 kW
63
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 10
8.10 POWER
Power is defined as the rate of doing work. Thus if W is the amount of work done in time t, then :
Power =
Work done W
=
…………….. 3.0
t
Time taken
If the work W is done by a constant force P moving through a distance s along its line of action, then:
Power =
Force x Dis tan ce moved
Ps
=
…………………… 3.1
t
Time taken
But s/t = v the velocity of the body
Hence,
Power = Force x velocity = P x v ………………………. 3.2
The S.I unit of power is the watt (W), which is defined as a rate of working of 1 joule per second, i.e. 1 newton
metre per second. Thus:
1Nm/s = 1J/s = 1W
In practice, the watt is often found to be inconveniently small; consequently the kilowatt (kW) is frequently used,
the kilowatt being 1000watts. For still larger powers, the megawatt (MW) is used, where:
1MW = 1000kW = 103kW = 1000000W = 106W
Similarly, when we are dealing with large amounts of work (or energy), it is usually more convenient to express the
latter in kilowatt hours (kW h),
where: 1kW h = 1000 watt hour
= 1000 x 3600 watt seconds or joules
= 3.6 x 106 J = 3.6 MJ
64
Example 4
A motor car of mass 0.8 metric tonnes is being propelled up a hill of gradient 1 in 15 at a steady speed of 63 km/h.
Find the work done against gravity per minute. Neglect frictional resistances. (1 metric tonne = 1Mg = 1000kg.)
Solution
The reader should note that a gradient of 1 in 15 means that a hill (or slope) rises vertically 1 m for every 15 m
measured along the road.
63 km/h =
63 x 1000
= 1050 m/min
60
The distance travelled (measured along the road ) by the car in one minute is then 1050 m.
Mass of car = 0.8 metric tones
= 0.8 x 1000 kg = 800kg
Gravitational force acting on a mass of 800 kg = 800 x 9.81 = 7848 N
1050m
B
15
1
Vertical distance
7848N
A
C
Fig. 1
In this case, since the work done against gravity, i.e , against the gravitational force of 7848 N is required, the
distance travelled in one minute must be measured along the line of action of this force; this is the vertical distance
BC, see Fig.3.1.
Thus: Vertical distance BC =
Dis tan ce AB
1050 m
=
= 70 m
Gradient
15
Hence,
Work done against gravity / min = Force x vertical distance moved = 7848 x 70 = 549.36kJ
Work done against gravity / min = 549.36KJ
Example 5
If an engine drives a car against a total resistance of 1.2kN over a distance of a quarter of a kilometer in half a
minute, what power is being developed at the crankshaft ?.
Solution
Force at crankshaft = P= 1.2 KN = 1.2 X 103 N
Distance travelled = S = 0.25 km = 0.25 X 103 m
Time taken = t = 0.5 min = 30 s
Power developed by crankshaft =
Force x Dis tan ce moved
Ps
=
= 1.2 x 103 x 0.25 x 103 / 30
t
Time taken
= 10 x 103 W = 10 kW
Example 6
A lorry hauls a trailer at 72 km/h when exerting a steady pull of 800 N on the tow – rope. Calculate:
(a) The work done in 20 min
(i)
In megajoules, and (ii)
in kilowatt hours;
(b) The power required.
Solution
(i) distance travelled in 20 min = velocity x time = 72 x
20 km
[
x h ] = 24 km = 24 x 103 m
60 h
65
work done in 20 min = Force x distance travelled
= 800 x 24 x 103 = 19200000J = 19.2 MJ
19.2
= 5.33 KWh
3.6
Work done in kilowatt joules 19200000
(a) Power required =
=
= 16000 W = 16 Kw
Time taken in sec onds
20 x 60
(ii)
Since 3.6 MJ = 1kWh
Then
19.2 MJ =
Alternatively,
Power required =
Work done in kilowatt hours
19.2 60
=
x
= 16 kW
3. 6
20
Time taken in hours
Example 7
A motor vehicle propeller shaft transmits 44 kW at 1000 rev/min. Calculate the torque applied to the shaft in newton
metre.
Solution
Power transmitted by shaft = 44 Kw = 44000W
Rotational speed of shaft = N = 1000 rev/min
Torque applied to shaft = T = ?
Now
2πNT
watts
60
1000 rev
44000 = 2 π T X (
)
60 s
44000 x 60
T=
= 420 Nm
2π x 1000
Power transmitted =
EXERCISES
(1) Explain what is meant by the terms ‘work’ and ‘power ‘. In what units may work be expressed?
(2) A piston moved at a uniform velocity of 7 m/s against a resistance of 250N. Find the power developed in
kilowatts.
(3) A car of mass 900 kg is lifted 2m on a garage lift in a time of 45s. Calculate the power required to perform
this task.
(4) An engine, running at 2100 rev/min, develops a torque of 150 Nm. Calculate:
( a) The work done in 20 min
(i)
In megajoules, and
(ii)
in kilowatt hours;
(b) The power delivered at the crankshaft in kilowatts.
66
8.11 Transmission of motion and power
8.12Belt drive
Motion and power can be transmitted from one rotating shaft to another by means of pulleys and
an endless belt, see Fig. 6.0. The pulley A, which gives motion to the belt, is called the driving
pulley, or simply driver; the pulley which receives the motion is called the driven pulley, or
simply driven. The application of belt drive to the modern motor vehicle is usually limited to the
dynamo, fan and water-pump.
B
A
A
Fig.2
8.13 Belt
A Belt is a looped strip of flexible material, used to mechanically link two or more rotating
shafts. They may be used as a source of motion, to efficiently transmit power, or to track relative
movement.
Belts are looped over pulleys. In a two pulley system, the belt can either drive the pulleys in the
same direction, or the belt may be crossed, so that the direction of the shafts is opposite.
As a source of motion, a conveyor belt is one application where the belt is adapted to continually
carry a load between two points.
8.14 Power Transmission
Power transmission is achieved by specially designed belts and pulleys. The demands on a belt
drive transmission system are large and this has led to many variations on the theme.
67
8.15 Flat belts
Fig. 3
Belts on a Yanmar 2GM20 marine diesel engine.
Pulleys were made with a slightly convex face (rather than flat) to keep the belts centered. The
flat belt also tends to slip on the pulley face when heavy loads are applied. In practice, such belts
were often given a half-twist before joining the ends so that wear was evenly distributed on both
sides of the belt (DB).
8.16 Round belts
Round belts are a circular cross section belt designed to run in a pulley with a circular (or near
circular) groove. They are for use in low torque situations and may be purchased in various
lengths or cut to length and joined, either by a staple, gluing or welding Early sewing machines
utilized a leather belt, joined either by a metal staple or glued, to great effect.
8.17 Vee belts
Vee belts (also known as V-belt or wedge rope) are an early solution that solved the slippage and
alignment problem. The V-belt was developed in 1917 by John Gates of the Gates Rubber
Company. The "V" shape of the belt tracks in a mating groove in the pulley (or sheave), with the
result that the belt cannot slip off. The belt also tends to wedge into the groove as the load
increases — the greater the load, the greater the wedging action — improving torque
transmission and making the vee belt an effective solution. They can be supplied at various fixed
lengths or as a segmented section, where the segments are linked (spliced) to form a belt of the
required length. For high-power requirements, two or more vee belts can be joined side-by-side
in an arrangement called a multi-V, running on matching multi-groove sheaves. The strength of
these belts is obtained by reinforcements with fibers like steel, polyester .
68
8.18 Timing Belts
Fig. 4 Timing belt
Timing belts, (also known as Toothed, Notch or Cog) belts are a positive transfer belt and can
track relative movement. These belts have teeth that fit into a matching toothed pulley. When
correctly tensioned, they have no slippage and are often used to transfer direct motion for
indexing or timing purposes (hence their name). Camshafts of automobiles and stepper motors
often utilize these belts.
Timing belts with a helical offset tooth design are available. The helical offset tooth design forms
a chevron pattern and causes the teeth to engage progressively. The chevron pattern design is
self-aligning. The chevron pattern design does not make the noise that some timing belts make at
idiosyncratic speeds, and is more efficient at transferring power (up to 98%).
[edit] Specialty Belts
Belts normally transmit power only on the tension side of the loop. However, designs for
continuously variable transmissions exist that use belts that are a series of solid metal blocks,
linked together as in a chain, transmitting power on the compression side of the loop.
A belt and pulley system is characterized by two or more pulleys in common to a belt. This
allows for mechanical power, torque, and speed to be transmitted across axes and, if the pulleys
are of differing diameters, a mechanical advantage to be realized.
A belt drive is analogous to that of a chain drive, however a belt sheave may be smooth (devoid
of discrete interlocking members as would be found on a chain sprocket, spur gear, or timing
belt) so that the mechanical advantage is given by the ratio of the pitch diameter of the sheaves
only (one is not able to count 'teeth' to determine gear ratio).Belt and pulley systems can be very
efficient, with stated efficiencies up to 98%.
69
Fig. 5 Belt and pulley systems
8.19 Pulley
A pulley (also called a sheave or block) is a wheel with a groove between two flanges around its
circumference. A rope, cable or belt usually runs inside the groove. Pulleys are used to change
the direction of an applied force, transmit rotational motion, or realize a mechanical advantage in
either a linear or rotational system of motion.
8.20 Rope and pulley systems
Also called block and tackles, rope and pulley systems (the rope may be a light line or a strong
cable) are characterized by the use of one rope transmitting a linear motive force (in tension) to a
load through one or more pulleys for the purpose of pulling the load (often against gravity.) They
are often included in the list of simple machines.
In a system of a single rope and pulleys, when friction is neglected, the mechanical advantage
gained can be calculated by counting the number of rope lengths exerting force on the load.
Since the tension in each rope length is equal to the force exerted on the free end of the rope, the
mechanical advantage is simply equal to the number of ropes pulling on the load. For example,
in Diagram 3 below, there is one rope attached to the load, and 2 rope lengths extending from the
pulley attached to the load, for a total of 3 ropes supporting it. If the force applied to the free end
of the rope is 10 lb, each of these rope lengths will exert a force of 10 lb. on the load, for a total
of 30 lb. So the mechanical advantage is 3.
The force on the load is increased by the mechanical advantage; however the distance the load
moves, compared to the length the free end of the rope moves, is decreased in the same
proportion. Since a slender cable is more easily managed than a fat one (albeit shorter and
stronger), pulley systems are often the preferred method of applying mechanical advantage to the
pulling force of a winch (as can be found in a lift crane).
Pulley systems are the only simple machines in which the possible values of mechanical
advantage are limited to whole numbers.
70
In practice, the more pulleys there are, the less efficient a system is. This is due to sliding friction
in the system where cable meets pulley and in the rotational mechanism of each pulley.
It is not recorded when or by whom the pulley was first developed. It is believed however that
Archimedes developed the first documented block and tackle pulley system, as recorded by
Plutarch. Plutarch reported that Archimedes moved an entire warship, laden with men, using
compound pulleys and his own strength.
8.21 Types of systems
Fixed pulley
Movable pulley
Fig. 6 These are different types of pulley systems:
Fixed A fixed or class 1 pulley has a fixed axle. That is, the axle is "fixed" or anchored
in place. A fixed pulley is used to change the direction of the force on a rope (called a belt).
A fixed pulley has a mechanical advantage of 1. A mechanical advantage of one means that
the force is equal on both sides of the pulley and there is no multiplication of force.
Movable A movable or class 2 pulley has a free axle. That is, the axle is "free" to
move in space. A movable pulley is used to multiply forces. A movable pulley has a
mechanical advantage of 2. That is, if one end of the rope is anchored, pulling on the other
end of the rope will apply a doubled force to the object attached to the pulley.
71
Compound A compound pulley is a combination of a fixed and a movable pulley
system.
Block and tackle - A block and tackle is a compound pulley where several pulleys are
mounted on each axle, further increasing the mechanical advantage.
How it works
Diagram 2 - A simple Diagram 2a - Another A practical
Diagram 1 - A basic
pulley system - a single simple pulley system compound pulley
equation for a pulley: In
equilibrium, the force F on movable pulley lifting similar to diagram 2, corresponding to
diagram 2a.
the pulley axle is equal and weight W. The tension but in which the
in each line is W/2,
lifting force is
opposite to the sum of the
yielding
an
advantage
redirected downward.
tensions in each line leaving
of 2.
the pulley, and these
tensions are equal.
The simplest theory of operation for a pulley system assumes that the pulleys and lines are
weightless, and that there is no energy loss due to friction. It is also assumed that the lines do not
stretch.
Fig.7
http://en.wikipedia.org/wiki/Image:Crane_pulley_4x.jpg
Fig. 9
Fig. 8 A crane using the compound pulley system yielding
an advantage of 4. The single fixed pulley is installed on
the crane. The two movable pulleys (joined together) are
72
attached to the hook. One end of the rope is attached to the crane frame, another - to the winch.
In equilibrium, the total force on the pulley must be zero. This means that the force on the axle of
the pulley is shared equally by the two lines looping through the pulley. The situation is
schematically illustrated in diagram 1. For the case where the lines are not parallel, the tensions
in each line are still equal, but now the vector sum of all forces is zero.
A second basic equation for the pulley follows from the conservation of energy: The product of
the weight lifted times the distance it is moved is equal to the product of the lifting force (the
tension in the lifting line) times the distance the lifting line is moved. The weight lifted divided
by the lifting force is defined as the advantage of the pulley system.
It is important to notice that a system of pulleys does not change the amount of work done. The
work is given by the force times the distance moved. The pulley simply allows trading force for
distance: you pull with less force, but over a longer distance.
In diagram 2, a single movable pulley allows weight W to be lifted with only half the force
needed to lift the weight without assistance. The total force needed is divided between the lifting
force (red arrow) and the "ceiling" which is some immovable object (such as the earth). In this
simple system, the lifting force is directed in the same direction as the movement of the weight.
The advantage of this system is 2. Although the force needed to lift the weight is only W/2, we
will need to draw a length of rope that is twice the distance that the weight is lifted, so that the
total amount of work done (Force x distance) remains the same.
A second pulley may be added as in diagram 2a, which simply serves to redirect the lifting force
downward, it does not change the advantage of the system.
Diagram 3 - A
simple compound
pulley system - a
movable pulley and
a fixed pulley
lifting weight W.
The tension in each
line is one W/3,
yielding an
advantage of 3.
Diagram 3a - A simple
compound pulley system a movable pulley and a
fixed pulley lifting weight
W, with an additional
pulley redirecting the
lifting force downward.
The tension in each line is
one W/3, yielding an
advantage of 3.
Diagram 4a - A more
complicated compound
pulley system. The
tension in each line is
W/4, yielding an
advantage of 4. An
additional pulley
redirecting the lifting
force has been added.
Figure 4b - A
practical block and
tackle pulley system
corresponding to
diagram 4a. Note
that the axles of the
fixed and movable
pulleys have been
combined.
Fig. 10
The addition of a fixed pulley to the single
pulley system can yield an increase of advantage. In
diagram 3, the addition of a fixed pulley yields a lifting advantage of 3. The tension in each line
is W/3, and the force on the axles of each pulley is 2W/3. As in the case of diagram 2a, another
73
pulley may be added to reverse the direction of the lifting force, but with no increase in
advantage. This situation is shown in diagram 3a.
This process can be continued indefinitely for ideal pulleys with each additional pulley yielding a
unit increase in advantage. For real pulleys friction among rope and pulleys will increase as more
pulleys are added to the point that no advantage is possible. It puts a limit for the number of
pulleys usable in practice. The above pulley systems are known collectively as block and tackle
pulley systems. In diagram 4a, a block and tackle system with advantage 4 is shown. A practical
implementation in which the connection to the ceiling is combined and the fixed and movable
pulleys are encased in single housings is shown in figure 4b.
Other pulley systems are possible, and some can deliver an increased advantage with fewer
pulleys than the block and tackle system. The advantage of the block and tackle system is that
each pulley and line is subjected to equal tensions and forces. Efficient design dictates that each
line and pulley be capable of handling its load, and no more. Other pulley designs will require
different strengths of line and pulleys depending on their position in the system, but a block and
tackle system can use the same line size throughout, and can mount the fixed and movable
pulleys on a common axle.
8.22 Single Pulley Systems
There are basically only two pulley configurations you can build with a single pulley. These
systems are illustrated below and analysed for cases of static equilibrium, or "massless" pulleys.
A single pulley can change the direction of a force. Force is transferred from
body to body by a rope (or string, cord, etc.) which can be bent around corners
with a pulley. In these configurations the pulley changes only the direction of the
force, not the magnitude. In all likelihood, this is the only way you've seen pulleys
used in your physics life, but pulleys are most helpful when they provide
mechanical advantage. This system does not provide mechanical advantage.
A single pulley can also be used to change the magnitude of a force. Again, a
rope is used to transfer the force to a pulley, which (this time without bending it)
doubles or halves the applied force. In these configurations, the pulley changes
only the magnitude of the force, but not the direction. Notice also that the
acceleration of the objects on either side of the pulley are doubled or halved as well.
Fig. 11 In this
moving.
on the left
diagram, you can see why the force is doubled if the pulley is not
Notice that the rope passing through the pulley is attached to a weight
and to the ground on the right. Let's look at that part of the system
74
Fig. 12
Simple pulley
Fig. 13
75
Single fixed pulley
Single movable pulley
Fig. 14
Observe the tension in a string or rope. Tie the upper end of a string to a support, and tie a brick
to the lower end. The string will be tight, i.e. have tension all along it.
Block and tackle
Pulleys
76
Fig. 15
This pulley system is used in cranes and lifts. In a car garage, the mechanics can lift a car engine
out of a car by hand using a block and tackle. You will notice that they pull down a long way
while the engine block moves up a short way. In the diagram the pulleys have been separated
here to show the path of the rope more clearly. Find the gain in force from the number of strings
supporting the load. The tension in the string remains constant and is one fourth of the upward
pull on the load.
77
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 11
9.1 Chain drive
The chain drive is a positive form of drive and consists sprocket and chain wheels which are
connected by endless chain. The sprocket is considered as the driver and the chain wheel as the
driven member.
Chain (object), series of links or rings, usually made of metal, joined to form a flexible cable.
Chains are used in ornaments and jewelry, and to perform a variety of mechanical functions,
including lifting, pulling, restraining heavy loads, and transmitting power. The shape of the link
determines the chain’s function. Stud-link chain, for example, has a metal stud running across
the inside width of each link to increase strength for heavyweight applications. The links of roller
chain are designed to fit over the teeth of a sprocket and are used on bicycles and industrial
machinery.
Fig. 1
9.2 Common Types of Chains
Chains come in varying shapes and sizes, depending on their intended use. Three common types
of chains are shown here.
On most motorcycles, the transmission delivers engine power to the rear wheel via a drive chain.
Chains stretch with age and require periodic adjustment. In some motorcycles, a cogged rubber
78
drive belt or an enclosed drive shaft replaces the drive chain. Belts and drive shafts do not
require adjustments and operate more quietly than chains.
9.3 Gear drive
Belt tend to slip, chains stretch and become noisy in operation. Therefore, to ensure a drive
which can not slip and is reasonably quite, the method of transmitting rotary motion by gear
wheels has been generally adopted for most types of motor – vehicle transmissions.
9.4 How Gears Work
Fig. 2 Gears are found in
everything from cars to
clocks.
Gears are used in tons of mechanical devices. They do several
important jobs, but most important, they provide a gear
reduction in motorized equipment. This is key because, often, a
small motor spinning very fast can provide enough power for a
device, but not enough torque. For instance, an electric
screwdriver has a very large gear reduction because it needs lots
of torque to turn screws, but the motor only produces a small
amount of torque at a high speed. With a gear reduction, the
output speed can be reduced while the torque is increased.
Another thing gears do is adjust the direction of rotation. For instance, in the differential between
the rear wheels of your car, the power is transmitted by a shaft that runs down the center of the
car, and the differential has to turn that power 90 degrees to apply it to the wheels.
There are a lot of intricacies in the different types of gears. In this article, we'll learn exactly how
the teeth on gears work, and we'll talk about the different types of gears you find in all sorts of
mechanical gadgets.
9.5 Spur Gears
Spur gears are the most common type of gears. They have
straight teeth, and are mounted on parallel shafts. Sometimes,
many spur gears are used at once to create very large gear
79
reductions.
Photo courtesy Emerson Power Transmission Corp.
Figure 3. Spur gears
Spur gears are used in many devices that you can see like the electric screwdriver, dancing
monster, oscillating sprinkler, windup alarm clock, washing machine and clothes dryer. But you
won't find many in your car.
This is because the spur gear can be really loud. Each time a gear tooth engages a tooth on the
other gear, the teeth collide, and this impact makes a noise. It also increases the stress on the gear
teeth.
To reduce the noise and stress in the gears, most of the gears in your car are helical.
9.6 Worm Gears
Worm gears are used when large gear reductions are needed. It is common for worm gears to
have reductions of 20:1, and even up to 300:1 or greater.
80
Photo courtesy Emerson Power Transmission Corp.
Figure 4. Worm gear
Many worm gears have an interesting property that no other gear set has: the worm can easily
turn the gear, but the gear cannot turn the worm. This is because the angle on the worm is so
shallow that when the gear tries to spin it, the friction between the gear and the worm holds the
worm in place.
This feature is useful for machines such as conveyor systems, in which the locking feature can
act as a brake for the conveyor when the motor is not turning. One other very interesting usage of
worm gears is in the Torsen differential, which is used on some high-performance cars and
trucks.
Virtually all motor vehicles have a transmission, whether it is automatic
or manual. The primary functions of the transmission are to transmit the
rotation of the engines crankshaft into a more usable rotational speed and
to multiply the torque (see torque and horsepower) of the engine's
output.
Fig. 5
81
DYNAMICS
MECHANICAL ENGINEERING SCIENCE
WEEK 12
10.1 Machines
The term ‘machine’ is applied to any device with which a large load can be moved by a
relatively small effort. Or it can be define as a device which receives energy and uses or converts
it to do work in a better way. The energy received by a machine may be in any form, and it is
usually called the power input, applied force, power or effort. The output of a machine is
called the load or power output.
Many forms of machine are found in common use, for example, the spanner which increases the
muscular effort of the engineer to tighten a nut. All lifting machines, the crane, jack and hoist,
are further example.
Three important factors arise when considering the usefulness of a machine.
10.2 Mechanical Advantage (M.A)
The normal function of a machine is to move a load that cannot be moved by the application of a
direct effort. Though the maximum load that can be lifted manually is relative small, by use of a
suitable machine, e.g. a block and tackle, a very large load can be lifted with a small effort
(Fig.8.1). the relationship between the load and the effort is called the mechanical advantage or
force ratio.
Load ( L)
Mechanical advantage (M.A.) =
……………………………… 1
Effort( E )
MACHINE
EFFORT
Fig.1
LOAD
The mechanical advantage of any machine is not constant. This is due to the friction forces in the
machine which must be overcome before any load is moved. Hence the mechanical advantage of
a machine is usually low at low speeds and loads and increases as the load or speed is increased.
10.3 Velocity ratio (V.R.)
As it is impossible to get more work out of a machine than was originally put in, then the amount
of work put in to the ‘effort ’ side of a machine must be greater than the amount produced at the ‘
load ‘ side. This is because friction in the machine absorbs some of the effort. In machines the
movement of the driving or input part bears a definite relationship to the movement of the output
or load.
This comparison between the distances moved is called the velocity or movement ratio.
82
Dis tan ce moved by effort
……………………….. 2
Dis tan ce moved by load
The velocity ratio of any simple machine is a constant because its value is fixed by the
dimensions of the machine.
Velocity ratio (V.R.) =
10.4 Mechanical Efficiency (M.E.)
The mechanical efficiency of any machine is given by :
Work output
…………………………………….. 3
M.E. =
Work input
From previous definitions
Work = force x distance
Hence Work output = Load x distance moved
And
Work input = Effort x distance moved
Load
Dis tan ce moved by the load
Therefore M.E. =
x
Effort
Dis tan ce moved by the effort
1
= M.A. X
V ..R.
M . A.
=
V .R.
Since efficiencies are normally expressed as a percentage then
M . A.
M.E. =
X 100% ……………………………….. 3
V .R.
Note that efficiency of any machine can never exceed 100%. Examples of simple machines
include:
Question
(a) Define the following terms as applied to a machine:
(i) Mechanical advantage (M.A)
(ii)
velocity ratio (V.R)
(iii) efficiency
(b) The law of a small lifting machine is given by:
E = 0.24W + 24
Where E is the effort and W the load, in N. The velocity ratio for the machine is
160. Calculate the power required to raise a load of 400N at a speed of 5m/min.
Example
A wheel 300 mm in diameter is rigidly attached to an axle 100 mm diameter. If an effort of
140 N is needed to raise a load of 360 N calculate the M.A. , V.R. , and Efficiency.
Answer
(c)
Mechanical advantage (M.A.) =
Load ( L)
360
=
= 2.73
Effort( E )
140
83
Velocity ratio (V.R.) =
M.E. =
M . A.
x 100%
V .R.
Dis tan ce moved by effort
Dis tan ce moved by load
=
R 150
=
= 3
r 50
2.73
x 100% = 91%
3
10.5 Screw
A screw is a simple type of machine, acting like an inclined plane rolled up in a helix. The pitch
of the screw is the distance between the threads. So for one revolution of the screw, it moves
laterally through a screwed nut a distance equal to the pitch. The screw is usually turned by a
force applied tangentially at one end of an arm. The other end of the arm is fixed to the screw.
Work done by the effort = effort X distance travelled by the effort = effort X 2pi X r, where r =
radius of the screw
Work done on the load = load X pitch
So load X pitch = effort X 2pi X r
Mechanical advantage = load / effort = 2pi X r / pitch
A screw may transfer between translation and rotation. The screw thread is a ridge in the form
of a helix on a cylindrical core. The ridge may be triangular (v-shape) or square or round. If the
screw thread is triangular, greater friction may be used. If the screw thread is square, larger loads
may be lifted. The distance between the adjacent threads of a screw or bolt is called pitch. When
you turn a screw through one full turn, it moves up or down a distance equal to the pitch of its
thread. Different standard screw threads include pre metric British Standard Whitworth (BSW)
with an angle of 55o, Sellers or USS screw thread (pre metric USA standard) with an angle of
60o, and various metric screw threads.
An Archimedes screw is a hollow inclined screw so that rotation of the screw raises water. It was
invented by Archimedes but apparently was not invented in China.
10.6 The lever
Fig. 2
Levers can be used to exert a large force over a small distance at one end by exerting only a
small force over a greater distance at the other.
In physics, a lever (from French lever, "to raise",) is a rigid object that is used with an
appropriate fulcrum or pivot point to multiply the mechanical force that can be applied to another
object. This is also termed mechanical advantage, and is one example of the principle of
moments. A lever is one of the six simple machines.
Theory of operation
84
The principle of the lever tells us that the above is in static equilibrium, with all forces balancing,
if F1D1 = F2D2.
The principle of leverage can be derived using Newton's laws of motion, and modern statics. It
is important to note that the amount of work done is given by force times distance. For instance,
to use a lever to lift a certain unit of weight with a force of half a unit, the distance from the
fulcrum of the spot where force is applied must be twice the distance between the weight and the
fulcrum. For example, to cut in half the force required to lift a weight resting 1 meter from the
fulcrum, we would need to apply force 2 meters from the other side of the fulcrum. The amount
of work done is always the same and independent of the dimensions of the lever (in an ideal
lever). The lever only allows to trade force for distance.
Archimedes was the first to explain the principle of the lever, stating:
"(equal) weights at equal distances are in equilibrium, and equal weights at unequal distances are
not in equilibrium but incline towards the weight which is at the greater distance."
Archimedes once famously remarked: ("Give me a place to stand and with a lever I will move
the whole world.")
The point where you apply the force is called the effort. The effect of applying this force is
called the load. The load arm and the effort arm are the names given to the distances from the
fulcrum to the load and effort, respectively. Using these definitions, the Law of the Lever is:
Load arm X load force = effort arm X effort force. When 2 things are balanced, when a 1 gram
feather for instance is balanced by a one kilogram rock on a lever the feather would go up and
the rock would go down, but if a 1 kilogram rock was balanced by a 1 kilogram rock, the lever
would be in the middle.
The three classes of levers
There are three classes of levers which represent variations in the location of the fulcrum
and the input and output forces.
85
10.7 First-class levers
A first-class lever is a lever in which the fulcrum is located between the input effort and the
output load. In operation, a force is applied (by pulling or pushing) to a section of the bar,
which causes the lever to swing about the fulcrum, overcoming the resistance force on the
opposite side. The fulcrum may be at the center point of the lever as in a seesaw or at any
point between the input and output. This supports the effort arm and the load.
Examples:
1. Beam engine although here the aim is just to change the direction in which the
applied force acts, since the fulcrum is normally in the centre of the beam (i.e. D1 =
D2)
2. Bicycle hand brakes
3. Can opener
4. Crowbar (curved end)
5. Curb bit
6. Hammer, when pulling a nail with the hammer's claw
7. Hand trucks are L-shaped but work on the same principle, with the axis as a fulcrum
8. Oars
9. Pliers (double lever)
10. Scissors (double lever)
11. Seesaw (also known as a teeter-totter)
12. Wheel and axle because the wheel's motions follows the fulcrum, load arm, and
effort arm principle
86
10.8 Second-class levers
In a second class lever the input effort is located at one end of the bar and the fulcrum is
located at the other end of the bar, opposite to the input, with the output load at a point
between these two forces. Examples:
1. Dental elevator
2. Door
3. Nutcracker
4. Paddle
5. Springboard (diving board)
6. Wheelbarrow
7. Wrench
8. bottle opener
9. Diving Board
10. Crowbar (flat end)
11. Push-up
10.9 Third-class levers
For the lever in this diagram to work correctly, one must assume that the fulcrum is attached
to the bar.
For this class of levers, the input effort is higher than the output load, which is different
from second-class levers and some first-class levers. However, the distance moved by the
resistance (load) is greater than the distance moved by the effort. Since this motion occurs
in the same length of time, the resistance necessarily moves faster than the effort. Thus, a
third-class lever still has its uses in making certain tasks easier to do. In third class levers,
effort is applied between the output load on one end and the fulcrum on the opposite end.
Examples:
1.
Arm
2.
Baseball bat
3.
Boat paddle
4.
Broom
5.
Chopsticks
6.
Door
7.
Fishing rod
8.
Hockey stick
87
9.
Tongs
10.
Mousetrap
11.
Nail clippers, the main body handle exerts the incoming force
12.
Shovel (the action of picking or lifting up sand or dirt)
13. Tools, such as a hoe
This is one of the earliest and most simple of machines, it operates base on the principle of
moments. When the lever is in equilibrium, i.e. before any work is carried out by the effort
Clockwise moments =Anticlockwise moments
Effort x Distance moved = Load x Distance moved
x W
=
P X x =W X y
y
P
W
Load
But
= Mechanical advantage
=
P
Effort
x
Hence
= Mechanical advantage ……………………………. 8.4
y
When movement occurs the position of the lever will change to that shown in Fig 8.2 then
Dis tan ce moved by effort
AB
Velocity ratio =
=
Dis tan ce moved by load
CD
88
y
A
D
W
C
F
x
B
Fig. 8.2
P
As triangles ABF and CDF are similar triangles then,
AB
x
x
=
then: Velocity ratio =
=
CD
y
y
Mechanical advantage
Hence ignoring friction losses
Efficiency =
89
M . A.
1
= = 100%
V .R.
1
MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 13
11.1 The screw jack
The screw jack is a method of turning rotary effort or torque in to a linear force. The efficiency
of this type of machine is usually low due to the high friction forces existing between the square
threaded thrust screw and the body of the jack. A high friction thrust screw is deliberately used
as it gives the advantage that the jack will not run back or overhaul under load.
During one revolution of the jack head the effort will move through a distance equal to the
circumference of a circle of radius r mm. In the same time the load will rise through a height
equal to the pitch of the screw thread (a) (Fig. 8.3).
W
Load
Mechanical advantage =
=
P
Effort
Fig. 1
Screw Jacks Series BD
Velocity ratio (V.R.) =
M.E. =
Dis tan ce moved by effort
Dis tan ce moved by load
M . A.
Load
a
=
x
V .R.
Effort 2 π r
=
2π r
a
…………………. 8.5
…………………………………….. 8.6
Example
A wheel 300 mm in diameter is rigidly attached to an axle 100 mm diameter. If an effort of
90
140 N is needed to raise a load of 360 N calculate the M.A. , V.R. , and Efficiency.
.
Solution
Mechanical advantage (M.A.) =
Velocity ratio (V.R.) =
M.E. =
M . A.
x 100%
V .R.
Load ( L)
360
=
= 2.73
Effort( E )
140
Dis tan ce moved by effort
Dis tan ce moved by load
=
R 150
=
= 3
r 50
2.73
x 100% = 91%
3
Exercise
(a) Define the following terms as applied to a machine:
(i) Mechanical advantage (M.A)
(ii)
velocity ratio (V.R)
(iii) efficiency
(b) The law of a small lifting machine is given by:
E = 0.24W + 24
Where E is the effort and W the load, in N. The velocity ratio for the machine is
160. Calculate the power required to raise a load of 400N at a speed of 5m/min.
91
11.2
Wheel and Axle
A wheel and axle is a lever that rotates in a circle around a center point or fulcrum.
The larger wheel (or outside) rotates around the smaller wheel (axle). Bicycle
wheels and gears are all examples of a wheel and axle. Wheels can also have a
solid shaft with the center core as the axle such as a screwdriver or drill bit or the
log in a log rolling contest.
Why is a wheel a lever? Read on.
- A wheel is a lever that can
turn 360 degrees and can have
an effort or resistance applied
anywhere on that surface. The
effort or resistance force can
be applied either to the outer
wheel or the inner wheel
(axle).
Be sure to read the explantions
below.
Fig. 2
In the first example the resistance is in the mass of the wheel itself, the axle and
whatever it might be connected to. The effort force is applied to the outer wheel.
Steering wheels and door knob are good examples. Remember EFR?
The second example (on the right) the effort comes from the axle, the fulcrum is
92
the core of the axle and resistance is on the road. (vehicle wheels are this way)
Remember FER?
Now list five of your own examples of wheel and axles. You may use the term
wheel only 3 times - be creative!
1.
2.
3.
4.
5.
Question 2 - Identify the effort, resistance and fulcrum of two of your
examples from above.
1.
2.
Question 3 - What type of lever is a steering wheel? A bicycle wheel?
93
Fig. 3 A well known application of the wheel and axle.
The wheel and axle is a simple machine.
The traditional form as recognised in 19th century textbooks is as shown in the image. This also
shows the most widely recognised application, ie lifting water from a well. The form consists of
a wheel that turns an axle and in turn a rope converts the rotational motion to linear motion for
the purpose of lifting.
By considering the machine as a torque multiplier, ie the output is a torque, items such as gears
and screwdrivers can fall within this category.
The doorknobs is an example of the same form as the water well, the mechanism uses the axle as
a pinion to withdraw the latch.
The simple Chain Fall is another example. Here the user pulls on the wheel using the input chain,
so the input motion is actually linear.
The Screwdrivers is an examples of the rotational form. The diameter of the handle gives a
mechanical advantage (compared to a screwdriver with no handle!)
Gears are examples of the rotational form.
21.2.0 Wheel and axle, screwdriver, windlass, crank handle, steering wheel
94
Fig 4 Wheel and axle
95
Fig. 5A
Fig. 5B
Fig. 5
A set of wheel and axle consists of two wheels having different radii. The large wheel has radius
R and the small wheel an axle, has radius r, such that R > r. Wind one rope around the wheel and
another rope in the opposite direction around the axle. Pulling on the wheel rope supplies the
effort. The wheel around the axle bears the load. When you pull on the wheel so that it make one
complete turn, a point on the circumference of the wheel has moves through 2 pi R and a point
on the circumference of the axle has moved through 2 pi r. So velocity ratio = 2 pi R/ 2 pi r =
R/r. Taking moments about the centre of the axle effort X R = load X r so R / r = load / effort =
mechanical advantage.
96
11.3 Windlass
Remove the cover from a pencil sharpener and tie a string tightly around the end of the shaft.
When you turn the handle you find the force needed to turn the handle is much less than the
force of gravity on the books. Feel the magnitude of the force lifting the heavy weight. Lift the
heavy weight directly. Compare the magnitudes of the forces at two conditions.
11.4 Simple belt drives, transmission belts
1. Drive two long nails into a block of wood. Place spools, one larger than the other, over the
nails so that these can be used as axles. Slip a rubber band over both spools. Rotate the larger
spool through one turn and note whether the smaller spool makes more or less than one full turn.
In which direction does the small spool turn? Try crossing the rubber band and observe the
result.
2. Use several spools with different diameters; a wooden block; two long nails; a piece of elastic.
Nail the two nails on the block. Cover the two spoons on the two nails to make the nails as axles.
Cover the elastic on the two spools. Tighten the elastic at fit degree, not too loose and not too
tight. Rotate the spool with a larger axle a circle and meanwhile observe the small spool's
rotating amount and direction. Again cover the elastic across on the two spools. Repeat the
experiment and observe the small spool's rotating amount and direction again. Compare the
97
above two conditions and find the difference. Redo the experiment but differently using two
spools with the same diameters and using two spools with very different diameters. Compare and
analyse the experiment data to find the relationship of the spool's diameter and the way of
covering the spool with elastic to the rotating amount and direction.
11.5 Inclined plane, ramp, screw, thread, wedge
Use a smooth board at an angle of 300 to the table. Weigh the trolley by suspending it from a
spring balance. This is the effort needed to lift the trolley from the table to the top of the board.
Put the trolley on the smooth board. Pull it slowly up the board noting the reading on the spring
balance. The effort will be about its weight. The smooth plank is twice as long as it is high at the
top. By taking a longer path, the slope will be less and the effort less. Inclined plane is a slope
that allows a load to be raised gradually using a smaller effort than would be needed if it lifted
vertically upwards. So it is a force multiplier. The ratio of the height of the top point of the
inclined plane to the length of the plane is called the gradient. The smaller the gradient the more
force is saved.
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11.6 Calculating mechanical advantage
11.6.1 Ideal mechanical advantage
The ideal mechanical advantage of a wheel and axle is calculated with the following formula:
M.A.= Radius of wheel/Radius of axle
The effort distance is the radius, diameter, or circumference of which ever part of the simple
machine, wheel or axle, is initially being rotated. The resistance distance is the same
measurement of the opposite part of the wheel and axle. For example, if the axle is initially
rotated and the wheel is rotated by the axle then the axle is the effort distance and the wheel
would be the resistance distance.
11.6.2 Actual mechanical advantage
The actual mechanical advantage of a wheel and axle is calculated with the following formula:
R
E
= AMA
ACTUAL
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MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 14
12.1 Simple harmonic motion
Idea: Any object that is initially displaced slightly from a stable equilibrium point will oscillate
about its equilibrium position. It will, in general, experience a restoring force that depends
linearly on the displacement x from equilibrium:
Hooke's Law:
Fs = - kx
(1)
Where the equilibrium position is chosen to have x -coordinate x = 0 and k is a constant that
depends on the system under consideration. The units of k are: N/m.
A body is said to be moving with simple harmonic motion (S.H.M.) when its
acceleration is always directed towards a fixed point in its path and is proportional to its
distance from that point.
Motion of the above kind is of fundamental importance, since it is the basis of all periodic and
oscillatory or vibratory motions, such as the periodic reciprocating motion of the piston of an
engine.
Consider a point P moving with a constant speed v, or with a uniform angular velocity ω rad/s,
around a circle of centre O and radius r (Fig.9.0).
C
V =ωr
P
ω A
θ
O
r
Q
B
x
D
Fig. 9.0
Let Q be the projected position of P on the diameter AB at any given instant. Then, while P
makes one complete revolution, point Q will make one complete oscillation or vibration across
the diameter AB, i.e. from B to A and back again to B. The time taken, called the periodic
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time (T), for one complete revolution of P is equal to the time taken for one complete
oscillation or vibration of Q. Thus:
Periodic time, T =
But since
Then
2π r
v
…………………………….. 9.0
v = ωr
2π r 2 π
T=
seconds ……………………….. 9.1
=
ωr
ω
12.2 Frequency
The frequency (n) of oscillation is defined as the reciprocal of the periodic time. The unit of
frequency is the hertz (Hz), which is one oscillation or vibration per second. Thus:
1
ω
Frequency, n =
hertz …………………………….. 9.2
=
T 2π
The characteristics of a wave are wavelength, amplitude, frequency and period.
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12.3 Amplitude
The distance that Q moves on either side of the centre of oscillation O is called the amplitude
of the motion or oscillation. It is simply the maximum displacement of Q from O, and is
therefore equal to the radius r of the circle.
The total distance 2r is called the travel or stroke.
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Example
A body of mass 12 kg moves with S.H.M. in a straight line over a distance of 400mm on each
side of its central position. If the frequency of the motion is 2.5Hz, Find:
(a) The maximum acceleration of the body;
(b) The maximum force acting on the body;
(c) The maximum velocity of the body;
(d) The acceleration and velocity of the body at a point 150mm from the central position.
Solution
ω
1
= 2.5Hz
=
T 2π
ω = 2π x 2.5 = 5π rad/s
Amplitude, r = 400mm = 0.4m
(a) Maximum acceleration of the body = rω2 = 0.4 x (5π)2 = 98.74m/s2
Frequency, n =
(b) Maximum force acting on the body = mrω2 = 12 x 98.74 = 1185N 1.185kN
(c) Maximum velocity of the body = ωr = 5π x 0.4 = 6.285m/s
(d) When the body is at a point 150mm from the central position,
Displacement, x = 150mm = 0.15m
Hence,
Acceleration of body at this point = ω2x = (5π)2 x 0.15 = 37m/s2
Exercise
The piston of an engine has a stroke of 80mm, the mass of the reciprocating parts is 0.8kg, and
the crankshaft revolves at 2100 rev/min. assuming the piston to travel with S.H.M., determine
the force to overcome the inertia of the reciprocating parts at the ends of the stroke. Determine
also the inertia force and the velocity of the reciprocating parts when the crank has turned
through 30o from top dead centre.
Answer : Force to overcome inertia of reciprocating parts at ends of stroke = 1.549 kN
Inertia force and velocity of reciprocating parts when crank has turned through 30o
from t.d.c. = 1.3264 kN; 4.4 m/s
Exercise
State under what conditions a vibrating body will execute simple harmonic motion and deduce
an expression for the period of oscillation of a body having this motion.
A body vibrates with S.H.M. along a straight line. Its maximum acceleration is 125 π2 mm/s2,
and when its distance from the equilibrium position is 100 mm the velocity of the body is 75 π
mm/s. calculate the amplitude and the period of oscillation of the body.
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MECHANICAL ENGINEERING SCIENCE
DYNAMICS
WEEK 15
12.4 Acceleration at any instant in terms of displacement
The centripetal acceleration of P is rω2 directed towards centre O. Thus, at the instant when Q is
at a distance x from O, the horizontal component acceleration of P towards O will be:
rω2cosθ
This will be the actual acceleration of Q, but from right angle POQ,
rcosθ = OQ = Displacement x
hence :
Acceleration of Q towards centre O = ω2x ……………………………….. 9.3
Thus point Q has simple harmonic motion along AB.
12.5 Velocity at any instant in terms of displacement
Now , the tangential velocity of point P is equal to ωr , therefore the actual velocity of Q at any
given instant, is :
Velocity of Q towards O = ω
( r
2
_x
2
)
………………………………… 9.4
12.6 Variable forces producing s.h.m
Suppose m is the mass of a body moving with simple harmonic motion. Now since at any
given displacement x the acceleration is given by ω2x , then it is clear , from Newton’s second
law , that a force must exist to produced this acceleration.
If F is the force acting at that displacement, then from newton’s second law,
F = m ω2x ………………………………………… 9.5
F is zero when the body passes through the central position of its path, where x = 0, and
maximum when it is at either end of its path, where x = r
The maximum force acting at is then given by:
F = mrω2 ………………………………. 9.6
Note that F will be in newtons if m is in kilogrammes, ω in radians/second, and x or r in metres.
Example
A body moving with S.H.M. has a velocity of 6m/s and an acceleration of 24m/s2 when, having
started from one end, it has traversed one quarter of the distance to the other end of its path. Find
the length of the path and the time taken to move from one end to the other.
Solution
When the body has traversed one quarter of the distance of its path from end B to end A
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Fig 9.0, Displacement, x = r/2
Now since the velocity of the body at this instant is 6m/s, then;
2
r
( r _  )
2
2
6=ω
 3 2
= ω  r 
4 
→
36 =
3 2 2
→
4ω r
2
ωr
2
= 48
Also , since the acceleration of the body at this instant is 24m/s2, then;
24 = ω2 x r/2 ,
ω2 x r = 48 dividing, we get:
2
ωr
ωr
2
2
=
48
48
r = 1m
Hence ,
Length of path AB = 2r = 2 x 1 = 2m
Now substituting the value of r = 1, we get;
ω2 X 1 = 48
→ ω = 6.928rad/s
2π
2π
Periodic time, T =
=
= 0.908s
ω
6.928
This is the time taken by the body in moving from B to A and back again to B. therefore,
T
0.908
Time taken to move from one end of the path to the other =
=
= 0.454s
2
2
Length of path = 2m
Time to move 2m = 0.454s
Exercise
The piston of an engine has a stroke of 80mm, the mass of the reciprocating parts is 0.8kg, and
the crankshaft revolves at 2100 rev/min. assuming the piston to travel with S.H.M., determine
the force to overcome the inertia of the reciprocating parts at the ends of the stroke. Determine
also the inertia force and the velocity of the reciprocating parts when the crank has turned
through 30o from top dead centre.
Answer : Force to overcome inertia of reciprocating parts at ends of stroke = 1.549 kN
Inertia force and velocity of reciprocating parts when crank has turned through 30o
from t.d.c. = 1.3264 kN; 4.4 m/s
Exercise
Explain what is meant by ‘simple harmonic motion’,
A piston moving with simple harmonic motion has a stroke of 800 mm and a frequency of 0.5
Hz. Calculate:
(a) The maximum velocity;
(b) The maximum acceleration;
(c) The velocity and acceleration when the displacement is 150 mm from the midposition.
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Exercise
A body has a mass of 10 kg and moves with S.H.M. at a displacement of 300 mm from the
centre of oscillation the velocity and acceleration of the body are 5 m/s and 33 m/s2 respectively.
Determine:
(a) The frequency of the oscillation;
(b) The amplitude of motion;
(c) The maximum velocity;
(d) The maximum force acting on the body.
Exercise
A body of mass 10 kg moves S.H.M. in a straight line over a distance of 1m on each
side of its central position. If the frequency of the motion is 2 Hz, find:
(a) The maximum acceleration of the body;
(b) The maximum force acting on the body;
( c) The velocity of the body at a point 500 mm from the central position.
106