This document discusses various types of brakes and dynamometers used in mechanical engineering. It describes shoe brakes, internally expanding shoe brakes, and how braking works when applied to rear wheels only, front wheels only, or all wheels of a vehicle. It also covers different types of dynamometers used to measure power including pony brake, rope brake, epicyclic train, belt transmission, and torsion dynamometers. Example problems are provided to calculate braking torque and distance required to stop a vehicle under different braking conditions.

1. Mechanics of Machines
Module - 5
Binil Babu
Department of Mechanical Engineering
SNMIMT, Maliankara

2. Contents
1 Basic terminologies 1
2 Shoe brake 1
3 Internally expanding shoe brake 3
4 Braking of a Vehicle 4
4.1 Brakes applied to rear wheels . . . . . . . . . . . . . . . . . . 5
4.2 The brakes are applied to front wheels only . . . . . . . . . . . 6
4.3 The brakes are applied to all the four wheels . . . . . . . . . . 7
5 Dynamometers 10
5.1 Pony brake dynamometer . . . . . . . . . . . . . . . . . . . . 10
5.2 Rope brake dynamometer . . . . . . . . . . . . . . . . . . . . 11
5.3 Epicyclic train dynamometer . . . . . . . . . . . . . . . . . . . 13
5.4 Belt transmission dynamometer . . . . . . . . . . . . . . . . . 15
5.5 Torsion dynamometer . . . . . . . . . . . . . . . . . . . . . . . 15
i

3. BRAKES AND DYNAMOMETERS
1 Basic terminologies
Dynamometer is a brake with additional device to measure frictional re-
sistance. by knowing frictional resistance we can find the torque transmitted
and hence the power of engine.
A brake is a device by means of which artificial frictional resistance is ap-
plied to a moving machine member, in order to retard or stop the motion of
a machine
2 Shoe brake
Figure 1: Single block Shoe brake
A single block or shoe brake is shown in figure. It consists of a block or
shoe which is pressed against the rim of a revolving brake wheel drum.The
friction between the block and the wheel causes a tangential braking force to
act on the wheel, which retard the rotation of the wheel.
P = Force applied at the end of the lever,
RN = Normal force pressing the brake block on the wheel,
r = Radius of the wheel,
2θ = Angle of contact surface of the block,
µ = Coefficient of friction, and
Ft = Tangential braking force or the frictional force acting at the contact
surface of the block and the wheel.
tangential braking force on the wheel, Ft = µRN
The braking torque, TB = Ft.r = µ.RN .r
The equation for braking torque varies with the position of line of action
of tangential force and fulcrum ”O”. In this brakes we can see that frictional
force acts along with applied force for braking. In other words frictional
1

4. force helps to apply brake, such type of brake is called self energizing
brakes.When the frictional force is great enough to apply the brake with no
external force, then the brake is said to be self-locking brake.
When angle of contact 2θ is greater than 60o
the equivalent co-efficient
of friction find using equation µ = 4µ sin θ
2θ+sin θ
Problem.1
A single block brake is shown in Fig. The diameter of the drum is 250 mm
and the angle of contact is 90o
. If the operating force of 700 N is applied at
the end of a lever and the coefficient of friction between the drum and the
lining is 0.35, determine the torque that may be transmitted by the block
brake.
Given data: d = 250 mm = 0.25 m ; P = 700 N ; 2θ = 90o
; µ = 0.35
µ = 4µ sin θ
2θ+sin θ
= 4×0.35 sin 90
π/2+sin 90
= 0.385
RN = Normal force pressing the block to the brake drum, and
Ft= Tangential braking force =µ .RN
Take moment about ”O” ;
(700×450)+(Ft × 50) = 200 × RN = µ
RN
= 520 Ft
Ft = 670 N
Braking torque TB = Ft × r = 670 × 0.125 = 83.75 N-m.
Problem.2
A bicycle and rider of mass 100 kg are travelling at the rate of 16 km/h on
2

5. a level road. A brake is applied to the rear wheel which is 0.9 m in diameter
and this is the only resistance acting. How far will the bicycle travel and
how many turns will it make before it comes to rest ? The pressure applied
on the brake is 100 N and µ = 0.05.
Given data: m =100 kg ; v = 16 km/hr = 4.44 m/s ; D = 0.9 m ; RN =
100 N ; µ = 0.5
Let x = Distance travelled (in metres) by the bicycle before it comes to
rest.
Tangential force Ft = µRN = 0.05 × 100 = 5 N
Work done = Force × distance/displacement = 5×x N-m
Kinetic energy of the bicycle = 1/2 mv2
= 1/2 ×100 × 4.442
= 986 N-m
In order to bring the bicycle to rest, the work done against friction must
be equal to kinetic energy of the bicycle.
986 N-m = 5x ; x = 197.2 m
N = Required number of revolutions. We know that distance travelled
by the bicycle (x),
197.2 =πDN ; 197.2 =π × 0.9N; N = 70.
3 Internally expanding shoe brake
The rim type internal expanding shoe is widely used for braking systems
in automotive applications and is generally referred as internal shoe drum
brake. The nonrotating unit may be placed inside the rotating drum with
the drum acting as a cover for the braking surfaces. This type of brake is
known as an internal-expanding brake because the nonrotating braking sur-
face is forced outward against the drum to produce braking action. This type
of brake is used on the wheel brakes of cars and trucks because it permits a
more compact and economical construction.
3

6. Figure 2: Internal expanding shoe brake
Construction and Working
• An internal expanding shoe brake consists of two shoes S1 and S2
• The outer surface of the shoes are lined with some friction material to
increase the coefficient of friction and to prevent wearing away of the
metal
• Each shoe is pivoted at one end about a fixed fulcrum O1 and O2 and
made to contact a cam at the other end.
• When the cam rotates, the shoes are pushed outwards against the rim
of the drum.
• The friction between the shoes and the drum produces the braking
torque and hence reduces the speed of the drum.
• The shoes are normally held in off position by a spring. The drum
encloses the entire mechanism to keep out dust and moisture.
4 Braking of a Vehicle
In a four wheeled vehicle we can applied brakes in three different methods:-
1. In rear wheels only
2. In front wheel only
3. Both rear and front wheel.
In this type of problems our aim is to find the retardation of vehicle due
4

7. to brake which we applied on wheels.Since the brake is applied to moving
vehicle its a dynamic problem.Here we approach the problem statically by
considering inertia force.Because,the inertia force is equal and opposite to
the braking force causing retardation.
4.1 Brakes applied to rear wheels
Figure 3: Motion of vehicle up the inclined plane and brakes are applied to
rear wheels only
Let,
α = Angle of inclination of the plane to the horizontal
h = Height of the C.G. of the vehicle above the road surface in metres.
m = Mass of the vehicle in kg
x = Perpendicular distance of C.G. from the rear axle in metres
L = Distance between the centres of the rear and front wheels (also called
wheel base) of the vehicle in metres.
µ = Coefficient of friction between the tyres and road surface.
a = Retardation of the vehicle in m/s2
RA = Total normal reaction between the ground and the front wheels in
newtons
RB = Total normal reaction between the ground and the rear wheels in new-
tons
FB = Total braking force acting at the rear wheels due to the application of
the brakes. And FB = µRB
To be in equilibrium the forces acting on vehicle should be balanced.
Consider the force parallel to the plane.
5

8. FB + mg sin α = m.a.......................(1)
Consider forces perpendicular to the plane
RA + RB = m.g cos α ...................(2)
Now, take moment about ”G”, the cog of vehicle,
FB × h + RB × x = RA(L − x) .................(3)
By substituting the value of FB = µRB and RA = m.g cos α − RB in
equation 3.
RB =
mg cos α(L − x)
L + µh
and RA =
mg cos α(x + µh)
L + µh
.
The value of ”a” from equation 1 obtained as
a =
µg cos α(L − x)
L + µh
± g. sin α, +ve when vehicle is moving upward and
-ve when vehicle moving downward. When the vehicle moves on level surface
then, a =
µg(L − x)
L + µh
4.2 The brakes are applied to front wheels only
Figure 4: Motion of the vehicle up the inclined plane and brakes are applied
to front wheels only.
FA = Total braking force acting at the front wheels due to the applica-
tion of brakes. Its maximum value is µ.RA.
6

9. By resolving forces as same as above we the the values of RA, RB and a
as follows:
RA =
mg cos α × x
L − µh
and RB = mg cos α −
mg cos α × x
L − µh
a =
µ.g cos α × x
L − µh
± g. sin α.+ve when vehicle moving up the plane and
-ve when the vehicle moving down the plane. When α = 0, a =
µgx
L − µh
4.3 The brakes are applied to all the four wheels
Figure 5: Motion of the vehicle up the inclined plane and the brakes are
applied to all the four wheels
FA = Brakingforceprovidedbythefrontwheels = µ.RA, FB = µRB
By resolving forces parallel and perpendicular to the plane as above we
get the value of RA and RB :-
RA =
m.g cos α(µ.h + x)
L
and RB = m.g cos α[
L − µh − x
L
]
The retardation ”a” is given by equation, a = g(µ cos α ± sin α), +ve
when vehicle is moving up the plane and -ve when vehicle moving down the
plane. When α = 0 ie; when vehicle moving on plane surface the retardation
a = µg
7

10. Problem.3
A car moving on a level road at a speed 50 km/h has a wheel base 2.8
metres, distance of C.G. from ground level 600 mm, and the distance of C.G.
from rear wheels 1.2 metres. Find the distance travelled by the car before
coming to rest when brakes are applied, 1. to the rear wheels, 2. to the front
wheels, and 3. to all the four wheels. The coefficient of friction between the
tyres and the road may be taken as 0.6.
Given data:
u = 50 km/hr = 13.89 m/s ; L = 2.8 m (Wheel base) ; h = 600 mm
= 0.6 m (distance of C.G from ground) ; x = 1.2 m (distance of C.G from
rear wheels) ; µ = 0.6 (coeffiecent of friction) ; s = distance travelled by car
before coming rest.
Since the vehicle moving on level road, the retardation of car,
a =
µg(L − x)
L + µ.h
; a =
0.6 × 9.81(2.8 − 1.2)
2.8 + 0.6 × 0.6
= 2.98 m/s2
for uniform retardation the distance traveled by the car is given by equa-
tion, s =
u2
2.a
=
(13.89)2
2 × 2.98
= 32.4 m.
When brakes applied to front wheel, the retardation;
a =
µ.g.x
L − µ.h
=
0.6 × 9.81 × 1.2
2.8 − 0.6 × 0.6
= 2.9 m/s2
Uniform retardation, s =
u2
2.a
=
(13.89)2
2 × 2.9
= 33.26 m
When brakes applied to all four wheels
the retardation a = g.µ = 9.81 × 0.6 = 5.886 m/s2
Uniformretardation s =
u2
2.a
=
(13.89)2
2 × 5.886
= 16.4 m
Problem.4
A vehicle moving on a rough plane inclined at 10o
with the horizontal at
a speed of 36 km/h has a wheel base 1.8 metres. The centre of gravity of the
vehicle is 0.8 metre from the rear wheels and 0.9 metre above the inclined
8

11. plane. Find the distance travelled by the vehicle before coming to rest and
the time taken to do so when 1. The vehicle moves up the plane, and 2. The
vehicle moves down the plane. The brakes are applied to all the four wheels
and the coefficient of friction is 0.5.
Given data: α = 10o
; u = 36 km/h = 10 m/s ; L = 1.8 m ; x = 0.8 m ;
h = 0.9 m ; µ = 0.5
s = distance travelled by vehicle before coming to rest.
t = time taken by the vehicle coming to rest.
1. When the vehicle moves up the plane and brakes are applied
to all the four wheels.
The retardation of vehicle given by equation, a = g(µ cos α + sin α)
= 9.81(0.5 × cos 10o
+ sin 10o
) = 6.53 m/s2
uniform retardation s =
u2
2.a
=
102
2 × 6.53
= 7.657 m
final velocity of vehicle (ν),
0 = u + a.t ; 0 = 10 - 6.653 * t ;(negative sign used due to retardation)
t =
10
6.53
= 1.53 s
2. When the vehicle moves down the plane and brakes are ap-
plied to all the four wheels
retardation a = g(µ cos α − sin α) = 9.81 × (0.5 cos 10o
− sin 10o
)
= 3.13 m/s2
distance travelled by vehicle after braking; s =
u2
2.a
=
102
2 × 3.13
= 16 m
final velocity of vehicle (ν),
9

12. 0 = u + a.t ; 0 = 10 - 3.13 * t ;(negative sign used due to retardation)
t =
10
3.13
= 3.2 s.
5 Dynamometers
Dynamometer is a brake with additional device to measure frictional resis-
tance. by knowing frictional resistance we can find the torque transmitted
and hence the power of engine.
Dynamometer classified into two types, (1) Absorption Dynamometer (2)
Transmission dynamometer.
In Absorption dynamometer the entire energy / power produced by en-
gine is absorbed by friction resistance of brake and is transformed to heat dur-
ing process of measurement.Eg.Pony brake dynamometer, rope brake
dynamometer.
In Transmission dynamometer, the energy produced by the engine is not
wasted in friction but used to doing work. ie; the power produced by the en-
gine is transmitted through dynamometer to some other machines where the
power developed can be measured easily.Eg. Epicyclic train dynamome-
ter, belt transmission dynamometer.
5.1 Pony brake dynamometer
Figure 6: Pony brake dynamometer
• Simplest form of absorption dynamometer
• It consists of two wooden blocks placed around a pulley fixed to a shaft
connected to engine,whose power is to be measured.
10

13. • Wooden blocks are clamped by means of nut and bolt.
• Helical spring provided between upper block and nut helps to adjust
the pressure on pulley to control speed.
• A lever is attached to the upper block which carries a weight W at the
outer end. A counter weight is provided at other end to balance when
the weight is removed.
• When the brake has to apply lever is loaded with weight W and nuts
are tightened until engine shaft rotates at constant speed.
• At this condition the moment due to weight must be balanced by fric-
tional resistance between shaft and pulley.
W = weight at outer end of lever in N
L = distance from centre of pulley to weight w
F = frictional resistance between pulley and wooden blocks
R = radius of pulley in metre.
N = speed of shaft in rpm.
Torque on the shaft, T = W.L = F.R
Work done in one revolution = torque × angle turned in radians = T × 2π
N-m
work done per minute = T ×2π×N N-m
Brake power of engine given by work done per second, ie; 2πNT
60
=W×L×2π×N
60
Watts. from this equation we can find power with out knowing frictional
resistance and radius of pulley.
5.2 Rope brake dynamometer
• It consists of two or more ropes wound around a flywheel or pulley
rigidly fixed to shaft of an engine
• Upper ends of rope connected to a spring balance and dead weight
applied at the lower end.
• To prevent slipping of ropes wooden blocks are provided around the
circumference of flywheel.
• The frictional torque must be equal to the torque transmitted by the
engine.
11

14. Figure 7: Rope brake dynamometer
W = Dead load in newtons
S = Spring balance reading in newtons,
D = Diameter of the wheel in metres,
d = diameter of rope in metres, and
N = Speed of the engine shaft in r.p.m.
Net load on the brake = (W – S) N
Distance moved in one revolution = π(D + d) m
Work done per Revolution = (W-S) π (D+d) N-m
Work done per minute = (W-S) π (D+d) N N-m
Brake power of the engine = (W−S)π(D+d)N
60
W
Problem.1
In a laboratory experiment, the following data were recorded with rope brake:
Diameter of the flywheel 1.2 m; diameter of the rope 12.5 mm; speed of the
engine 200 r.p.m.; dead load on the brake 600 N; spring balance reading 150
N. Calculate the brake power of the engine.
Given data: D = 1.2 m ; d = 12.5 mm = 0.0125 m ; N = 200 r.p.m ;
W = 600 N ; S = 150 N
12

15. BP of engine = workdoneperminute
60
= (W−S)π(D+d)N
60
= (600−150)π(1.2+0.0125)600
60
= 5715 W = 5.715 kW
5.3 Epicyclic train dynamometer
Figure 8: Epicyclic train dynamometer
• An epicyclic-train dynamometer consists of a simple epicyclic train of
gears, i.e. a spur gear, an annular gear (a gear having internal teeth)
and a pinion
• The spur gear is keyed to the engine shaft (i.e. driving shaft) and
rotates in anticlockwise direction. The annular gear is also keyed to
the driving shaft and rotates in clockwise direction.
• The pinion or the intermediate gear meshes with both the spur and
annular gears. The pinion revolves freely on a lever which is pivoted to
the common axis of the driving and driven shafts.
• A weight w is placed at the smaller end of the lever in order to keep it
in position
• The tangential effort P exerted by the spur gear on the pinion and the
tangential reaction of the annular gear on the pinion are equal.
• These efforts act in the upward direction as shown, therefore total
upward force on the lever acting through the axis of the pinion is 2P
• This force tends to rotate the lever about its fulcrum and it is balanced
by a dead weight W at the end of the lever.
13

16. Take moment about fulcrum F,
2P × a = W.L or W.L/2a
Let, R = Pitch circle radius of the spur gear in metres,
N = Speed of the engine shaft in r.p.m.
Torque transmitted, T = P.R
Power transmitted = 2πNT
60
= P.R×2πN
60
W
14

17. 5.4 Belt transmission dynamometer
Figure 9: Belt transmission dynamometert
• In belt transmission dynamometer belt is transmitting power from one
pulley to another, the tangential effort on the driven pulley is equal to
the difference between the tensions in the tight and slack sides of the
belt.
• The dynamometer measure directly the difference between the tensions
of the belt,while it is running
Take moment about E
2T1 × a = 2T2 × a + W.L
T1 − T2 = W.L
a
D = diameter of the pulley A in metres
N = Speed of the engine shaft in r.p.m.
Work done in one revolution = (T1 − T2) × πD N-m
Work done per minute = (T1 − T2) × πDN N-m
Brake power, BP = (T1−T2)×πDN
60
W
5.5 Torsion dynamometer
Problem.2
A torsion dynamometer is fitted to a propeller shaft of a marine engine. It
is found that the shaft twists 2o
in a length of 20 metres at 120 r.p.m. If
the shaft is hollow with 400 mm external diameter and 300 mm internal di-
ameter, find the power of the engine. Take modulus of rigidity for the shaft
15

18. material as 80 GPa.
Given data: θ = 2o
= 0.035 radians ; l = 20 m ; N = 120 rpm ; D = 400 mm
= 0.4 m ; d = 300 mm = 0.3 m ; C = 80 Gpa = 80 × 109
N/m2
.
We know torsion equation T
θ
= CJ
l
J is the polar mass moment of inertia, J = π
32
(D4
− d4
)
= π
32
(0.44
− 0.34
) = 0.0017m4
From torsion equation T = CJ
l
× θ
= 80×109×0.0017
20
× 0.035 = 238 × 103
N-m
The power of engine, P = 2πNT
60
= 2π×120×238×103
60
= 2990 × 103
W
= 2990 kW
16