DJJ5113
MECHANICS OF MACHINES
by Wan Siti Fatimah Bt Wan Ab Rahman
DJJ5113 Mechanics of Machines
Wan Siti Fatimah Bt Wan Ab Rahman
 Copyright 2016 JWSF
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CHAPTER 1 : HOIST
Hoist
A hoist is the device used for lifting and lowering loads. This is done using a barrel or drum around
which a rope or chain can be wrapped. It may be manually operated, electrically or pneumatically
driven and may use chain, fiber or wire rope as its lifting medium. Some example of lifting
machines:
Construction lift
Portable cylinder lifter
Crane
Lift
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Component of Lifting Machines
Angular Motion
Torque in Angular motion:
i. Inertia Couple (Iα)
 Opposite direction of angular acceleration
ii. Torque Drive (T)
 Torque to drive the pulley
 Power transmitted from motor
 Same direction with motion
iii. Torque Brake (Tb)

Opposite direction with motion

Act when brake is apply
iv. Frictional force (Tf)
 Friction between cable and pulley
 Opposite direction with motion
v. Torque couple (Pr)
 Torque that produced by the action of force
from cable that rotate the pulley

Direction of Pr is downward
 Torque couple = rope tension (P) x radius (r)
Hoist/ Drum
Rope/ Cable
Linear Motion
Force in linear motion:
i.
Inertia force (ma)
 Opposite direction with linear acceleration
ii.
Gravitational force (mg)

Weight of the mass

Direction of mg is downward
iii.
Tension of the rope (P)

Direction of P is upward
Load
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Conditions of Lifting Machines
1. Ascending & descending load with acceleration
2. Ascending load with uniform velocity
3. Ascending & descending load and stop with brake
4. Descending load with acceleration without driven torque (Freely fall)
1. Ascending & descending load with acceleration
Ascending load with acceleration:
Angular Motion
Fccw = Fcw
T = Tf + Iα + Pr
Linear Motion
F = F
P = mg + ma
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Descending load with acceleration:
I
T
r
Angular Motion
Fcw = Fccw
T + Pr = Tf + Iα
T = Tf + Iα - Pr
Tf
Pr
P
Linear Motion
ma
F = F
P + ma = mg
P = mg - ma
mg
2. Ascending load with uniform velocity
Uniform velocity is a motion with zero acceleration. So, a = 0 and α =0.
0
0
Angular Motion
Fccw = Fcw
T = Tf + Pr
0
Linear Motion
F = F
P = mg
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3. Ascending & descending load and stop with brake
Ascending load and stop with brake:
Angular Motion
Fcw = Fccw
Tb + Tf + Pr = Iα
Tb = Iα - Tf - Pr
Linear Motion
F = F
P + ma = mg
P = mg - ma
Descending load and stop with brake:
Angular Motion
Fccw = Fcw
Tb + Tf = Pr + Iα
Tb = Pr + Iα - Tf
Linear Motion
F = F
P = ma + mg
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4. Descending load with acceleration without driven torque (Freely fall)
Objects that are said to be undergoing free fall are falling under the sole influence of gravity. No
external power is considered in freely fall motion. So, the driven torque, T is zero (T=0).
0
Angular Motion
Fcw = Fccw
Pr = Tf + Iα
P
Linear Motion
ma
F = F
P + ma = mg
P = mg - ma
mg
Example
Simple hoisting machine is used to lift up 5 ton of mass with 1.2 m/s2 acceleration. The pulley
mass is 1.5 ton, diameter 1.8 m and radius of gyration is 630 mm. Find the torque to lift up the
mass if the friction couple between cable and pulley is 1.9 kNm. What is the power to move the
mass from rest until 5 second.
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Solution
For linear motion:
F = F
P = ma + mg
= (5000)(1.2) + (5000)(9.81)
= 55050 N
Where;
I = mk2
= (1500)(0632)
= 595.35 kgm2
For angular motion:
a
r
1.2
α
0.9
Fccw = Fcw
T = Tf + Iα + Pr
= 1900 + (595.35)(1.333) + (55050)(0.9)
= 52238.60 Nm
α
α  1.333 rad/s 2
Given; a  1.2 m/s 2 , u  0 m/s, t  5 s
 v  u  at
v  0  (1.2)(5)
v  6 m/s
ω
v
6

 6.667 rad/s
r 0.9
Power  Tω
 52239  6.667
 348260 watt
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Lifting machine with balancing mass
A counterweight (balancing mass) is an equivalent counterbalancing weight that
balances a load. Its purpose is to make lifting the load more efficient, which saves energy and is
less taxing on the lifting machine. Counterweights are often used in traction lifts (elevators),
cranes and funfair rides.
In traction (non-hydraulic) elevators, a heavy counterweight counterbalances the load of
the elevator carriage, so the motor lifts much less of the carriage's weight. The counterweight also
increases the ascending acceleration force and decreases the descending acceleration force to
reduce the amount of power needed by the motor. The elevator carriage and the counterweights
both have wheel roller guides attached to them to prevent irregular movement and provide a
smoother ride for the passengers.
For linear motion:
Linear Motion
Linear Motion
F = F
P1 = m1a + m1g
F = F
P2 + m2a = m2g
P2 = m2g – m2a
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For angular motion:
Angular Motion
Fccw = Fcw
T + P2r = Tf + Iα + P1r
T = Tf + Iα + P1r - P2r
REMEMBER!!!!!!!
1. If the hoisting system
move with uniform
velocity, a = 0 and α
=0
2. If the hoisting system
is freely fall, T = 0
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Exercises
1. A lifting machine has a moment of inertia 85 kgm2 is using to raising 1 tonne load with 1.5
m/s2. The diameter of drum is 1 metre. Determine:
a. Free body diagram
b. Required of torque
c. Power after accelerates for 3 seconds from rest.
2. A steel drum of lifting machine has a mass of 25 kg, 1.5 m diameter and 0.53 m radius of
gyration. A mass of 280 kg is suspended to the one end of the rope and it would like to
downward with 3 m/s2, calculate the torque at drum.
3. 5 tonne of mass has to be raising by using lifting machine with acceleration 1.2 m/s2. The
drum of the machine has a mass 1.5 tonne, the diameter is 1.8 metre and radius of gyration is
630 milimeter. Calculate torque is require raising the mass. How much power was produced
after the mass moved for 5 seconds from rest? Given, friction at drum is 1.9 kNm.
4. A 800 kg of mass is tied up to the drum of lifting machine. The drum has a mass 400 kg,
diameter is 800 mm and the radius of gyration is 700 mm. The mass is freely falls. If the
friction at the drum is 1.8 kNm, determine:
a. Free body diagram
b. Acceleration of the mass.
5. A drum of lifting machine has a moment of inertia 70 kgm2. It was using to downward 1.2
tonne of load with 1.5 m/s2. The diameter of drum is 1 meter. Determine:
a. Free body diagram
b. Tension of rope
c. Torque require to lowering a load
d. Power after the load accelerates for 3 seconds from rest
6. A 3500 kg of box is being lifting up with 2 m/s2 by using hoisting machine. The drum of the
hoisting has a diameter of 1.8 m, a mass is 900 kg and radius of gyration is 550 mm. Friction
torque is 2.7 kNm. Calculate the power to lifting up a box with 4 m/s.
7. 42 kg of load raising by a rope on the drum with a radius of 760 mm. The mass of the drum is
50 kg and its radius of gyration is 360 mm. If the load moves with 2 m/s2 upward, find the
torque of the drum to drive the load.
8.
A drum of lifting machine has a mass of 26 kg, 2.5 m diameter and the radius of gyration of
0.21 m. A mass of 80 kg is tied to the one end of the rope and the other end is tied with a
mass of 30 kg. Determine:
a. The torque to lift up the mass of 80 kg with acceleration 2.5 m/s2.
b. The linear velocity of mass when the power is 1.8 kW.
c. The acceleration when the drum is freely released.
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9.
500 kg mass of the cage is lift up by the acceleration of 2 m/s2 using hoisting drum with a
diameter of 1.8 m. The drum has a mass 0f 900 kg and a radius of gyration of 550 mm. The
friction torque is 2.7 kNm. Calculate:
a. The power required to reach a speed cage of 4 m/s
b. The power required if the cage moves with uniform velocity of 4 m/s.
10. Tied at two ends of the rope is two mass of 15 kg and 5 kg, respectively. Rope wound on a
drum with a moment of inertia of 0.86 kgm2 and 920 mm in diameter. Ignore any friction,
calculate:
a. Acceleration of the system and the tensions of the rope when the drum is allowed
to fall freely.
b. Torque of the driver if the mass of 15 kg is required to lift up with acceleration of
1.2 m/s2.
11. A hoisting drum wound with a rope in which both ends tied up to the load of 900 kg and 300
kg respectively. Hoisting drum has a mass of 125 kg, diameter of 2 m and a radius of gyration
of 0. 44 m. Calculate:
a. Tension of the rope to raise the load of 900 kg with an acceleration of 0.7 m/s2
b. The torque on the drum to raise the load of 900 kg with the friction torque of
1.3 kNm
c. The output power of the drum at a velocity of 1.9 m/s
12. A load with a mass of 800 kg is tied up to the hoisting drum with a diameter of 800 mm and a
radius of gyration of 700 mm. The mass of the drum is 400 kg. The body is allowed to fall
freely. If the friction torque of the drum is 1.8 kNm,
a. Find the acceleration of the body.
b. When a balancing load of 850 kg is used to lift up of the body of 800 kg with
acceleration of 1.2 m/s2, calculate the power generated when the angular velocity
of the drum is 4.25 rad/s.
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