CHAPTER 24 SIMPLE MACHINES
EXERCISE 106, Page 240
1. A simple machine raises a load of 825 N through a distance of 0.3 m. The effort is 250 N and
moves through a distance of 3.3 m. Determine: (a) the force ratio, (b) the movement ratio, (c) the
efficiency of the machine at this load.
Force ratio =
load
825 N
= 3.3

effort 250 N
Movement ratio =
Efficiency =
dis tan ce moved by the effort 3.3m
= 11

dis tan ce moved by the load 0.3m
3.3
force ratio
 100 = 30%
 100% =
11
movement ratio
2. The efficiency of a simple machine is 50%. If a load of 1.2 kN is raised by an effort of 300 N,
determine the movement ratio.
Force ratio =
Efficiency =
load
1200 N
=4

effort 300 N
force ratio
force ratio
4
4
from which, movement ratio =
=8


50
movement ratio
efficiency
0.5
100
3. An effort of 10 N applied to a simple machine moves a load of 40 N through a distance of
100 mm, the efficiency at this load being 80%. Calculate: (a) the movement ratio, (b) the
distance moved by the effort.
Force ratio =
load
40 N
=4

effort 10 N
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(a) Efficiency =
force ratio
force ratio
4
4
from which, movement ratio =
=5


movement ratio
efficiency 80 0.8
100
(b) Movement ratio =
dis tan ce moved by the effort
dis tan ce moved by the load
from which, the distance moved by the effort = movement ratio  distance moved by the load
= 5  100 = 500 mm
4. The effort required to raise a load using a simple machine, for various values of load is as
shown:
Load F l (N)
2050
4120
7410
8240
10300
Effort F e (N)
252
340
465
505
580
If the movement ratio for the machine is 30, determine (a) the law of the machine, (b) the
limiting force ratio, (c) the limiting efficiency.
The load/effort graph is shown below.
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(a) The law of the machine is F e = a F l + b
where gradient of curve, a =
AB 570  170
400
4



= 0.04
BC
10000
10000 100
and intercept, b = 170.
Hence, the law of the machine is: F e = 0.04 F l + 170
(b) Limiting force ratio =
(c) Limiting efficiency =
1
1

= 25
a 0.04
1
1

100% = 83.3%
a  movement ratio 0.04  30
5. For the data given in question 4, determine the values of force ratio and efficiency for each value
of the load. Hence plot graphs of effort, force ratio and efficiency to a base of load. From the
graphs, determine the effort required to raise a load of 6 kN and the efficiency at this load.
Load F l (N)
2050
4120
7410
8240
10300
Effort F e (N)
252
340
465
505
580
8.13
12.12
15.94
16.32
17.76
27.1%
40.4%
53.1% 54.4%
Force ratio =
Efficiency =
load
effort
force ratio
movement ratio
59.2%
Graphs of load/effort, load/force ratio and load/efficiency are shown below.
From the graph, when the load is 6 kN, i.e. 6000 N
effort = 410 N and efficiency = 48%
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EXERCISE 107, Page 242
1. A pulley system consists of four pulleys in an upper block and three pulleys in a lower block.
Make a sketch of this arrangement showing how a movement ratio of 7 may be obtained. If the
force ratio is 4.2, what is the efficiency of the pulley.
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Efficiency =
force ratio
4.2

 100% = 60%
movement ratio
7
2. A three-pulley lifting system is used to raise a load of 4.5 kN. Determine the effort required to
raise this load when losses are neglected. If the actual effort required is 1.6 kN, determine the
efficiency of the pulley system at this load.
Load = 4.5 kN and movement ratio = n = 3
When losses are neglected, efficiency = 100% =
from which,
force ratio =
load
= movement ratio
effort
4.5 kN
=3
effort
i.e.
and
force ratio
movement ratio
effort =
4.5kN
= 1.5 kN
3
If the actual effort required is 1.6 kN, efficiency =
force ratio
 100%
movement ratio
load
4.5
effort
100%  1.6 100%
=
movement ratio
3
= 93.75%
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EXERCISE 108, Page 243
1. Sketch a simple screw-jack. The single-start screw of such a jack has a lead of 6 mm and the
effective length of the operating bar from the centre of the screw is 300 mm. Calculate the load
which can be raised by an effort of 150 N if the efficiency at this load is 20%.
A simple screw-jack is shown below, where lead, L = 6 mm and radius, r = 300 mm
Movement ratio =
Efficiency =
from which,
Force ratio =
2r 2(300)

 100
L
6
force ratio
movement ratio
i.e.
20 force ratio

100
100
force ratio =
20(100)
 20
100
load
from which, load = force ratio  effort = 20  150 N
effort
= 3000 = 9425 N = 9.425 kN
2. A load of 1.7 kN is lifted by a screw-jack having a single-start screw of lead 5 mm. The effort is
applied at the end of an arm of effective length 320 mm from the centre of the screw. Calculate
the effort required if the efficiency at this load is 25%.
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Movement ratio =
Efficiency =
from which,
Force ratio =
2r 2(320)

 128
L
5
force ratio
movement ratio
i.e.
25 force ratio

100
128
force ratio =
25(128)
 32
100
load
load
1.7 kN 1700
from which, effort =


effort
force ratio
32
32
= 16.91 N
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EXERCISE 109, Page 245
1. The driver gear of a gear system has 28 teeth and meshes with a follower gear having 168 teeth.
Determine the movement ratio and the speed of the follower when the driver gear rotates at 60
revolutions per second.
Movement ratio =
teeth on follower 168
=6

teeth on driver
28
Also, movement ratio =
speed of driver
speed of follower
from which,
the speed of the follower =
i.e. 6 =
60 rev / s
speed of follower
60
= 10 rev/s
6
2. A compound gear train has a 30-tooth driver gear A, meshing with a 90-tooth follower gear B.
Mounted on the same shaft as B and attached to it is a gear C with 60 teeth, meshing with a gear
D on the output shaft having 120 teeth. Calculate the movement and force ratios if the overall
efficiency of the gears is 72%.
The speed of D = speed of A 
Movement ratio =
TA TC

TB TD
90 120
speed of A TB TD

=
=6


30 60
speed of D TA TC
The efficiency of any simple machine =
from which,
force ratio
 100%
movement ratio
force ratio = efficiency  movement ratio
=
72
 6 = 4.32
100
300
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3. A compound gear train is as shown on page 223. The movement ratio is 6 and the numbers of
teeth on gears A, C and D are 25, 100 and 60, respectively. Determine the number of teeth on
gear B and the force ratio when the efficiency is 60%.
Movement ratio =
i.e.
speed of A TB TD


speed of D TA TC
6=
TB 60

25 100
from which, number of teeth on B, TB =
Efficiency =
from which,
force ratio
movement ratio
i.e.
6  25 100
= 250
60
60 force ratio

100
6
force ratio =
60  6
= 3.6
100
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EXERCISE 110, Page 246
1. In a second-order lever system, the force ratio is 2.5. If the load is at a distance of 0.5 m from the
fulcrum, find the distance that the effort acts from the fulcrum if losses are negligible.
Force ratio =
dis tan ce of effort from fulcrum
dis tan ce of load from fulcrum
i.e. 2.5 =
x
0.5
Hence, the distance that the effort acts from the fulcrum, x = 2.5  0.5 = 1.25 m
2. A lever AB is 2 m long and the fulcrum is at a point 0.5 m from B. Find the effort to be applied
at A to raise a load of 0.75 kN at B when losses are negligible.
Clockwise moment = anticlockwise moment
 Fe 1.5   0.5 0.75
Hence,
effort at A, Fe =
(0.5)(0.75)
= 0.25 kN or 250 N
1.5
3. The load on a third-order lever system is at a distance of 750 mm from the fulcrum and the effort
required to just move the load is 1 kN when applied at a distance of 250 mm from the fulcrum.
Determine the value of the load and the force ratio if losses are negligible.
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Clockwise moment = anticlockwise moment
 Fl  750   1 250 
Fl =
i.e.
Force ratio =
250 1
 kN = 333.3 N
750 3
1
dis tan ce of effort from fulcrum 250
=

3
dis tan ce of load from fulcrum 750
EXERCISE 111, Page 246
Answers found from within the text of the chapter, pages 238 to 246.
EXERCISE 112, Page 247
1. (b)
2. (f) 3. (c) 4. (d) 5. (b) 6. (a)
7. (b)
8. (d) 9. (c) 10. (d)
11. (d) 12. (b)
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