Chapter 12 – Simple Machines
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
SIMPLE MACHINES are used to perform a variety
of tasks with considerable efficiency. In this
example, a system of gears, pulleys, and levers
function to produce accurate time measurements.
Photo Vol. 1 PhotoDisk/Getty
Objectives: After completing this
module, you should be able to:
• Describe a simple machine in general terms and
apply the concepts of efficiency, energy
conservation, work, and power.
• Distinguish by definition and example between the
concepts of the ideal and actual mechanical
advantages.
• Describe and apply formulas for the mechanical
advantage and efficiency of the following devices:
(a) levers, (b) inclined planes, (c) wedges, (d)
gears, (e) pulley systems, (f) wheel and axel, (g)
screw jacks, and (h) the belt drive.
A Simple Machine
In a simple machine,
input work is done by
the application of a
single force, and the
machine does output
work by means of a
single force.
A simple
machine
Fin
sin
sout
Fout
W
Win= Finsin
Wout= Foutsout
Conservation of energy demands that the work
input be equal to the sum of the work output
and the heat lost to friction.
A Simple Machine (Cont.)
Input work = output work + work against friction
Efficiency e is defined
as the ratio of work
output to work input.
e
A simple
machine
sin
W ork output
Win= Finsin
W ork input
e
Fininin
Fout s out
Fin s in
FFout
out
W
W
sout
Wout= Foutsout
Example 1. The efficiency
of a simple machine is
80% and a 400-N weight
is lifted a vertical height of
2 m. If an input force of
20 N is required, what
distance must be covered
by the input force?
A simple
machine
Fin = ?
sin
sout
W
W
Efficiency
e
Fout s out
Fin s in
TheThe
efficiency
is 80%
e = 0.80,
therefore
advantage
is a or
reduced
input
force,
but it is Fat the
expense ofFdistance.
The
s
s
out
out out
 outmust
or s inagreater
input eforce
move
distance.
Fin s in
s in 
(400 N )(2 m )
(0.80)(20 N )
eFin
sin = 5.0 m
Power and Efficiency
Since power is work per
unit time, we may write
P 
W ork
A simple
machine
e
W in

sin
or W ork  P t
t
W out
Fin = ?
P0 t
Pi t
e
sout
P0
W
W
Pi
Efficiency is the ratio
of the power output
to the power input.
e
Efficiency
e
P ow er out
P ow er in
Pout
Pin

P0
Pi
Example 2. A12-hp winch motor lifts a 900-lb load
to a height of 8 ft. What is
the output power in ftlb/s if A simple Fin = ?
the winch is 95% efficient? machine
s
in
First we must
find the power
output, Po:
e
P0
sout
Pi
P0  ePi
W
W
Efficiency
e
Pout
Pin
Po = (0.95)(12 hp) = 11.4 hp
(1 hp = 550 ft/s):
 550 ft  lb/s 
Po  (11.4 hp) 
  6600 ft  lb/s
1 hp


Po = 6270 ftlb/s
Ex. 2 (cont.) A12-hp winch motor lifts a 900 lb load
to a height of 8 ft. How
much time is required if
the winch is 95% efficient?
A simple
machine
We just found that
Po = 6270 W
Po 
W ork out
t

Fin = ?
sin
sout
e
W
W
Fo s o
Efficiency
Pout
Pin
t
Now we solve for t :
t
Fo s o
Po

(900 lb)(8 ft)
6270
Time required: t = 1.15 s
Actual Mechanical
Advantage
The actual mechanical
advantage, MA, is the
ratio of Fo to Fi.
M
A

80 N
output force
input force

Fo
Fin = ?
A simple
machine
Fout
sin
sout
W
W
Actual
Mechanical
Advantage
MA
Fi
For example, if an
40 N input force of 40 N
lifts an 80 N weight,
the actual mechanical
advantage is:
M
M
80 N
A

A
 2 .0
40 N
An Ideal Machine
Conservation of energy demands that:
Input work = output work + work against friction
Fi s i  Fo s o  (W ork ) f
An ideal or perfect machine is 100%
efficient and (Work)f = 0, so that
Fi s i  Fo s o
or
Fo
si

si
so
The ratio si/so is the ideal mechanical advantage.
Ideal Mechanical
Advantage
The ideal mechanical
advantage, MI, is the
ratio of sin to sout.
M
A

in distance
out distance

Fin = ?
A simple
machine
2m
Fout
si
sout
W
W
6m
sin
Ideal
Mechanical
Advantage
MI
so
For example, if an input force
moves a distance of 6 m while
the output force moves 2 m, the
ideal mechanical advantage is:
MI 
6 m
2m
M I  3.0
Efficiency for an Ideal Engine
For 100% efficiency MA = MI. In other words,
in the absence of friction, the machine IS an
ideal machine and e = 1.
IDEAL EXAMPLE:
A
simple
machin
e
Fout=
400 N
W
W
M
A

Fin = 80 N
Sin = 8 m
Sout= 2 m
e = 100%
MI 
e
Fo

80 N
4
Fi
20 N
si
8m

so
M
2m
A
Mi
 1.0
4
Efficiency for an Actual Engine
The actual efficiency is always less than the
ideal efficiency because friction always exits.
The efficiency is still equal to the ratio MA/MI.
The efficiency of any
engine is given by:
e
M
A
Mi
In our previous example, the ideal mechanical
advantage was equal to 4. If the engine was only
50% efficient, the actual mechanical advantage
would be 0.5(4) or 2. Then 160 N (instead of 80
N) would be needed to lift the 400-N weight.
The Lever
A lever shown here
consists of input and
output forces at different
distances from a fulcrum.
Fout
rout
rin
Fulcrum
The input torque Firi is equal
to the output torque Foro.
The actual mechanical
advantage is, therefore:
Fin
Fi ri  Fo ro
M
A

Fo
Fi

ri
ro
The Lever
Friction is negligible
so that Wout = Win:
Fi s i  F s o or
Fo
Fi

si
Fout
sout
rout
sin
q
rin
q
Fin
so
Note from figure that angles are the same and
arc length s is proportional to r. Thus, the ideal
mechanical advantage is the same as actual.
The ideal MI is:
MI 
Fo
Fi

ri
ro
and M
I
M
A
Example 3. A 1-m metal lever is used to lift a
800-N rock. What force is required at the left end
if the fulcrum is placed 20 cm from the rock?
1. Draw and label sketch:
2. List given info:
800 N
ri
r2
Fo = 700 N; r2 = 20 cm
F=?
r1 = 100 cm - 20 cm = 80 cm
3. To find Fi we recall the definition of MI :
MI 
ri
MI 
and
ro
Thus,
M
A

Fo
Fi
80 cm
4 ;
20 cm
4
and
For lever: MA = MI
Fi 
800 N
4
 200 N
Other Examples of Levers
Wheel and Axel:
Application of Lever Principle:
Fi
R
r
With no friction MI = MA and
For Wheel
and Axel:
M
A

Fo
Fi

ri
ro
Fo
Wheel and Axel
For example, if R = 30 cm and r = 10 cm, an
input force of only 100 N will lift a 300-N weight!
If the smaller radius is 1/3 of the larger radius,
your output force is 3 times the input force.
Single Fixed Pulleys
Single fixed pulleys serve only to change the
direction of the input force. See examples:
Fin = Fout
Fout
W
Fin
Fout
Fin
Single Moveable Pulley
Fin
2m
1m
80 N
Fin
Fin
Fout 80 N
Fin + Fin = Fout
40 N + 40 N = 80 N
A free-body diagram shows an actual mechanical
advantage of MA = 2 for a single moveable pulley.
Note that the rope moves a distance Fofo M  s in  2
2 Fin  Fout or
MA 
2 I
2 m while the weight is lifted only 1 m.
s out
Fi
Block and Tackle
Arrangement
Fi
We draw a free-body diagram:
Fi
Fi Fi F
i
4 Fin  Fout
M
Fo
W
Fo
A

Fo
4
Fi
The lifter must pull 4 m of rope
in order to lift the weight 1 m
The Belt Drive
A belt drive is a device used to transmit torque
from one place to another. The actual mechanical
advantage is the ratio of the torques.
Fo
M
ro
A

output torque
input torque
o
i
Since torque is defined as Fr,
the ideal advantage is:
MI  M
Belt
Drive

ri
Fi
Belt Drive:
A
Fo ro

Fi ri
M
I

ro
ri

Do
Di
Angular Speed Ratio
The mechanical advantage
of a belt drive can also be
expressed in terms of the
diameters D or in terms of
the angular speeds w.
Belt Drive:
MI 
Do
Di

wi
wo
Note that the smaller pulley
diameter always has the
greater rotational speed.
Do
wo
Belt
Drive
wi
Di
Speed ratio:
wi
wo
Example 4. A 200 Nm torque
is applied to an input pulley
12 cm in diameter. (a) What
should be the diameter of the
output pulley to give an ideal
mechanical advantage of 4?
(b) What is the belt tension?
To find Do we use the fact that
MI 
Do
Di
Fo
M
 4; D o  4 D i
Do = 4(12 cm) = 48 cm
Now, i = Firi and ri = Di/2.
Belt tension is Fi and ri is
equal to ½Di = 0.06 m.
MI = 4
ro
I

Do
Di
ri
Fi
 i  Fi ri  200 N  m
Fi 
200 N  m
0.06 m
 3 3 30 N
Gears
Mechanical advantage
of gears is similar to
that for belt drive:
Gears:
MI 
Do
Di

No
Ni
In this case, Do is the
diameter of the driving
gear and Di is diameter
of the driven gear. N is
the number of teeth.
Ni
No
If 200 teeth are in the
input (driving) gear, and
100 teeth in the output
(driven) gear, the mechanical advantage is ½.
Example 5. The driving gear on a bicycle has 40
teeth and the wheel gear has only 20 teeth. What
is the mechanical advantage? If the driving gear
makes 60 rev/min, what is the rotational speed of
the rear wheel?
MI 
No

Ni
22
44
; M I  0.5
Remember that the
angular speed ratio is
opposite to the gear ratio.
MI 
No
Ni

wi
wo
;
wi
wo

1
2
wo = 2wi  2(60 rpm)
No = 20
Ni = 40
Output angular speed:
wo = 120 rpm
The Inclined Plane
The Inclined Plane
si
q
Fo = W
Ideal Mechanical
Advantage
Fi
so
M
I

slope
height

si
so
A ctual A dvantage: M
A

W
Fi
Because of friction, the actual mechanical
advantage MA of an inclined plane is usually much
less than the ideal mechanical advantage MI.
Example 6. An inclined plane has a slope of 8 m
and a height of 2 m. What is the ideal mechanical advantage and what is the necessary input
force needed to push a 400-N weight up the
incline? The efficiency is 60 percent.
Si = 8 m
MI 
Fi
2m
q
Fo = 400 N
M
A
 2.4 
Fo
Fi
e
MA
MI
Fi 
Fo
2.4

si
so

8 m
2m
;
MI 4
; M A  eM I  (0.60)(4)
400 N
2.4
Fi = 167 N
The Screw Jack
Fo
R
Fi
p
MI 
2 R
p
Screw
Jack
An application of the
inclined plane:
Input distance: si = 2R
Output distance: so = p
Screw Jack
MI 
si
so

2 R
p
Due to friction, the screw jack is an inefficient
machine with an actual mechanical advantage
significantly less than the ideal advantage.
Summary for Simple Machines
Efficiency e is defined
e
W ork input
as the ratio of work
output to work input.
Efficiency is the ratio
of the power output
to the power input.
W ork output
e
Fout s out
Fin s in
e
P ow er out
P ow er in

P0
Pi
Summary
The actual mechanical
advantage, MA, is the
ratio of Fo to Fi.
M
A

output force
input force

A simple
machine
Fin = ?
sin
Fo
sout
Fi
The ideal mechanical
advantage, MI, is the
ratio of sin to sout.
M
A
Efficiency
W
W
e

in distance
Pout
Pin
out distance

si
so
Summary (Cont.)
The actual mechanical
advantage for a lever:
M
A

Fo
Fi
Application of lever principle:
With no friction MI = MA
For Wheel
and axel:
M
A

Fo
Fi

ri
ro

ri
ro
Summary (Cont.)
Belt Drive:
M
A

MI 
Do

Di
output torque
Belt Drive:
M
I

ro
ri
wo

input torque

Fo
wi
ro
MI = 4
o
i
Do
Di
Belt
Drive
ri
Fi
Summary
Gears:
MI 
Do
Di

No
Ni
The Inclined
Plane
si
Ideal Mechanical
Advantage
Fi
q
Fo = W
No
Ni
so
M
I

slope
height

si
so
A ctual A dvantage: M
A

W
Fi
Summary (Cont.)
Fo
R
Fi
p
MI 
2 R
p
Screw
Jack
An application of the
inclined plane:
Input distance: si = 2R
Output distance: so = p
Screw Jack
MI 
si
so

2 R
p
CONCLUSION: Chapter 12
Simple Machines