Updated 6/30/03
+---------------------------------------------------+
| 746 W =
550 ft*lb/s = 1 hp
1,000 W = 1 kW |
+---------------------------------------------------+
|
DIST. | SPEED | ACCEL | g
_ |
MASS | FORCE | WORK | POWER | ENERGY
.
METRIC
| meter | m/s |
m/s2 | 9.8 m/s2
| kg
| Newton| Joule| Watt |
Joule or Cal
ENGLISH| foot | ft/s
| ft/s2
| 32 ft/s2
| slug | Pound | ft lb| hp
| Btu
1
a] ENERGY is the ability to do work.
b] ENERGY is
measured in Joules or calories.
c] LIST 7
FORMS OF ENERGY.
1]
Mechanical: Potential & Kinetic
[2] Chemical [3] Electrical
[4] Heat
5] Light
[6] Sound [7] Nuclear
2
WHAT TYPE OF ENERGY CONVERSIONS TO THE FOLLOWING DEVICES MAKE?
ELECTRIC
MOTORS CONVERT
Electrical ENERGY INTO Mechanical, Heat, Sound ENERGY
CAR ENGINES
CONVERT Chemical ENERGY INTO Mechanical, Heat,
Sound ENERGY
ELECTRIC
GENERATORS CONVERT Mechanical
ENERGY INTO Electrical ENERGY.
ELECTRIC
DRILLS CONVERT Electrical ENERGY INTO Mechanical, Heat, Sound
ENERGY
BATTERIES CONVERT
Chemical ENERGY INTO Electrical ENERGY.
STARS
CONVERT Nuclear ENERGY INTO Light
ENERGY.
3
KINETIC ENERGY is energy due to motion:
K.E. = mV2/2
POTENTIAL
ENERGY is stored energy due to position in a force field.
Gravitational
P.E. = WeightxHeight moved. = P.E. = mgh
A good
example of this is water stored behind a dam.
The higher the dam, not the
bigger the
lake, the more power it can generate.
Similarly,
electrical potential energy is stored in an electrical force field.
A good
example of this is not only the energy stored in a battery but energy stored in
food or
gasoline. All chemical reactions
are governed by electrical forces between the
outer
electrons and the positive nuclei of the various atoms in compounds.
4
WORK PUT INTO A SYSTEM WILL CHANGE ITS kinetic & potential energy.
5
| MIN PE & MAX KE | MAX PE & MIN KE |
a] PENDULUM
| Bottom of Swing | Top
of Swing |
b] BALL
& RAMP | Bottom of
Ramp | Top of
Ramp |
c] SPRING
& WEIGHT| Rest Position | Max Stretch
|
6
A 70 kg TEACHER IS DROPPED OUT A CLASSROOM WINDOW, 4 m HIGH.
a] DRAW AND LABEL A DIAGRAM
WITH THE EQUATIONS FOR THE TOTAL ENERGY AT EACH METER OF FALL.
PE
+ KE
= constant
mgh
+ mV2/2 = constant
4 m | 70 kg x 9.8 m/s2 x 4 m + 70 kg (0)2/2 =
| 2744 Joules
+ 0
= 2744 Joules
-----------------------------------------------------------
3 m | 70 kg x 9.8 m/s2 x 3 m + ____________ = 2744 Joules
| 2058 Joules
+ ____________ = 2744 Joules
-----------------------------------------------------------
2 m | 70 kg x 9.8 m/s2 x 2 m + ____________ = 2744 Joules
| 1372 Joules
+ ____________ = 2744 Joules
-----------------------------------------------------------
1 m | 70 kg x 9.8 m/s2 x 1 m + ____________ = 2744 Joules
| 686 Joules
+ ____________ = 2744 Joules
-----------------------------------------------------------
0 m | 70 kg x 9.8 m/s2 x 0 m + ____________ = 2744 Joules
| 0 Joules
+ ____________ = 2744 Joules
-----------------------------------------------------------
b] FIND HIS
SPEED JUST BEFORE CONTACT WITH THE SIDEWALK.
mV2/2 = 2744 ---> V2 = 2*2744/70 ---> V = \/2*2744/70 = 8.85 m/s
7
As an apple falls from a tree, its potential energy increase and its
kinetic energy
decrease by
an identical amount, so that its total energy is constant.
This is due to
the law of
conservation of energy
8
A 2-kg vase falls from a height of 2 meters.
Its
potential energy just before it reaches the ground is zero since its height is
zero: PE = mgh
Its kinetic
energy = its PE at the start: mgh =
2 kg 9.8 m/s2 2 m =
39.2 J
9
A 10-kg cart in a frictionless environment rolls from the top of a
15-meter-high
hill to the bottom of the track, then continues to roll.
Find the
cart's velocity at the bottom of the hill.
Its kinetic
energy = its PE at the start: mgh =
10 kg 9.8 m/s2 15
m = 1470 J
mV2/2
= 1470 J-->V2 =
2x1470/m = 2940/10 kg--->V = √294 = 17.1 m/s
10
In a frictionless environment, a ball rolls down a steep slope from a
height of
5 meters. It rolls to
the bottom, then up a gentle slope on the other side.
The ball
will attain a height sufficient to restore the potential energy it had at the
start, or 5
m. This is only true in the absence
of friction.
WORK
11
It is used to keep track of energy changes.
That is energy is conserved only if no
external
force acts on the object. In an
equation: W = ΔKE
+ ΔPE
OR Work
causes an increase in Kinetic and Potential Energy.
Since a
force on something can increase its Kinetic Energy (speed) only if the force is
in the
direction of motion
Work is
the component of force in the direction of motion x the distance moved.
Work can be
positive or negative.
It is
positive if the force is in the direction of motion--->it speeds up an object
and
increases
its KE. Ex: the motor of a car.
It is
negative if the force opposes the motion--->it slows down an object and
decreases
its KE. Ex: the breaks
on a car.
The car does
work on the breaks causing an increase in energy in the form of heat.
Tot Car
Energy After = Tot Car Energy Before + Work done on the Car by break
Since the
work on the car is negative it looses energy.
Tot Break
Energy After = Tot Break Energy Before + Work done on the Break by car
Since the
force on the break is in the direction of motion it is positive and the
break gains energy in the form of heat.
Adding these
two equations and taking the sign into account:
Work done on
the Car by break = Work done on the Break by car
Power is
the work per unit time = Work/Time = Force x Speed
12
| WORK
|
POWER
.
WORD | Work = Force x Distance | Power = Work/Time
SYMBOL | W = FxD
|
P = FxD/t
UNIT | Joule = Newton Meter
| Watt = Newton Meter/Second
13
FIND THE WORK DONE IN THE FOLLOWING SITUATIONS:
a] A GIRL
PUSHES A CAR WITH A 200 N FORCE FOR 4 METERS.
W = FxD W = 200
N x 4 m = 800 Nm = 800 Joules
b] A BOY
PUSHES WITH 150 N ON A WALL FOR 10 s, IT DOESN'T MOVE.
Distance moved = zero--->W = Fxd = 150 N x 0 m = 0 Joules
c] A STUDENT
CARRIES A 30 kg BOX 20 m HORIZONTALLY.
D = 20 m horizontal
F = 30 kg x 9.8 m/s2 = 29.4 N vertical-->Not in
direction of motion
F = 0 N in direction of motion (assume no friction)
W = FxD = 0 x 20 m
d] A PERSON
DROPS A 2 kg BALL 10 m TO THE GROUND.
The person does no work for he/she exerts no force.
However the planet works because it exerts the force of gravity.
W = FxD---> W = 2 kg x 9.8 m/s2 x 10 m = 196 J
e] A MAN
PULLS A ROPE AT 35o, WITH 230 N, TO MOVE A 450 N BOX 7 m.
F in direction of motion = 230 N x cos(35) (Note:
use degree mode)
W = FxD = 230 N x cos(35) x 7 m = 1376 J
f] A
GENERATOR RUNS A 100 WATT LIGHT BULB 8 hr A DAY FOR 30 DAYS.
P = 100 watts = 100 J/s
P = W/t --->W = Pt
W = ?
W = 100 watt x 240 h
t = 8 h/dx30d = 240 h
W = 24,000 Watt h = 24 kilowatt hrs
Note:
kilowatt hr is the unit of energy PECO uses.
POWER
THE
WATT IS NAMED FOR JAMES WATT, THE INVENTOR OF THE STEAM ENGINE.
IT IS THE SAME UNIT OF POWER USED BY ELECTRIC COMPANIES.
14
a] FIND THE POWER IF A HORSE RAISES 2,487 N, 3 m IN 10 s.
P = Fd/t = 2,487 N x 3 m/10 s = 746 Watts = 1 hp
b] WHAT SIZE
MOTOR [hp] IS NEEDED TO LIFT 400 lb, 20 ft IN 5 s.
P = ?
P = FD/t
F = 400 lb P
= 400 lb x 20 ft/ 5s
D = 20 ft P
= 1600 lb ft/s x 1 hp/550 ft lb/s
t = 5 s
P = 2.9 hp---> Ans:
buy 3 hp motor.
c]
ASSUMPTIONS: The force is in the
direction of motion &
The motor is 100% efficient.
15
a] HOW LONG (min & sec) DOES IT TAKE A 1 hp MOTOR TO LIFT A 1100 lb
BOX 131 ft.
t = ?
P
= FD/t
P = 1 hp = 550 ft lb/s
550 ft lb/s = 1100 lb 131 ft/t
F = 1100 lb
550 ft lb/s = 144,100 lb ft/t
D = 131 ft
t = 144,100 lb ft/550 ft lb/s
t = 262 s x 1 min/60 s
t = 4 min .36666666 min x 60 s/min
t = 4 min 22 sec
c]
ASSUMPTIONS: The force is in the
direction of motion &
The motor is 100% efficient.
16
A CAR IS DRIVEN AT A STEADY SPEED OF 21 m/s DOWN A ROAD.
THE CAR'S
ENGINE DELIVERS 48,000 WATTS OF POWER.
FIND THE
AVERAGE FORCE OF AIR FRICTION ON THE CAR.
V =
21 m/s
P = FxV
P = 48,000
Watts 48,000
Watts = F 21 m/s
F = ?
F = 48,000 N m/s = 2286 N
21 m/s
Updated 5/26/03
Top
+---------------------------------------------------+
| 746 W = 550 ft*lb/s = 1 hp
1,000 W = 1 kW |
+---------------------------------------------------+
1
a] DEFINE WORK & POWER. Work
is the component of force in the direction
of motion x the distance moved.
Power is the work per unit time = Work/Time = Force x Speed
b] WORD, SYMBOL & UNIT EQUATIONS FOR:
WORK & POWER?
Work |
Power
|
| Time
Symbol | W = FxD
| P = FxD/t
Unit | Joule = Newton x Meter | Watt = Newton Meter/Sec
2 FIND THE WORK DONE IN THE FOLLOWING SITUATIONS:
a] A GIRL PUSHES A CAR WITH A 200 N FORCE FOR 4 METERS.
b] A BOY PUSHES WITH 150 N ON A WALL FOR 10 s, IT DOESN'T MOVE.
c] A STUDENT CARRIES A 30 kg BOX 20 m HORIZONTALLY.
d] A PERSON DROPS A 2 kg BALL 10 m TO THE GROUND.
e] A MAN PULLS A ROPE AT 35o, WITH 230 N, TO MOVE A 450 N BOX 7 m.
f] A GENERATOR RUNS A 100 WATT LIGHT BULB 8 hr A DAY FOR 30 DAYS.
3A a] FIND THE POWER IF A HORSE RAISES 2,487 N, 3 m IN 10 s.
b] WHAT SIZE MOTOR [hp] IS NEEDED TO LIFT 400 lb, 20 ft IN 5 s.
c] WHAT ARE THE 2 ASSUMPTIONS?
The force is in the direction of motion & The motor is 100% efficient.
3B a]
How long (min & sec) DOES IT TAKE A 1 hp MOTOR TO LIFT A 1100 lb
ELEVATOR
131 ft.
b] THE 2 ASSUMPTIONS:
Force is in the direction of motion & no friction
c]
3 ASSUMPTIONS FOR A CAR:
Force is in the direction of motion, no
friction
4 A CAR IS DRIVEN AT A STEADY SPEED OF 21 m/s DOWN A ROAD.
THE CAR'S ENGINE
DELIVERS 48,000 WATTS OF POWER.
FIND THE AVERAGE FORCE OF AIR FRICTION ON THE
CAR.
5 USE A TABLE FORMAT TO WRITE THE ENGLISH & METRIC UNITS FOR DISTANCE,
SPEED,
ACCEL, g, MASS, FORCE, WORK, POWER, ENERGY
6 a] WHAT ARE THE 4 CHARACTERISTICS OF A MACHINE?
1] INCREASE FORCE BY LOOSING SPEED--INCREASING INPUT DISTANCE--LEVER
2] INCREASE SPEED BY LOOSING FORCED--INCREASING INPUT FORCE--BIKE
3] CHANGE THE DIRECTION OF A FORCE. EX--PULLEY
4] FRICTION CHANGES THE USEFUL WORK INTO WASTED HEAT
b] DIAGRAM A SIMPLE MACHINE & WRITE THE EQUATIONS FOR AMA & EFFICIENCY
c] WHAT IS AN IDEAL MACHINE?
d] DERIVE THE EQUATION FOR IDEAL MECHANICAL ADVANTAGE.
e] STATE 2 MORE EQUATIONS FOR IMA. IMA= Fout/Min Fin, IMA=
MaxFout/Fin
7 A 16 ft RAMP NEEDS A 90 lb FORCE TO LIFT 270 lb BOX 4 ft.
a] DRAW & LABEL THE DIAGRAM.
b] FIND THE AMA & IMA
c] ACTUAL FORCE NEEDED TO LIFT A 100 lbs
d] MINIMUM POSSIBLE FORCE NEEDED TO LIFT A 100 lbs
e] FRICTION FORCE FOR A 100 lb LOAD
f] ACTUAL WEIGHT THAT CAN BE LIFTED BY A 100 lb FORCE
g] MAX WEIGHT THAT CAN BE LIFTED BY A 100 lb FORCE.
h] EFFICIENCY.
8 a] DESIGN A RAMP TO LIFT 350 lb BOX 4.5 ft WITH 75 lbs OF FORCE.
ASSUME 90% EFFICIENCY. SHOW DIAGRAM
WITH ALL 3 DISTANCES (ft & in).
b] WHAT IS THE LOGIC CHECK? Din
> Dout & Din > Base
9 a] DIAGRAM & FIND THE IMA FOR A LEVER, AND A WHEEL & AXLE
b] DRAW & LABEL A DIAGRAM FOR A BICYCLE AND DETERMINE ITS
IMA.
IMA = IMA 1 x IMA 2 = PEDAL R/FRONT GEAR R x REAR GEAR R/WHEEL R
ANS:
2a) 800 J; (b) 0 J; (c) 0 J; (d) 196 J; (e) 1,319 J; (f) 24 kW*hr
3A(a) 746 W; (b) 2.91 hp; 3B(b)
4 min 22.5 s; (P) 1.0 min 11.6 s;
4] 2,286 N
7b] 3,4 [c] 33.3 lb [d] 25 lb [e] 8.3 lb [f] 300 lb [g] 400 lb [h] 75%
Pb] 4,5 [c] 25 lb [d] 20 lb
[e] 5 lb [f] 400 lb [g] 500
lb [h] 80%
8] Din= 23'4",B= 22'10.7" [P] Din = 26'7.1", B= 26'1.5"
Updated 7/01/03
Top
OBJECTIVE:
STATE THE 4 CHARACTERISTICS OF MACHINES.
1
a] WHAT ARE THE 4 CHARACTERISTICS OF A MACHINE?
1] INCREASE FORCE BY LOOSING SPEED--INCREASING INPUT DISTANCE--LEVER
2] INCREASE SPEED BY LOOSING FORCED--INCREASING INPUT FORCE--BIKE
3] CHANGE THE DIRECTION OF A FORCE. EX--PULLEY
4] FRICTION CHANGES THE USEFUL WORK INTO WASTED HEAT
2
a] DIAGRAM A SIMPLE MACHINE
FORCE IN
+--------+
FORCE OUT
7
lb --->| MA = 5 |----> 35 lb
+--------+
b] DEFINE MECHANICAL ADVANTAGE IN WORDS AND EQUATIONS.
Mechanical Advantage--is the number of times a machine multiplies a force.
FORCE
OUT
OBJECT WEIGHT
Mechanical Advantage =
--------- = -------------------------------
FORCE IN
FORCE DUE TO GRAVITY + FRICTION
OR
RESISTANCE FORCE
MECHANICAL ADVANTAGE =
----------------
EFFORT FORCE
c] WHEN DOES A MECHANICAL DISADVANTAGE OCCUR.
Mechanical disadvantage occurs if a machine is run in a reverse direction for a
speed advantage, the force is decreased [= mechanical advantage < 1].
Ex.: bicycle
gears [Pulley
demo]
d] A machine used 10 lb to move a 100 lb object.
Find its mechanical advantage?
M.A. = Fout/Fin = 100 lb/10 lb = 10 (Unitless!)
e] IF 5, 8, 15, 100 lb IS APPLIED TO THIS SAME MACHINE,
HOW MUCH FORCE DO YOU GET OUT?
Fout = M.S. x Fin = 10 x (5, 8, 15, 100 lb) = 50, 80, 150, & 1000 lbs
3
a] WHAT ARE THE WORD AND SYMBOL EQUATIONS FOR EFFICIENCY.
WORK OUT
FORCE OUT x DISTANCE OUT
EFFICIENCY = -------- x
100% =
------------------------ x 100%
WORK IN
FORCE IN x DISTANCE IN
Fout Dout
Eff = ---------100%
Fin
Din
Work out is always less than work in because work is lost overcoming friction.
b] WHAT IS AN IDEAL MACHINE?
AN IDEAL MACHINE IS 100% EFFICIENT, I.E., Wout = Win, IT HAS NO FRICTION
c] DERIVE THE EQUATION FOR IDEAL MECHANICAL ADVANTAGE.
Wout = Win
Fout Dout = Fin Din
Divide by Dout
Din
Fout = Fin ----
Dout
Divide by Fin
Fout =
Din = IMA Note:
this is true only for ideal machines
.
Fin Dout
Note: This is a no friction M.A. OR
the greatest possible Mechanical Advantage.
It is determined by the machine geometry.
d] STATE 2 MORE EQUATIONS FOR IMA.
IMA= Fout/Min Fin, IMA= MaxFout/Fin
4
a] NAME THE 6 SIMPLE MACHINES &
1) lever, 2) inclined plane, 3)
pulley, 4) wedge, 5) wheel & axle, AND 6) screw
b] DEFINE COMPOUND MACHINES.
Compound machines--consists of more than one simple machine.
Ex: egg beater; bicycle.
5
A 16 ft RAMP NEEDS A 90 lb FORCE TO LIFT 270 lb BOX 4 ft.
a] DRAW & LABEL THE DIAGRAM.
b] FIND THE AMA & IMA
AMA = Actual Mechanical
Advantage = Fout/Fin = 270 lb/90 lb = 3
IMA = Ideal Mechanical
Advantage = Din/Dout = 16 ft/ 4 ft = 4
c] EFFICIENCY
Fout Dout
270 lb 4ft
Eff = ---------100% = ----------- 100% = 75%
Fin
Din 90 lb 16 ft
d] ACTUAL FORCE NEEDED TO LIFT A 100 lbs
Fout = 100 lb
AMA = Fout/Fin
Ans Actual Fin = 33.3 lb
Fin
= ?
Cross Multiply
AMA = 3 Fin
= Fout/AMA = 100 lb/3 = 33.3 lb
e] MINIMUM POSSIBLE FORCE NEEDED TO LIFT A 100 lbs
Min possible force required
an ideal machine
Fout = 100 lb
AMA = Fout/Fin
Ans Min Fin = 25 lb
Fin
= ?
Cross Multiply
IMA = 4 Fin
= Fout/IMA = 100 lb/4 = 25 lb
f] FRICTION FORCE FOR A 100 lb LOAD
Actual Fin = Min Fin +
Friction
33.3 lb =
25 lb + Ff
-25.0 lb
-25 lb
.
8.3 lb = Ff
g] ACTUAL WEIGHT THAT CAN BE LIFTED BY A 100 lb FORCE
Fout = ? AMA =
Fout/Fin
Ans Actual Fin = 33.3 lb
Fin
= 100 lb Multiply
by Fin
AMA = 3 Fout
= Fin AMA = 100 lb 3 = 300 lb
Fout = ?
h] MAX WEIGHT THAT CAN BE LIFTED BY A 100 lb FORCE.
Max possible weight requires
an ideal machine
Fout = ? AMA =
Fout/Fin
Ans Min Fin = 25 lb
Fin
= 100 lb Multiply
by Fin
IMA = 4 Fout
= Fin IMA = 100 lb 4 = 400 lb
6
a] DESIGN A RAMP TO LIFT 350 lb BOX 4.5 ft WITH 75 lbs OF FORCE.
ASSUME 90% EFFICIENCY.
SHOW DIAGRAM WITH ALL 3 DISTANCES (ft & in).
Fout = 350 lb
Fout Dout
Dout = 4.5 ft
Eff = --------- 100%
Fin =
75 lb
Fin Din
Eff = 90%
350 lbx4.5 ft
Din =
?
90% = --------------100%
Base = ?
75 lb x Din
Multiply by Din
350 lbx4.5 ft
Din90% = --------------100%
75
lb
Divide by 90%
350 lbx4.5 ft
Din = --------------100% = 23.33333 ft
75 lb 90%
Don’t Round Off
Din = 23 ft + 0.333333ft x 12 in/ft
Din = 23 ft + 4 in
Pythagorean Theorem c2
= a2
+ b2
OR L2
= Dout2
+ Base2--->
Base2
= L2
- Dout2
Base
= √Din2
- Dout2
= √23.332
– 4.52
= 22.89189595 ft
Don’t Round Off
Base
= 22 ft + 0.89189595 ft x 12in/ft =
= 22 ft 10.7 in
b] WHAT IS THE LOGIC CHECK?
Din >
Dout
&
Din >
Base
23.3 ft >
4.5 ft & 23.3 ft>
22.9 ft
7
a] DIAGRAM & FIND THE IMA FOR A LEVER, AND A WHEEL & AXLE
b] DRAW & LABEL A DIAGRAM FOR A BICYCLE AND DETERMINE ITS IMA.
IMA = IMA 1 x IMA 2 = PEDAL R/FRONT GEAR R x REAR GEAR R/WHEEL R
Updated 5/26/03
Top
Common
Power/Weight Ratios
Cruz Ship (Nordic Empress): 22,000 hp/48,563tons = 0.45 hp/ton
Ratio/SCar
Car: Small (Toyota Corolla): 130
hp/2,590 lb = 0.0502 hp/lb 1.0000
Large (Cadillac
DeVille): 300 hp/4,070 lb = 0.0737
hp/lb 1.4681
Sport
(Corvette):
405 hp/3,295 lb = 0.1229 hp/lb 2.4482
Race (Formula
I):
800 hp/1,325 lb = 0.6038 hp/lb
12.0279
Planes: Small ( Piper Cub):
65 hp/1,170 lb = 0.0556 hp/lb 1.1076
Ratio/Liner
Liner (Boing 737-300):
22,000 lb/138,500 lb = 0.1588 lb/lb 1.000
Fighter (F15):
46,000 lb/68,000 lb = 0.6765 lb/lb 4.2601
Rocket (Shuttle):
6,929,134/4,500,000 lb = 1.54 lb/lb 9.6977
Large
(Saturn V): 7,750,000 lb/6,100,000
lb = 1.27 lb/lb 7.9975
Small (Redstone): 88,000 lb/38,500lb =
2.29 lb/lb
14.4207
1 STATE THE 4 CHARACTERISTICS OF MACHINES.
1] INCREASE FORCE BY LOOSING SPEED--INCREASING INPUT DISTANCE--LEVER
2] INCREASE SPEED BY LOOSING FORCE--INCREASING INPUT FORCE--BIKE
3] CHANGE THE DIRECTION OF A FORCE. EX--PULLEY
4] FRICTION CHANGES THE USEFUL WORK INTO HEAT
2 +----------------------+
a] MACHINE: ---Force In--->|
Mechanical Advantage |---Force Out--->
+----------------------+
b] Actual Mechanical Advantage = Force Out/Force In
AMA = Fout/Fin
Work Out
Fout Dout
c] EFFICIENCY = --------100% = ---------100%
Work In
Fin Din
3
A MACHINE USES 10 lb TO MOVE A 100 lb OBJECT.
4 THE 4 CHARACTERISTICS OF AN IDEAL MACHINE
1] THERE IS NO FRICTION
2] EFFICIENCY = 100%
3] WORK OUT = WORK IN
4] THE M.A. CAN BE CALCULATED FROM THE MACHINE GEOMETRY.
5 DERIVE THE EQUATION FOR THE MECHANICAL ADVANTAGE OF AN IDEAL MACHINE
(IMA).
[Fout/Fin = ?]
Ideal---> Work In = Work Out
Fin Din = Fout Dout
[Divide by Din]---> Fin = Fout Dout
Din
[Divide by Fout]--->Fin = Dout = Mechanical advantage of an
Fout Din
Ideal Machine (IMA)
5 A 20 ft RAMP NEEDS A 150 lb FORCE TO LIFT 300 lb BOX 4 ft.
a] DIAGRAM & FIND THE:
b] AMA & IMA: SIZE CHECK:
IMA > AMA
c] ACTUAL FORCE NEEDED TO LIFT A 100 lbs [Diagram]. Act Fin < 100 lb
d] MIN POSSIBLE FORCE TO LIFT A 100 lbs [Diagram]. Min Fin < Act Fin
e] FRICTION FORCE FOR A 100 lb LOAD. Ff
< Act Fin
f] ACTUAL WEIGHT LIFTED BY A 100 lb FORCE [Diagram] Act Fout > 100 lb
g] MAX WEIGHT LIFTED BY A 100 lb FORCE [Diagram]. Max Fout > Act Fout
h] EFFICIENCY. Eff < 100%
6 DESIGN A RAMP TO LIFT 300 lb, 5 ft WITH A 100 lb FORCE.
ASSUME 92% EFFICIENCY.
[a] DIAGRAM THE PROBLEM [b] DIAGRAM THE SOLUTION (3 sides in ft & in)
7
DRAW 5 PULLEY SYSTEMS (SEE BOARD) & DETERMINE THE IMA.
8 FOR A 3 CORD
PULLEY SYSTEM:
a] DIAGRAM & FIND THE IMA.
b] THE MINIMUM POSSIBLE FORCE NEEDED TO LIFT A 100 LB WEIGHT
c] IF IT IS USED FOR A SPEED ADVANTAGE, DIAGRAM & FIND THE
IMA.
d] THE MINIMUM POSSIBLE FORCE NEEDED TO LIFT A 100 LB WEIGHT.
9 WHAT SIZE PULLEY SYSTEM WOULD A 100 lb PERSON NEED TO LIFT A
1000 lb ENGINE? [ASSUME 100%
EFFICIENCY]
HMWK: A 150 lb PERSON NEED TO LIFT A 2000 lb ENGINE?
10 a] WHEN THE PULLEY IS RUN FORWARDS YOU GAIN Force AT THE COST OF Speed
b] WHEN RUN BACKWARDS YOU GAIN Speed AT THE COST OF Force
c] WHAT IS THE EQUATION FOR IMA WHEN A PULLEY IS RUN BACKWARDS?
d] Friction CAUSES ENERGY LOSES & MAKES IT LESS THAN 100% EFFICIENT
e] TO INCREASE THE EFFICIENCY oil the axel bearings.