Download Chapter 10 and 11 Work and Energy

Chapter 10 and 11
Work and Energy
Work Done by a Constant Force
The work done by a constant force is defined as
the distance moved multiplied by the component
of the force in the direction of displacement:
Work for Force at an Angle
If the force is at an angle to the displacement:
Only the horizontal component of the force
does any work (horizontal displacement).
Negative and Positive Work
The work done may be positive, zero, or
negative, depending on the angle between the
force and the displacement:
Work Done by a Constant Force
In the SI system, the units of work are joules:
As long as this person does
not lift or lower the bag of
groceries, he is doing no work
on it. The force he exerts has
no component in the direction
of motion.
Typical Work
Example: Pulling a Suitcase
A rope inclined upward at 45o pulls
a suitcase through the airport. The
tension on the rope is 20 N.
How much work does the tension
do, if the suitcase is pulled 100 m?
W = T (∆x) cos θ
W = (20 N)(100 m) cos 45o = 1410 J
Note that the same work could have been done by a tension of just
14.1 N by pulling in the horizontal direction.
Example: Perpendicular Force and
Work
A car is traveling on a curved
highway. The force due to friction fs
points toward the center of the
circular path.
How much work does the frictional
force do on the car?
Zero!
General Result: A force that is
everywhere perpendicular to the
motion does no work.
Work Done by a Constant Force
If there is more than one force acting on an
object, we can find the work done by each force,
and also the work done by the net force:
Gravitational Work
In lifting an object of weight mg by a height h, the
person doing the lifting does an amount of work W = mgh.
If the object is subsequently allowed to fall a
distance h, gravity does work W = mgh on the object.
Example: Loading with a Crane
A 3,000 kg truck is to be loaded onto a ship by a
crane that exerts an upward force of 31 kN on the
truck. This force, which is large enough to
overcome the gravitational force and keep the truck
moving upward, is applied over a distance of 2.0 m.
(a) Find the work done on the truck by the crane.
(b) Find the work done on the truck by gravity.
(c) Find the net work done on the truck.
Wapp = Fapp y ∆y = (31 kN)(2.0 m) = 62 kJ
Wg = mg y ∆y = (3000 kg)(−9.81 m/s 2 )(2.0 m) = −58.9 kJ
Wnet = Wapp + Wg = (62.0 kJ) + (−58.9 kJ) = 3.1 kJ
Positive & Negative
Gravitational Work
When positive work is done on an object, its
speed increases; when negative work is done, its
speed decreases.
Work Done by a Constant Force
Solving work problems:
1. Draw a free-body diagram.
2. Choose a coordinate system.
3. Apply Newton’s laws to determine any
unknown forces.
4. Find the work done by a specific force.
5. To find the net work, either
find the net force and then find the work it
does, or
find the work done by each force and add.
Work Done by a Constant Force
Work done by forces that oppose the direction
of motion, such as friction, will be negative.
Centripetal forces do no
work, as they are always
perpendicular to the
direction of motion.
Work Done by a Varying Force
For a force that varies, the work can be
approximated by dividing the distance up into
small pieces, finding the work done during
each, and adding them up. As the pieces
become very narrow, the work done is the area
under the force vs. distance curve.
Kinetic Energy, and the Work-Energy
Principle
Energy was traditionally defined as the ability to
do work. We now know that not all forces are
able to do work; however, we are dealing in
these chapters with mechanical energy, which
does follow this definition.
Kinetic Energy, and the Work-Energy
Principle
If we write the acceleration in terms of the
velocity and the distance, we find that the
work done here is
We define the kinetic energy:
Kinetic Energy, and the Work-Energy
Principle
This means that the work done is equal to the
change in the kinetic energy:
• If the net work is positive, the kinetic energy
increases.
• If the net work is negative, the kinetic energy
decreases.
Kinetic Energy, and the Work-Energy
Principle
Because work and kinetic energy can be
equated, they must have the same units:
kinetic energy is measured in joules.
Question 1
Car 1 has twice the mass of Car 2, but
they both have the same kinetic energy.
If the speed of Car 1 is v, approximately
what is the speed of Car 2?
a) 0.50 v
b) 0.707 v
c) v
d) 1.414 v
e) 2.00 v
Example: A Dogsled Race
During your winter break, you enter a
“dogsled” race across a frozen lake, in which
the sleds are pulled by people instead of dogs.
To get started, you pull the sled (mass 80 kg)
with a force of 180 N at 40° above the
horizontal. The sled moves ∆x = 5.0 m,
starting from rest. Assume that there is no
friction.
(a) Find the work you do.
(b) Find the final speed of your sled.
Wtotal = Wyou = Fx ∆x = F cos θ∆x
= (180 N)(cos 40°)(5.0 m) = 689 J
Wtotal = mv − mv = mv
1
2
2
f
1
2
2
i
1
2
2
f
2Wtotal
v =
m
2
f
2Wtotal
2(689 J)
vf =
=
= 4.15 m/s
m
(80 kg)
Example: Work and Kinetic Energy
in a Rocket Launch
A 150,000 kg rocket is launched straight
up. The rocket engine generates a thrust of
4.0 x 106 N.
What is the rocket’s speed at a height of
500 m? (Ignore air resistance and mass loss
due to burned fuel.)
Wthrust = Fthrust ∆y cos θ = (4.0 ×106 N )(500m) cos 0° = 2.0 ×109 J
Wgrav = Fgrav ∆y cos θ = mg∆y cos θ = (1.5 ×105 kg )(9.80 m s 2 )(500m) cos180° = −0.74 ×109 J
∆K = 12 mv 2 − 0 = Wthrust + Wgrav
2∆K
2 m
9
=
=
1
.
30
×
10
v
s
= 1.26 × 10 J
m
Example: Pushing a Puck
A 500 g ice hockey puck slides across
frictionless ice with an initial speed of
2.0 m/s. A compressed air gun is used to
exert a continuous force of 1.0 N on the
puck to slow it down as it moves 0.50 m.
The air gun is aimed at the front edge of
the puck, with the compressed air flow
30o below the horizontal.
What is the puck’s final speed?
W = F (∆r ) cos θ = (1.0 N)(0.5 m) cos150o = −0.433 J
∆K = 12 mv12 − 12 mv0 2 = W
v1 = v0 2 +
2W
2(−0.433 J)
= (2.0 m/s) 2 +
= 1.51 m/s
m
(0.5 kg)
Example: Work on an Electron
In a television picture tube, electrons are accelerated
by an electron gun. The force that accelerates the
electron is an electric force due to the electric field
in the gun. An electron is accelerated from rest by an
electron gun to an energy of 2.5 keV (2,500 eV) over a distance
of 2.5 cm. (1 eV = 1.60 x 10-19 J)
Find the force on the electron, assuming that it is constant and in the
direction of the electron’s motion.
Fx ∆x = K f − K i
Wtotal = ∆K
Fx =
K f − Ki
∆x
(2,500 ev)(1.6 ×10−19 J/eV) − 0
=
(0.025 m)
= 1.6 ×10−14 N
Potential Energy
An object can have potential energy (PE or U) by
virtue of its surroundings.
Familiar examples of potential energy:
• A wound-up spring
• A stretched elastic band
• An object at some height above the ground
PE is due to an object position
Potential Energy
In raising a mass m to a height
h, the work done by the
external force is
(6-5a)
We therefore define the
gravitational potential energy:
(6-6)
Potential Energy
This potential energy can become kinetic energy
if the object is dropped.
Potential energy is a property of a system as a
whole, not just of the object (because it depends
on external forces).
If
, where do we measure y from?
It turns out not to matter, as long as we are
consistent about where we choose y = 0. Only
changes in potential energy can be measured.
Potential Energy Value
• Discovery Ride Packet due Friday
Example: Pike’s Peak or Bust
An 82.0 kg mountain climber is in the
final stage of the ascent of Pike’s Peak,
which 4,301 m above sea level.
(a) What is the change in gravitational
potential energy as the climber gains the
last 100.0 m of altitude? Use PE=0 at sea
level.
(b) Do the same calculation with PE=0 at
the top of the peak.
∆U a = mgy f − mgyi = mg ( y f − yi )
= (82.0 kg)(9.81 m/s 2 ) [ (4301 m) − (4201 m) ] = 80, 400 J
∆U b = mgy f − mgyi = mg ( y f − yi )
= (82.0 kg)(9.81 m/s 2 ) [ (0) − (−100 m) ] = 80, 400 J
Example: A Mountain Bar
A candy bar called the Mountain Bar
has an energy content when metabolized
of 212 Cal = 212 kcal. This is equivalent
to 8.87 x 105 J.
If an 81.0 kg mountain climber eats a
Mountain Bar and magically converts all
of its energy content into gravitational
potential energy, how much altitude ∆y
should he be able to gain?
∆U = mgy f − mgyi = mg ∆y
∆U
(8.87 ×105 J)
∆y =
=
= 1,120 m
2
mg (81.0 kg)(9.81 m/s )
Example: A Falling Bottle
A 0.350 kg bottle falls from rest from a shelf that is 1.75 m
above the floor.
(a) Find the potential energy of the bottle-Earth system when
the bottle is on the shelf.
(b) Find the kinetic energy of the bottle-Earth system just
before impact with the floor.
Wtotal = Wg = ∆K
Wg = mg ( yi − y f ) = mgh
mgh = ∆K = K f − K i
K f = K i + mgh = 0 + (0.350 kg)(9.81 m/s 2 )(1.75 m) = 6.01 J
Elastic Solids & Restoring Forces
An “elastic” material is one
that exhibits a restoring
force, a force that acts so
that it restores a system to an
equilibrium position. Examples
are springs and rubber bands.
An elastic material stores
potential energy when it is
deformed and restores it when
it returns to equilibrium.
Microscopically, elastic
solids depend on the springlike bonds that bind atoms in a
solid.
rubber
band
Potential Energy
Potential energy can also be stored in a spring
when it is compressed; the figure below shows
potential energy yielding kinetic energy.
Potential Energy
The force required to
compress or stretch a
spring is:
where k is called the
spring constant, and
needs to be measured for
each spring. This is
Hooke’s Law
Hooke’s Law
The linear proportionality between
force and displacement is found to be
valid whether the spring os stretched
or compressed, and the force and
displacement are always in opposite
directions.
Therefore, we write the forcedisplacement relation as:
( Fsp ) s = − k ∆s
This relation for the restoring
force of a spring is sometimes called
Hooke’s Law, named after Robert
Hooke, a contemporary of Newton. It
is not really a law or nature, but rather
a rule of behavior for most springs.
Potential Energy
The force increases as the spring is stretched or
compressed further. We find that the potential
energy of the compressed or stretched spring,
measured from its equilibrium position, can be
written:
(6-9)
Work and Springs
The force needed to stretch a spring an
amount x is F = kx.
Therefore, the work
done in stretching
the spring is
Stretching a Spring
The unloaded spring has a length L0. Hang a weight of
mass m on it and it stretches to a new length L. Repeat
the process and measure ∆s=L-L0 vs. the applied force
Fsp=mg.
We find that Fsp=k∆
∆s, where k is the “spring
constant”.
Hooke’s Law and Work
Wby spring = A1 + A2 = A1 − A2
= kx − kx
1
2
1
2
2
1
W = kx
1
2
2
2
2
f
Question 1
The force vs. displacement curves of three springs are measured.
Which spring has the largest spring constant?
a) Spring 1
b) Spring 2
c) Spring 3
d) They are all the same
Question 2
4.0 m/s
A spring-loaded gun shoots a plastic ball with a speed of 4.0 m/s.
If the spring is compressed twice as far, what is the ball’s
speed?
a) 2.0 m/s
b) 4.0 m/s
c) 8.0 m/s
d) 16.0 m/s
e) 32.0 m/s
Example: Work Done
on a Block by a Spring
A 4.0 kg block on a frictionless surface is
attached to a horizontal spring with k = 400
N/m. The spring is initially compressed to 5.0
cm.
(a) Find the work done on the block by the
spring as the block moves from x = x1 = -5.0 cm
to its equilibrium position of x = x2 = 0 cm.
(b) Find the speed of the block at x2 = 0 cm.
W = ∑ Fx ∆x = −k ∑ x∆x = 12 k x12 − 12 k x22
= 12 (400 N/m)[(−0.05 m) 2 − (0 m) 2 ] = 0.50 J
W = 12 m v 2f − 12 m vi2
vf =
⇒ v 2f = vi2 +
2W
m
2W
2(0.50 J)
=
= 0.50 m/s
(4.0 kg)
m
Example: Dragging a Block (1)
A spring is attached to a 2 kg
block. The other end is pulled by a
motorized toy train that moves
forward at 5.0 cm/s. The spring
constant is k=50 N/m and the
coefficient of static friction
between the block and the surface
is µs=0.6. The spring is in
equilibrium at t=0 s when the train
starts to move.
At what time does the block
start to slip?
Example: Dragging a Block (2)
∑ (F
) = ( Fsp ) x + ( f s ) x = Fsp − f s = 0
net x
f s = µ s mg = Fsp = k ∆x
µ s mg
(0.60)(2.0 kg)(9/80 m/s 2 )
∆x =
=
k
(50 N/m)
= 0.235 m = 23.5 cm
t=
∆x (23.5 cm)
=
= 4.7 s
v (5.0 cm/s)
This is an example of “stick-slip motion”, which is common in nature.
Example: behavior of rocks during seismic activity and earthquakes.
Conservation of Mechanical Energy
Definition of mechanical energy:
(8-6)
Using this definition and considering only
conservative forces, we find:
Or equivalently:
Conservation of
Mechanical Energy
Emech = K + U
∆Emech = ∆K + ∆U = 0
(Conservation of Mechanical Energy)
Note that ∆K ≠ ∆U ; the correct statement is ∆K = −∆U
Conservative and Nonconservative Forces
If friction is present, the work done depends not
only on the starting and ending points, but also
on the path taken. Friction is called a
nonconservative force.
Nonconservative Forces
In the presence of nonconservative forces, the
total mechanical energy is not conserved:
Solving,
(8-9)
Conservative and Nonconservative Forces
Potential energy can
only be defined for
conservative forces.
Conservative and Nonconservative Forces
Therefore, we distinguish between the work
done by conservative forces and the work done
by nonconservative forces.
We find that the work done by nonconservative
forces is equal to the total change in kinetic and
potential energies:
Mechanical Energy and Its Conservation
If there are no nonconservative forces, the sum
of the changes in the kinetic energy and in the
potential energy is zero – the kinetic and
potential energy changes are equal but opposite
in sign.
This allows us to define the total mechanical
energy:
And its conservation:
Conservation of Mechanical Energy
Energy conservation can make kinematics
problems much easier to solve:
Problem Solving Using Conservation of
Mechanical Energy
In the image on the left, the total
mechanical energy is:
The energy buckets (right)
show how the energy
moves from all potential to
all kinetic.
Problem Solving Using Conservation of
Mechanical Energy
If there is no friction, the speed of a roller
coaster will depend only on its height
compared to its starting height.
Problem Solving Using Conservation of
Mechanical Energy
For an elastic force, conservation of energy tells
us:
Other Forms of Energy; Energy
Transformations and the
Conservation of Energy
Some other forms of energy:
Electric energy, nuclear energy, thermal energy,
chemical energy.
Work is done when energy is transferred from
one object to another.
Accounting for all forms of energy, we find that
the total energy neither increases nor
decreases. Energy as a whole is conserved.
Energy Conservation with Dissipative
Processes; Solving Problems
If there is a nonconservative force such as
friction, where do the kinetic and potential
energies go?
They become heat; the actual temperature rise of
the materials involved can be calculated.
Energy Conservation with Dissipative
Processes; Solving Problems
Problem Solving:
1. Draw a picture.
2. Determine the system for which energy will
be conserved.
3. Figure out what you are looking for, and
decide on the initial and final positions.
4. Choose a logical reference frame.
5. Apply conservation of energy.
6. Solve.
Example: Graduation Fling
At the end of a graduation
ceremony, the graduates fling their
caps into the air. Suppose a 0.120 kg
cap is thrown straight upward with a
speed of 7.85 m/s and that frictional
forces can be ignored.
(a) Use kinematics to find the speed
of the cap when it has risen 1.18 m
above the fling point.
(b) Show that the total mechanical
energy of the cap is unchanged.
v y2 = v02y + 2a y ∆y
v y = v02y + 2a y ∆y = (7.85 m/s) 2 + 2(−9.81 m/s 2 )(1.18 m) = 6.02 m/s
Ei = U i + K i = mgyi + 12 mvi2 = 0 + 12 (0.120 kg)(7.85 m/s) 2 = 3.70 J
E f = U f + K f = (0.120 kg)(9.81 m/s 2 )(1.18 m) + 12 (0.120 kg)(6.20 m/s) 2 = 3.70 J
Example: Catching a Home Run
At the bottom of the 9th inning,
a player hits a 0.15 kg baseball over
the outfield fence. The ball leaves
the bat with a speed of 36.0 m/s
and a fan in the bleachers catches
it 7.2 m above the point where it
was hit. Neglect air resistance.
(a) What is the kinetic energy Kf of the ball when caught?
(b) What is the speed vf of the ball when caught.
E = U i + K i = mgyi + 12 mvi2 = 0 + 12 (0.15 kg)(36 m/s)2 = 97 J
U f = mgy f = (0.15 kg)(9.81 m/s 2 )(7.2 m) = 11 J
K f = E − U f = (97 J) − (11 J) = 86 J
vf =
2K f
2(86 J)
=
= 34 m/s
(0.15 kg)
m
Speed and Path
Energy is a scalar.
The speed of the cap
is vi at height yi and
its speed is vf at
height yf, independent
of the path between
the two heights.
Thus the angle at
which the cap is
launched does not
change this result, as
long a vi is large
enough to carry the
cap to height yf.
Question 1
When a ball of mass m is dropped
from height h, its kinetic energy just
before landing is K.
If a 2nd ball of mass 2m is dropped
from height h/2, what is its kinetic
energy just before landing?
(a) K/4
(b) K/2
(c) K
(d) 2K
m
h
(e) 4K
2m
Basic Energy Model
1. There are (at least) two kinds of energy, the
kinetic energy K associated with motion of a
particle and the potential energy U associated
with its position .
2. Kinetic energy can be transformed into
potential energy, and potential energy can be
transformed into kinetic energy.
3. Under some circumstances, the mechanical
energy Emech = K + U is a conserved quantity.
Its value at the end of a process is the same
as at the beginning. (Energy loss≈0)
Q1: Under what circumstances is Emech conserved?
Q2: What happens to the energy when Emech is not conserved?
Q3: How do you calculate U for forces other than gravity?
Example: Find the Diver’s Depth
A 95.0 kg diver steps off a diving
board and drops into the water, 3.00 m
below. At some depth d below the
water’s surface, the diver comes to
rest.
If the nonconservative work done
on the diver is Wnc = −5,120 J, what is
the depth d?
Ei = mgh + 0 = mgh
E f = mg (−d ) + 0 = −mgd
Wnc = ∆E = E f − Ei = − mgd − mgh
d = −(Wnc + mgh) / mg = 2.49 m
Example: Judging a Putt
A golfer badly misjudges a putt,
giving the ball an initial speed v1,
which sends the ball a distance d
that is only one quarter of the
distance to the hole.
1
If the nonconservative force F
due to the resistance of the grass
is constant, what initial speed v2
would have been needed to putt the
ball from its initial position to the
hole?
∆E1 = K f − K i = 0 − 12 mv12 = − Fd
∆E2 = − mv22 = − F (4d )
∴ v2 = 2v1
1
2
Example: Landing with a Thud
A block of mass m1 = 2.40 kg is
on a horizontal table with a
coefficient of friction µk = 0.450
between them and is connected to
a hanging block of mass m2 = 1.80 kg
as shown. When the blocks are
released, they move a distance d =
0.50 m, and then m2 hits the floor.
Find the speed of the blocks
just before m2 hits.
U i = m1 gh + m2 gd ; K i = 0; Ei = m1 gh + m2 gd
∆E = E f − Ei =
1
(
2
m1 + m2 )v 2 − m2 gd
U f = m1 gh + m2 g (0); K f = 12 m1v 2 + 12 m2 v 2 ; E f = m1 gh + 12 m1v 2 + 12 m2 v 2
Wnc = − f k d = − µk m1 gd ; ∆E = Wnc ⇒
1
(
2
m1 + m2 )v 2 − m2 gd = − µk m1 gd
2(9.81 m/s 2 )(0.50 m) [ (1.80 kg) − (0.45)(2.40 kg)]
2 gd ( m2 − µk m1 )
v=
=
= 1.30 m/s
m1 + m2
(2.40 kg) + (1.80 kg)
Example: Marathon Man
An 80.0 kg jogger starts
from rest and runs uphill into
a stiff breeze. At the top of
the hill the jogger has done
work Wnc1 = +18,000 J, air
resistance has done work
Wnc2 = −4420 J, and the
jogger’s speed is 3.50 m/s.
Find the height of the hill.
Ei = U i + K i = 0; E f = U f + K f = mgh + 12 mv 2
Wnc = Wnc1 + Wnc 2 = ∆E = mgh + 12 mv 2
(
)
h = Wnc1 + Wnc 2 − 12 mv 2 / mg
= (18000 J) + (−4420 J) − 12 (80.0 kg)(3.50 m/s) 2  / (80.0 kg)(9.81 m/s 2 )
= 16.780.0 m
Potential Energy Curves
and Equipotentials
The curve of a hill or a roller coaster is
itself essentially a plot of the gravitational
potential energy:
Potential Energy Curve
The potential energy U and kinetic energy K add to the total
energy E0 (dashed line) at all x values. K vanishes at A and B,
which are the turning points of the motion.
Example: A Potential Problem
A 1.60 kg object in a
conservative system moves
along the x axis, where the
potential energy is as shown.
A physical example would be
a bead sliding along a wire
shaped like the red curve.
If the object’s speed at
x = 0 is 2.30 m/s, what is its
speed at x = 2.00 m?
U i = 9.35 J; K i = 12 mvi2 = 12 (1.60 kg)(2.30 m/s) 2 = 4.23 J; Ei = E f = 13.58 J
U f = 4.15 J; K f = 12 mv 2f = E f − U f
vf =
2( E f − U f )
m
=
2 [ (13.58 J) − (4.15 J) ]
(1.60 kg)
= 3.43 m/s
Power
Power is the rate at which work is done –
(6-17)
In the SI system, the units of
power are watts:
The difference between walking
and running up these stairs is
power – the change in
gravitational potential energy is
the same.
Power
Power is a measure of the rate at which work
is done:
SI power unit: 1 J/s = 1 watt = 1 W
1 horsepower = 1 hp = 746 W
Power and Velocity
If an object is moving at a constant speed in
the presence of friction, gravity, air resistance,
and so forth, the power exerted by the driving
force can be written:
(7-13)
Power
Power is also needed for acceleration and for
moving against the force of gravity.
The average power can be written in terms of the
force and the average velocity:
(6-17)
Power
Example:
The Power of a Motor
A small motor is used to operate a lift that raises a load of
bricks weighing 500 N to a height of 10 m in 20 s at constant
speed. The lift weighs 300 N.
What is the power output of the motor?
r r
P = F ⋅ v = Fv cos φ = Fv cos ( 0 ) = Fv
P = (500 N + 300 N)(10 m/20 s) = 400 W = 0.54 hp
Energy
• Energy (W) lost or gained by any system
is determined by:
W = Pt
• Since power is measured in watts (or
joules per second) and time in seconds,
the unit of energy is the wattsecond (Ws)
or joule (J)
Energy
• The watt-second is too small a quantity for
most practical purposes, so the watt-hour
(Wh) and kilowatt-hour (kWh) are defined as
follows:
Energy (Wh) = power (W) × time (h)
power (W) × time (h)
Energy (kWh) =
1000
• The killowatt-hour meter is an instrument
used for measuring the energy supplied to a
residential or commercial user of electricity.
Applications
• Household wiring
• Most older homes, without electric heating,
have a 100 A service.
• Power is broken down into different circuits
utilizing 15 A, 20 A, 30 A and 40 A protective
breakers.
• Maximum load on each breaker should not
exceed 80% of its rating (12 A of a 15 A
circuit breaker).
Typical wattage ratings of some common
household items
Insert Table 4.1
Mechanical Advantage
…but you don’t
move very far.
If you hold the wrench
here, you need a lot of
force...
…but your hand
moves a long way.
If you hold the wrench
here, you don’t need as
much force...
Mechanical Advantage
It takes the same amount of work to turn the bolt.
You can opt for a lot of force and little distance.
W=Fxd
Or you can choose a little force but a lot of distance.
W=fxD
In many of our machines, we want to increase our
force, so we don’t mind going the extra distance.
Mechanical Advantage
Question: Which ramp would you prefer to use to
move a heavy weight to the top of the box?
Answer: This ramp requires less force, but you have
to move the weight a longer distance.
Mechanical Advantage
With a gear box,
you were able to
create a large
torque here...
…but you had to turn this handle many times.
Remember: W = τ x θ
Machines
• A machine eases the load by changing
either the magnitude to the direction of a
force to match the force to the capability of
the machine or person.
What is a Simple Machine?
• A simple machine has
few or no moving
parts.
• Simple machines
make work easier
Wheels and Axles
• The wheel and axle
are a simple machine
• The axle is a rod that
goes through the
wheel which allows
the wheel to turn
• Gears are a form of
wheels and axles
Pulleys
• Pulley are wheels and
axles with a groove
around the outside
• A pulley needs a
rope, chain or belt
around the groove to
make it do work
Inclined Planes
• An inclined plane is a
flat surface that is
higher on one end
• Inclined planes make
the work of moving
things easier
Wedges
• Two inclined planes
joined back to
back.
• Wedges are used
to split things.
Screws
• A screw is an
inclined plane
wrapped around a
shaft or cylinder.
• The inclined plane
allows the screw to
move itself when
rotated.
Levers-First Class
• In a first class lever
the fulcrum is in the
middle and the load
and effort is on either
side
• Think of a see-saw
Levers-Second Class
• In a second class
lever the fulcrum is at
the end, with the load
in the middle
• Think of a
wheelbarrow
Levers-Third Class
• In a third class lever
the fulcrum is again at
the end, but the effort
is in the middle
• Think of a pair of
tweezers
Simple Machines
• Simple Machines can
be put together in
different ways to
make complex
machinery
• Machines make work easier by changing
the size and/or the direction of the force
– In order for work to be done:
• A force must be exerted
• There must be movement in direction of the force
• The two most common forces being
overcome are friction and gravity
• There are 2 general forces involved when
using a machine to do work
– Effort (Input) Force (FE) – This is the force that
is applied to the machine
– Resistance (Output) Force (FR) – This is the
force that the machine applies
• There are also two types of work involved
– Work In (Win) – The work done on the
machine (energy put into the machine)
– Work Out (Wout) – The work done by the
machine (energy put out by the machine)
Since work is force x distance, than we can find the value
of the work in and the work out
Wout=Fr x dr
Win=Fe x de
In an “ideal machine”, (no friction) no energy would be lost
so work in and work out would be equal
Win=Wout
Fe x de = Fr x dr
In reality, work out will always be less than work in because
energy is lost through friction
Mechanical Advantage
• Most machines multiply or increase the force applied to it
so that the resistance force is greater than the effort
force (Fr>Fe)
• In order for the work in to still equal the work out (Win =
Wout), the resistance force must travel a shorter distance
than the effort force (dr<de)
• The number of times that a machine multiplies the effort
force is called the Mechanical Advantage (MA).
MA=Fr/Fe
• the Ideal Mechanical Advantage (IMA). IMA=de/dr
Machine Efficiency
• Most machines not all input work goes to output work
• The efficiency of a machine is the percentage ratio of
output to input work (value is less than 100%)
•
e = (Wout / Win ) * 100
• It can also be defined as the percentage ratio of the
meacnical advantage to the ideal mechanical advantage
•
e = (MA / IMA ) * 100
Compound Machines
• A compound machine is a
combination of simple
machines that work together.
Human Walking Machine
• A human is a combination of
simple machines that work
together.
•
•
•
•
Rigid Bar (bone)
Source of Force (muscle)
Pivot (joint)
Resistance (body or weight
lifted)
Summary of Chapter 10 and 11
• Work:
•Kinetic energy is energy of motion:
• Potential energy is energy associated with forces
that depend on the position or configuration of
objects.
•
•The net work done on an object equals the change
in its kinetic energy.
• If only conservative forces are acting, mechanical
energy is conserved.
• Power is the rate at which work is done.