### SO-Machines 2014 - Marshall Middle School

```What do Machines do?
Do they allow one to do more work?
Not really, at best they make completing a task easier.
So then what do Machines do?
• Multiply the force.
• Multiply the distance.
• Change the direction of the force.
Work = Force x Distance an object moves
while the force is applied.
W=Fxd
In SI Units:
Force is measured in newtons (N)
distance is measured in meters (m)
Work in N.m which is a joule (J).
Named after James Prescott Joule
What does work do?
Work causes a change in Energy. In
other words, it can do any of the
following:
• Make something move faster.
• Lift something up.
• Move something against friction.
• A combination of the above.
Examples of Work:
A cart is pushed to the right as illustrated.
50.0 N
distance cart is moved 4.00 m.
How much work is done on the cart?
W = F x d = (50.0 N)(4.00 m) = 200. J
Examples of Work:
W=Fxd
W = (80.0 N)(20. m)
W = 1600 J
80.0 N
How much work is done on
the box?
distance box is moved 20. m.
A box is lifted as illustrated.
Now let’s apply this to
some of our machines.
The simplest is most
likely levers.
Class 1 Lever: load on one side of the
fulcrum and the effort on the other side.
Effort
The Fulcrum is the pivot point.
The Load is what we are trying to
lift or the output of the machine.
The Effort is the force that is applied to
lift the load or the input of the machine.
Load
Fulcrum
Class 1 Lever: Load on one side of the
fulcrum and the Effort on the other side.
Effort
Load
Fulcrum
Class 1 Lever: More terminology
Note that in lifting the load the Effort moved
much farther than the Load.
Effort
With a smaller Effort we could lift a
Load that is heavier.
Load
Fulcrum
Class 1 Lever: More terminology
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N The Effort moved 60 cm
while the Load moved only
20 cm. We moved 3 times
farther than the LOAD.
Load distance, dL = 20 cm
With an Effort of 300 N we were able to lift a Load of
900 N. We multiplied the input force by 3.
Class 1 Lever: More terminology
Work, W = F x d
WIN = E x dE = (300 N)(0.60 m) = 180 J
Fulcrum
Load = 900N
Note: WorkIN = WorkOUT
We didn’t do more work,
we just did it with less
Effort than if I tried to lift
it without the lever.
Load distance, dL = 20 cm
Effort distance, dE = 60 cm
WOUT = L x dL = (900 N)(0.20 m) = 180 J
Effort = 300 N
Class 1 Lever: More terminology
MA = Load/Effort
MA = (900 N)/(300 N)
MA = 3
MA = dE/dL
MA = (60 cm)/(20 cm)
MA = 3
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N
Load distance, dL = 20 cm
We say that we have a Mechanical Advantage, MA.
We can lift 3 times more than our input Effort.
Class 1 Lever: More terminology
MA = Load/Effort
MA = (900 N)/(300 N)
MA = 3
MA = dE/dL
MA = (60 cm)/(20 cm)
MA = 3
Fulcrum
Load = 900N
Effort distance, dE = 60 cm
Effort = 300 N
Load distance, dL = 20 cm
We triple our Effort (input force) at the expense of
moving the Load ⅓ as much.
Class 1 Lever: More terminology
We can also analyze the lever by measuring the
distance from the Effort to the fulcrum (pivot point)
and the distance from the Load to the fulcrum.
This is called the lever arm or just arm and is often
given the variable name “x.”
Effort arm, xE = 3.0 m
Load arm, xL = 1.0 m
Fulcrum
Class 1 Lever: More terminology
This can also be used to calculate the
Mechanical Advantage.
MA = xE/xL = (3.00 m)/(1.00 m) = 3
Looks familiar doesn’t it.
Effort arm, xE = 3.0 m
Load arm, xL = 1.0 m
Fulcrum
More terminology:
Often we use the terms, Ideal Mechanical
Advantage, IMA and Actual Mechanical
Advantage, AMA
IMA = xE/xL or dE/dL
AMA = L/E with the load being just what you
ultimately wanted to move, excluding anything
else that may have to be moved with it.
This will become clearer when we look at a 2nd class lever.
More terminology:
Often we use the terms, Ideal Mechanical
Advantage, IMA and Actual Mechanical
Advantage, AMA
We want to find out how well the particular
machine does its work. This is called
Efficiency, Eff.
Efficiency, Eff = (AMA/IMA) x 100%
2nd class lever:
Notice that the Effort and the Load
are on the same side of the Fulcrum
and the Load is between the Effort
and the Fulcrum.
Effort
Load
Fulcrum
2nd class lever:
Again the Effort moves much farther than the Load. We
are getting more force out than what we put in, but the
load only moves a short distance. We are also lifting
the lever along with the load.
Fulcrum
2nd class lever:
IMA = xE/xL = (2.8 m)/(0.35 m) = 8
The load is 900 N and since the Effort has
to lift the Load and the lever, let’s say that
the Effort is 150 N.
Effort
Effort arm, xE = 2.8 m
Load arm, xL = 0.35 m
Load
Fulcrum
2nd class lever:
IMA = xE/xL = (2.8 m)/(0.35 m) = 8
AMA = L/E = (900 N)/(150 N) = 6
Eff = AMA/IMA = 6/8 x 100% = 75%
E = 150N
L = 900 N
Effort arm, xE = 2.8 m
Load arm, xL = 0.35 m
Fulcrum
3rd class lever:
Notice that the Effort and the Load
are on the same side of the Fulcrum
and the Effort is between the Load
and the Fulcrum.
Load
Effort
Fulcrum
3rd class lever:
IMA = xE/xL = (0.50 m)/(2.0 m) = ¼ = 0.25
AMA = L/E = (200 N)/(1000 N) = ⅕ = 0.2
Eff = AMA/IMA = (0.2/0.25) x 100% = 80%
L = 200 N
E = 1000 N
xL = 2.0 m
xE = 50. cm
Fulcrum
There is a lab part of the competition.
Let’s look at some of the basic concepts
for a lever that is in static equilibrium.
The easiest lever to analyze is the first class
lever (seesaw), that is balanced by itself.
The center of gravity of the lever is on the
fulcrum.
c.g.
If a lever is not moving (rotating) then it is said to
be at static equilibrium. When an object is at
static equilibrium the following is true:
ΣF = 0, that is netF = 0, no unbalanced forces.
Στ = 0, that is there are no unbalanced torques.
If you place a seesaw so that its center of gravity
is on the fulcrum, it will balance. That is, the left
side balances the right side.
Torque is the tendency of a force to
cause an object to rotate around an
axis. In the case of a lever, the axis is
the fulcrum.
Force
In this case the force would make the left
side of the lever go down or rotate the
lever counterclockwise, ccw.
Torque is the tendency of a force to
cause an object to rotate around an
axis. In the case of a lever, the axis is
the fulcrum.
Force
In this case the force would make the left
side of the lever go down or rotate the
lever counterclockwise, CCW.
What if the force is 24 N, what torque is applied?
Earlier we talked about the lever arm or arm being
the distance from the fulcrum (axis) to the force. We
will use the letter “x” as the symbol for lever arm.
F = 24 N
The symbol for torque is the Greek letter tau, τ
x = 1.2 m
τ = (F)(x) = (24 N)(1.2 m)
τ = 28.8 N.m = 29 N.m
A torque of 29 N.m will rotate the lever CCW.
The weight of the seesaw on the left creates a torque
that tries to rotate the seesaw counter-clockwise, CCW,
so that the left side would go down.
The weight of the seesaw on the right creates a torque
that tries to make it rotate clockwise,CW, so that the right
side would go down.
F
F
The two balance each other
and it does not rotate.
Another way to look at this is that we can place all
the weight of the seesaw ( FL ) at its center of
gravity.
The center of gravity of the seesaw is at the axis of
rotation (fulcrum) so the value of the lever arm is
zero and the force creates no torque.
c.g.
Note: The center of
gravity may not be at
the geometric center.
FL
Especially when
using wooden
meter sticks!
Sample problem: Two identical 40.0 kg twin girls
are sitting on opposite ends of a seesaw that is 4.0
m long and weighs 700 N, so that the center of
gravity of the seesaw is on the fulcrum.
How do we analyze this situation?
c.g.
First, we need to draw a torque diagram of the seesaw.
This is a free body diagram which includes the lever arms.
FN = 1500 N
Next, we define the axis of
rotation (circle with a dot
in the middle) and the
lever arms.
We place all the forces at
their proper location.
x2 = 2.0 m
x1 = 2.0 m
c.g.
.
F1 = 400 N
FL = 700 N
F2 = 400 N
Στ = 0 or ΣτCCW = ΣτCW
F1x1 = F2x2
FN = 1500 N
(400 N)(2.0 m) = (400 N)(2.0 m)
FL and FN both act through
the axis of rotation, so their
lever arm is zero, making
their torque 0.
The torques balance so the
seesaw can be in static
equilibrium.
800 Nm = 800 Nm
x2 = 2.0 m
x1 = 2.0 m
c.g.
.
F1 = 400 N
FL = 700 N
F2 = 400 N
It is important that the seesaw be
level, so that the force applied by each
of the girls is acting downward and is
perpendicular to the lever arm. If the
Force and the Lever Arm are not
perpendicular, then the equation for
the torque becomes complex. It is
better that we avoid that situation.
So, what do you do to balance the seesaw if the two
people are not the same weight (mass)?
One 400 N girl sits on one end of a seesaw that is
centered on the fulcrum, is 4.0 m long, and weighs 700 N.
Where must her 650 N brother sit in order for the seesaw
to be in static equilibrium?
?
c.g.
Option #1, move the
heavier person closer to
the fulcrum.
FL and FN both act through
the axis of rotation, so their
lever arms are zero.
Στ = 0 or ΣτCCW = ΣτCW
FGxG = FBxB
FN = 1750 N
xB = (800 Nm)/(650 N)
(400 N)(2.0 m) = (650 N)xB
xB = 1.23 m
800 = 650x
B
xG = 2.0 m
xB = ?? m
c.g.
.
FG = 400 N
FL = 700 N
FB = 650 N
Option #2, move the center of gravity of the seesaw so that
more of the seesaw is on the side of the lighter person,
One 400 N girl sits on one end of a 4.0 m long seesaw
weighing 700 N That has moved the center of gravity of
the lever 0.2 meters towards her. Where must her 650 N
brother sit in order for the seesaw to be in static
equilibrium?
?
c.g.
Now the weight of the
seesaw creates a torque
helping the girl.
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FGxG + FLxL = FBxB
FN = 1750 N
(400)(2.2) + (700)(0.2) = (650 N)xB
xB = (1020 Nm)/(650 N)
880 + 140 = 650xB
xB = 1.57 m
xG = 2.2 m
xB = ?? m
c.g.
.
xL = 0.2 m
FG = 400 N
FL = 700 N
FB = 650 N
For the Middle School
(Division B) Competition,
you will need to build a
simple first class lever
system. The lever may not
be longer than 1.00 meter.
Simple Machines. (Simple case.)
Given small mass placed on one side.
Given unknown large mass on the other.
Unless the values are too extreme, you may be able to
move the large mass close enough to the fulcrum.
c.g.
If this is the setup, you
don’t have to worry about
the weight of the lever.
Simple Machines. (Simple case.)
c.g.
Big Mass
Small Mass
FN
xS
xB
c.g.
.
FS
FL
FB
Simple Machines. (Simple case.)
In this case the torque equation is: τCCW = τCW
(FS)(xS) = (FB)(xB) and you can solve for any value.
FN
xS
xB
c.g.
.
FS
FL
FB
So far we have been dealing with the force applied by the
hanging mass. This force is known as the weight of the
object or the force of gravity (Fg) acting on the object.
The force of gravity acting on an object is the product of
the mass of the object multiplied by the gravity constant
on the planet Earth (9.8 N/kg).
Fg = mg = m(9.8 N/kg) so, mass, m = Fg/(9.8 N/kg)
This gets quite confusing because weight is measured in
newtons and mass is measured in grams or kilograms (kg).
You may have been told to weigh something but you
actually measured its mass in grams.
Simple Machines. (Simple case.)
Knowing that Fg = mg
(FS)(xS) = (FB)(xB) This equation can be written:
(msg)(xS) = (mBg)(xB) Dividing by g we get:
(msg)(xS)/g = (mBg)(xB)/g
FN
xS
(ms)(xS) = (mB)(xB)
xB
c.g.
.
FS = mSg
FL
F B = m Bg
Simple Machines. (Simple case.)
We can now solve for a mass using this equation
and modify our torque diagram as shown:
(ms)(xS) = (mB)(xB)
FN
xS
xB
c.g.
.
mS
FL
mB
Simple Machines. (Simple case.)
Suppose that you were given a small mass of 125
grams and an unknown large mass. You set up
your lever so it balances as shown:
FN
xB = 10.0 cm
xS = 47.6 cm
c.g.
.
mS = 125 g
FL
mB = ??
Simple Machines. (Simple case.)
(ms)(xS) = (mB)(xB)
(125 g)(47.6 cm) = (mB)(10.0 cm)
mB = (125 g)(47.6 cm)/(10.0 cm) = 595 grams
FN
xB = 10.0 cm
xS = 47.6 cm
c.g.
.
mS = 125 g
FL
mB = ??
Simple Machines (More realistic case)
If the difference between the unknown mass and
the known mass is large, move the fulcrum near
one end of the lever.
Place the unknown large mass on the short side.
Then place the small mass on the long side.
c.g.
Now the lever helps
balance the large weight.
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FSxS + FLxL = FBxB
FN
xB = cm
xS = cm
c.g.
.
xL = cm
FS =
N
FL = N
FB = ?? N
FN acts through the axis of
rotation, so its lever arm is zero.
Στ = 0 or ΣτCCW = ΣτCW
FSxS + FLxL = FBxB
FN
(mSg)xS + (mLg)xL = (mBg)xB
If you divide through by “g” you get:
xS = cm
xB = cm
c.g.
.
mSxS + mLxL = mBxB
FS = mSg
FL = mLg
xL = cm
FB = mBg
Sample: Suppose that you have set up
your 1.00 meter long lever of mass 83.4
grams so that the center of gravity is
FN
20.0 cm from the fulcrum. You have also
determined that the big unknown mass
will be placed 10.0 cm from the fulcrum.
xB = 10.0 cm
xS = cm
c.g.
.
xL = 20.0 cm
FS = mSg
FL = mLg
FB = mBg
You are given a known mass of 76.2
grams and a big mass. Putting the
masses on the lever, you find that it
balances when the little mass is 62.8
centimeters from the fulcrum.
FN
xB = 10.0 cm
xS = 62.8 cm
c.g.
.
xL = 20.0 cm
mS = 76.2 g
mL = 83.4 g
mB = ??
mSxS + mLxL = mBxB
(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm) = mB(10.0 cm)
FN
mB = [(76.2 g)(62.8 cm) + (83.4 g)(20.0 cm)]/(10.0 cm)
mB = 645 grams
xS = 62.8 cm
xB = 10.0 cm
c.g.
.
xL = 20.0 cm
mS = 76.2 g
mL = 83.4 g
mB = ??
As long as the levers are horizontal
and in static equilibrium, you can
use the equations with mass
instead of force or weight. Your
Physics teacher probably will not
be too happy, but the equation is
mathematically correct for the
situation.
c.g.
c.g.
B
First Class Lever
Second Class Lever
For the High School Competition you will need to
build a compound lever system made up of a first
class lever connected to a second class lever. Each
lever may not be longer than 50.0 cm.
c.g.
First part of lever system
mSxS + mL1xL1 = ExE1
xE1 = cm
xS = cm
c.g.
.
xL1 = cm
mS =
g
mL1 =
g
E = ?? g
Second part of lever system
c.g.
ExE2 = mL2xL2 + mBxB
B
xL2 = cm
xE2 = cm
c.g.
.
E=
g
mL2 =
g
xB = cm
mB = ?? g
Sample:
c.g.
xE1 = 5.0 cm
xE2 = 35.0 cm
c.g.
xB = 10.0 cm
B
Suppose that you built your lever system so that
the fulcrum in the class 1 lever was 5.0 cm from
the string connecting the levers.
Also suppose that in the second class lever the
string connecting the levers is 35.0 cm from its
fulcrum and you set up the lever so that the
unknown big mass is 10.0 cm from the fulcrum.
Sample
xE1 = 5.0 cm
c.g.
xE2 = 35.0 cm
c.g.
xS = 32.7 cm
xB = 10.0 cm
B
You are given a small mass of 86.0 grams and an
unknown large mass that you place at the 10.0
cm mark.
You slide the small mass along the class 1 lever
and manage to get the levers to balance when
the small mass is 32.7 cm from its fulcrum.
xE1 = 5.0 cm
c.g.
xS = 32.7 cm
First part of lever system
You also measured the mass of the
lever as 27.4 g and arranged the
lever so that its center of gravity is
16.0 cm from the fulcrum.
xE1 = 5.0 cm
xS = 32.7 cm
c.g.
.
xL1 = 16.0 cm
mS = 86.0 g
mL1 = 27.4 g
mSxS + mL1xL1 = ExE1
E = ?? g
First part of lever system
mSxS + mL1xL1 = ExE1
(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm) = E(5.0 cm)
E = [(86.0 g)(32.7 cm) + (27.4 g)(16.0 cm)]/(5.0 cm)
E = 650 g
xE1 = 5.0 cm
x = 32.7 cm
S
c.g.
.
xL1 = 16.0 cm
mS = 86.0 g
mL1 = 27.4 g
E = ?? g
xE2 = 35.0 cm
Suppose the mass of the lever is
31.4 g and its center of gravity is
18.0 cm from the fulcrum.
c.g.
xB = 10.0 cm
Second part of lever system
B
xE2 = 35.0 cm
xL2 = 18.0 cm
c.g.
.
xB = 10.0 cm
E = 650 g
mL2 = 31.4 g
ExE2 = mL2xL2 + mBxB
mB = ?? g
Second part of lever system
ExE2 = mL2xL2 + mBxB
(650 g)(35.0 cm) = (31.4 g)(18.0 cm) + mB(10.0 cm)
mB = [(650 g)(35.0 cm) - (31.4 g)(18.0 cm)]/(10.0 cm)
mB = 2218 g = 2.218 kg
xL2 = 18.0 cm
xE2 = 35.0 cm
c.g.
.
xB = 10.0 cm
E = 650 g
mL2 = 31.4 g
mB = ?? g
Things to note:
• You must build your own lever system.
• You may want to have two set places to have
your fulcrum depending on the given masses.
• You may want to have the unknown mass at a
predetermined spot and thus notching the lever
at that point.
• Make sure that you know the mass of your lever
and have marked the location of its center of
gravity.
Fixed Pulley
A fixed pulley is basically a First
Class Lever that can rotate. The
mechanical advantage is 1. All a
fixed pulley does is change the
direction of the force.
Effort
Load
Fulcrum
Effort
Load
Movable Pulley
A movable pulley is basically a
Second Class Lever that can rotate.
The mechanical advantage is 2.
Effort
Effort
Load
Load
Fulcrum
Block & Tackle
A Block and Tackle is a combination
of a movable pulley connected to a
fixed pulley. In this case the
mechanical advantage of the
movable pulley is 2 and the MA of
the fixed pulley is 1. Combined the
mechanical advantage is 2.
In order to calculate the Ideal
Mechanical Advantage, IMA, of a
Block and Tackle, you count the
number of supporting ropes.
Effort
Load
Block & Tackle
A Block and Tackle is a combination
of a movable pulley connected to a
fixed pulley. In this case the
mechanical advantage of the
movable pulley is 2 and the MA of
the fixed pulley is 1. Combined the
mechanical advantage is 2.
Effort
Load
Work is done to lift the Load, but you also
must lift the movable pulley with it. The
AMA will be less, not only because of
friction in the system but because the
weight of the movable pulley also has to
be lifted by the Effort.
dE
Wheel and Axle
dL
A Wheel and Axle is two different
diameter cylinders on the same shaft. This
also is a first class lever that can rotate.
The Ideal Mechanical Advantage is the
ratio of the diameters. IMA = dE/dL.
Load
Effort
Effort
Load
Fulcrum
All these are examples of a wheel and Axle.
Inclined Plane
x
h
An inclined plane is basically a ramp that is stationary.
A load is pushed up the ramp instead of being lifted
straight up. The Ideal Mechanical Advantage of a
ramp is the ratio of the length of the ramp (x) to the
height of the ramp (h). IMA = x/h
Wedge
A wedge is like an inclined plane, but instead of being
stationary the wedge is driven into something or
between things.
All these are examples of wedges.
LEVERS.
Class 1 Levers.
Class 1 Levers. One of these can also be used
as a class 2 Lever. Which one? How?
COMPOUND
MACHINES.
This is actually a compound machine.
The teeth are wedges that cut into the wood.
The whole blade is a wheel and axle.
These too are compound machines.
The teeth are wedges that cut the wires.
The handles make a class 1 lever.
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