Physics
PHS 5043 Forces & Energy
Machines
Machine: Device or set of devices
used to accomplish a particular task
Machines are used to:
 Make our work easier
 Multiply force
 Multiply speed
 Change direction of force
PHS 5043 Forces & Energy
Machines
Machines can be:
 Simple
 Compound
(combination of simple machines)
Mechanical advantage (simple): Efficiency
of a simple machine
Mechanical advantage (compound): Product
of mechanical advantage of each simple
machine
PHS 5043 Forces & Energy
Machines
Simple machines:
 Levers
 Wheel & axle
 Winch
 Pulley
 Inclined plane
PHS 5043 Forces & Energy
Machines
Levers: Rigid bar that pivots around a fulcrum
PHS 5043 Forces & Energy
Machines
Law of levers:
Fe le = Fr lr
Mechanical advantage of
levers:
MA = Fr / Fe
Fe: Effort force (N)
Fr: Resistance force (N)
le: Effort arm (m). Distance
between effort and fulcrum
lr: Resistance arm (m). Distance
between load and fulcrum
or
MA = le / lr
PHS 5043 Forces & Energy
Machines
Class 1 levers:
_Fe and Fr point in the same direction
_le is larger than lr
_Fe and Fr on opposite sides of fulcrum
_MA > 1
PHS 5043 Forces & Energy
Machines
Practice:
You place a child
(m = 30 kg) on a
seesaw, 1.5 m
away from the
fulcrum. Where
should you sit so
that the seesaw is
perfectly balance,
if your mass is 60
kg?
F e le = Fr lr
le = Fr lr / F e
le = Fg lr / F’g
le = mg lr / m’g
le = (30 kg)(9.8 N/kg)(1.5 m)/(60 kg)(9.8 N/kg)
le = 0.75 m (75 cm)
PHS 5043 Forces & Energy
Machines
Practice:
How far would
you sit for the
seesaw to have a
mechanical
advantage of 2?
What would then
be the value of the
effort force?
MA = le / lr
le = MA lr
le = 2 (1.5 m)
le = 3 m
MA = Fr / Fe
Fe = Fr / MA
Fe = Fg / MA
Fe = mg / MA
Fe = (30 kg)(9.8 N/kg) / 2
Fe = 147 N
PHS 5043 Forces & Energy
Machines
Class 2 levers:
_Fe and Fr point in opposite directions
_le is larger than lr
_Fe and Fr on same sides of fulcrum
_MA > 1
PHS 5043 Forces & Energy
Machines
Practice:
You transport a 100 kg
load of sand in a
wheelbarrow. If the
effort arm extends 1.5
meters away from the
fulcrum, and the
resistance arm is fifty
centimeters long:
a) How much force
must you apply to lift
the load?
b) What is the
mechanical advantage
of this machine?
Fe l e = Fr lr
Fe = Fr lr / l e
Fe = Fg lr / l e
Fe = mg lr / l e
Fe = (100 kg)(9.8 N/kg)(0.5 m) / (1.5 m)
Fe = 327 N
MA = Fr / Fe
MA = 980 N / 327 N
MA = 3
PHS 5043 Forces & Energy
Machines
Class 3 levers:
_Fe and Fr point in opposite directions
_lr is larger than le
_Fe and Fr on same sides of fulcrum
_MA < 1
PHS 5043 Forces & Energy
Machines MA =
Practice:
Suppose that your
forearm measures 30
cm, and your bicep
muscle is attached 4
cm from your elbow
(fulcrum).
a) What is the MA
of your forearm?
b) What force must
the bicep muscle
exert to lift 1 kg?
MA = le / lr
MA = 4 cm / 30 cm
MA = 0.13
MA = Fr / Fe
MA = Fg / Fe
MA = mg / Fe
Fe = mg / MA
Fe = (1 kg) (9.8 N/kg) / 0.13
Fe = 75 N
PHS 5043 Forces & Energy
Machines
The wheel & axle:
_Simple machine
_Similar to first class lever
_Fe applied to the wheel’s
circumference
_Fr applied on the axle’s
circumference
_Fulcrum is the center of the axle
_le: radius of wheel (R)
_lr: radius of axle (r)
_MA > 1 (MA = Fr/Fe = le/lr = R/r)
PHS 5043 Forces & Energy
Machines
The winch:
_Simple machine
_Family of wheel & axle
_Similar to first class lever
_Fe applied to the handle
_Fr applied on the cylinder
_Fulcrum is the center of the
cylinder
_le: length of handle (R)
_lr: radius of cylinder (r)
_MA > 1 (MA = Fr/Fe = le/lr = R/r)
(see Fig. 6.20, page 6.23)
PHS 5043 Forces & Energy
Machines MA =
Practice:
A winch has a
cylinder with
diameter 15 cm and
handle measuring
30 cm
a) What is the MA
of this machine?
b) What force
must be exerted
to lift 10 kg?
MA = le / lr = R / r
MA = 30 cm / 7.5 cm
MA = 4
MA = Fr / Fe
MA = Fg / Fe
MA = mg / Fe
Fe = mg / MA
Fe = (10 kg) (9.8 N/kg) / 4
Fe = 24.5 N
PHS 5043 Forces & Energy
Machines
The pulley:
_Simple machine
_Used to change direction of force
_Used to reduce friction
_Suspended pulley has no MA (MA = 1)
_Movable pulley’s MA = # ropes around
movable pulley
_Tension is the same at all points of rope
_Fr weight applied on the pulley’s rope
_Fe tension on the pulley’s rope
_Fulcrum is the center of the pulley
_MA > or = 1 (MA = Fr / Fe = Fg / T)
PHS 5043 Forces & Energy
MA =
Machines
Practice:
According to the pulley system below:
a) What is the MA of the system?
b) What mass must be suspended from
the fixed pulley for the system to be in
equilibrium?
c)What would then be the value of the
effort force?
MA = (MA)f (MA)m
MA = 1 * 2
MA = 2
MA = Fr / Fe
MA = Fg / T
MA = mg / m’g
MA = m / m’
m’ = m / MA
m’ = 10 kg / 2
m’ = 5 kg
Fe = T
T = mg
T = (5 kg) (9.8 N/kg)
T = Fe = 49 N
PHS 5043 Forces & Energy
Machines
The inclined plane:
_Simple machine
_MA = Fr / Fe
_MA = (1 / sin θ)
_MA = l / h
_l: length of plane (m)
_h: height of plane (m)
*The smaller the angle of the
plane, the greater its
mechanical advantage
sin θ = h / l
PHS 5043 Forces & Energy
Machines
Practice:
A builder wants to
install an access ramp
with mechanical
advantage of 8, what
should be the length if it
is designed to reach a
level of 75 cm high?
What would be the
angle of said ramp?
MA = l / h
l = MA * h
l = 8 * 0.75 m
l = 6m
MA = 1 / sin θ
sin θ = 1 / MA
sin θ = 1 / 8
θ = 7.2°
PHS 5043 Forces & Energy
Machines
Compound machines:
_Combination of simple machines
_MA = Product of all MA’s
_MA = Fr / Fe
PHS 5043 Forces & Energy
Machines
Practice:
A compound machine
consists of a class 2 lever
(lr = 30 cm, le = 90 cm) and
a winch (R = 20 cm,
r = 10 cm) designed to lift
an object of mass 100 kg.
a) What is the MA of this
device?
b) What is the value of Fe
at the end of the handle?
MA = (MA)l *(MA) w
MA = (le/lr)(R/r)
MA = (90 cm / 30 cm)(20 cm / 10 cm)
MA = (3)(2)
MA = 6
MA = Fr / Fe
Fe = Fr / MA
Fe = Fg / MA
Fe = mg / MA
Fe = (100 kg)(9.8 N/kg) / 6
Fe = 163N