Download Mass Flow

Mass and Force
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Mass
• SI unit kg
• Weight is a force, unit N
• Mass is involved in multiple
engineering principles!
What is the mass of your calculator?
How about its weight on Earth?
And on Mars?
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Density
• Density=mass/volume
•
•
m

V
SI units Kg/m3
US units slugs/ft3 [ρ] = M/L3
• Important in e.g. material selection
• Note: Density can change due to e.g. temperature and pressure
Which one of these three materials would you choose for
a part (size 0.001m3) of an airplane interior decor?
How about as a counter weight for an elevator?
Material
Wood (oak)
Density (kg/m3)
750
Cement
1920
Aluminum
2740
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Lava Lamp Example
How does the lava lamp work?
http://www.youtube.com/watch?v=DL3Ez9bxMTo&feature=related
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Mass Flow
• Mass flow = mass/time
• m’ = dm/dt = Δm/Δt
• m’ units Kg/sec (SI) slugs/sec (US)
• Volume flow = volume/time
• V’ = dV/dt = ΔV/Δt
• V’ units m3/sec ft3/sec
• m’ = dm/dt = ρdV/dt = ρV’
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
• M’
m’
• V’
v’
• For an incompressible fluid, M’ = m’
• V’ = v’
• Therefore velocity is greater in narrow
pipe and slower in fat pipe.
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Volume or mass flow?
• Depends on application
• Volume flow preferred
– when filling a tank of a specific volume with liquids
of different densities
– when a process can accept only a limited volume at
a time
• Mass flow preferred
– in chemical reactions, where the number of
reactant molecules (mass) is important
– when measuring gas flow
– When goods sold based on weight
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Mass Flow - Task
Estimate the mass flow of a gas
pump. Density of regular gasoline
is 720 kg/m3.
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Mass Moment of Inertia
• Measure of how hard it is to
rotate something with
respect to the center of
rotation, or resistance to
rate of change of rotation
• For a single mass
particle: I z  z  r 2 m
• For a system of mass
2
particles: I z  z   ri mi
z-axes
r
m
m2
r2
i
EGR 102 03/24/2009
I z  z   r 2 dm
© Katja Hölttä-Otto 2009
Mass moment of inertia - example
z-axis
R=5 cm
h1=30 cm
Which one of the following
object is harder to rotate
around the z-axes? Both are
made of steel (=7860 kg/m3).
h1=4 cm
Ø=20 cm
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
How are these related to mass
moment of inertia?
flywheel
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Momentum
• Momentum p (or L)
• p = mv
L = mv
• Momentum is directional
Velocity (a component of momentum) is
directional
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Momentum - Example
www.aerospaceweb.org/question/investigations/columbia/foam-impact.jpg
Investigators into the Columbia accident have estimated that the dislodged
foam was about 48 x 29 x 14 cm (19 x 11.5 x 5.5 in) , weighed about 0.75 kg
(1.7 lb) and impacted the Shuttle at nearly 850 km/h (530 mph). For the sake
of a rough comparison, this block of foam would be about the same size and
weight as a large loaf of bread. (www.aerospaceweb.org/question/investigations/q0131.shtml)
p (or L) = mv = 0.75kg * 850,000m/3600s = 177 kg m/s
Same momentum as a 5 kg (11 lb) brick hitting you driving 127.5km/h (80mph) !
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Momentum - Task
Which has greater momentum?
A)An Olympic 100m runner at
speed 10 m/s
B)A 1000kg car pulling out of a
parking lot at 2 km/h
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Mass and Weight
• Mass - scalar (SI unit kg)
• Weight – vector, it’s force (SI unit N)
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Basics
• Force is the interaction of two objects, typically
one pushes or pulls the other
– Direct contact: you pulling a door open
– No direct contact: gravity pulling you toward the
center of the earth
• Force causes objects to move, lengthen,
shorten, twist, bend, etc.
• SI Unit: Newton [F]=N
– F=ma  N=kg·m/s2
• U.S. Customary unit: pound force lbf
– F=ma  lbf = 1slug·ft/s2 (1lbf=4.448 N)
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Applications
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Force is a vector quantity(on whiteboard)
F1
EGR 102 03/24/2009
F2
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Spring Forces
• Springs widely used in
engineering
– Store energy
– Return to original position
– Dampen vibration
www.motorsportscenter.com/uploads/suspension.jpg
• Spring types:
– Linear, torsional
www.pleasanthillgrain.com/bag_clip_bag_clips_stainless.asp
www.pharma-pen.com
http://rclsgi.eng.ohio-state.edu/~gnwashin/me481/mech_sys.html
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Hooke’s law
F  kx
F = applied force (N)
k = spring constant (N/mm, N/m)
x = deformation of the spring (mm, m)
• Applicable in the elastic range of the spring
– Elastic means there is no permanent deformation
after the force is removed
x
EGR 102 03/24/2009
F
© Katja Hölttä-Otto 2009
Whiteboard example
• A compression spring is 10 cm long
when no force is applied. When a force is
applied, the deformed length is 8 cm.
The spring has a spring constant of 10
N/m. Calculated the applied force.
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Determining the spring constant
F  kx
F = applied force (N)
k = spring constant (N/mm, N/m)
x = deformation of the spring (mm, m)
• In-class task: determine the spring
constant of one of the scales in the back
of the room
• Plot your data in Excel, explain all the
steps you take
• Prepare to present in front of the class
• See EF example 10.1 for help
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Homework and Teardown
• Homework – due Thursday 04/01 before the class
– From the course book: 9.5 (10p), 9.6 (less than 1 page
typed)(20p), 9.12 (10p), and 9.23 (10p)
• Look for (broken) products to take apart later
– good product will replace one bad assignment grade
– Bring products early!
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Friction Forces
• “Frictionless” systems,
commonly used in
physics, do not really
exist
• Friction can be useful
• Types of friction:
– Dry friction
www.garageboy.com
F friction  N
• Static friction
• Dynamic (kinetic) friction
– Viscous friction (fluid
friction)
EGR 102 03/24/2009
www.respo.net/respo_school/respo_school_006/pics
/pour_oil_01.jpg
© Katja Hölttä-Otto 2009
Friction force (N)
Applied force and friction
Maximum static friction force
Dynamic friction force
Applied force (N)
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Whiteboard Example
• The static coefficient of friction between
an object and a horizontal surface is
0.85. The object’s mass is 0.550 kg. If
the object is pushed on the surface (force
horizontal) of 5N, will the object move?
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Friction In-Class Task
How would you calculate the static
coefficient of friction of your
calculator starting to slide on your
course book?
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Free body diagrams
• Free body diagram shows all external forces
acting on the body.
• Commonly used in statics, dynamics, and
mechanics of materials
• Steps to draw the free body diagram
– Make a simplified drawing of the body in question
– Draw all force vectors acting on it
• Do not forget weight, unless gravitational forces are
ignored
– Label all forces
– Define fore coordinate system
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Practice
(white board)
• Steps to draw the free body diagram
– Make a simplified drawing of the body in question
– Draw all force vectors acting on it
• Do not forget weight, unless gravitational forces are ignored
– Label all forces
– Define fore coordinate system
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
In-class task
• Draw a free body diagram for the two
pipes in a v-shaped channel.
40
EGR 102 03/24/2009
20
© Katja Hölttä-Otto 2009
Newton’s Laws
I.
Every object in a state
of uniform motion
tends to remain in
that state of motion
unless an external
force
is
applied
to
it


F  ma
II.
III. For every action there
is an equal and
opposite reaction
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Newton’s Laws
I.
Every object in a state of uniform
motion tends to remain in that state
of motion unless an external force
is applied to it
This also applies
Remember what
happened to the
cannon ball in both
x and y-directions?
11m
v0
a
to an object in
rest – they will not
move unless acted
upon by an
unbalanced force
vy0
vx0
1.5m
a
EGR 102 03/24/2009
100m
© Katja Hölttä-Otto 2009
Newton’s Laws


d
v

II. F  ma , F  m
dt

a
I.
Notice the relation
between the
magnitudes and
directions of F and a!
m
EGR 102 03/24/2009

F
© Katja Hölttä-Otto 2009
Newton’s Laws


d
v

II. F  ma , F  m
dt

a

F2
I.
m

F1
  
F  F1  F2  ma
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Newton’s Laws
I.
II.
S
s
III. For every action there
is an equal and
opposite reaction
N
N  mg
N mg
m
mg
EGR 102 03/24/2009
Both the magnitude
and direction of the
two forces are equal
Why are the
absolute value
signs on N and g,
not mg or m?
© Katja Hölttä-Otto 2009
Force Basics - revisited
• Force is the interaction of two objects,
typically one pushes or pulls the other
– Direct contact: you pulling a door (from the
handle) open  the door pulling the handle
so it does not come off
– No direct contact: gravity pulling you toward
the center of the earth  the surface of the
earth (pavement?) pushing you so you do not
sink in the earth
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Team Assignment – Due Tue 4/7
8:00 am
• Problem/Design project:
– Design a mass-spring system that can be taken to
Mars to measure the acceleration due to gravity at the
surface of Mars.
• Explain the basis of your design
– The governing equations & law’s of physics and how they relate to your design
– Decisions on materials, components, attachment methods or working principle
– Decisions that relate to the ability to take it to Mars
• Include a drawing of your design
– Include rough dimensions
– Include a parts list. The level of detail can be “spring, glue, screw, metal plate”, so no
need to find the actual part numbers and exact materials for the components.
• Explain how your design should be calibrated and used
• No need to build the system.
• Hand in a report including equations and figures. The length can be
anything from 2-4 pages typed. The length will not be graded. Only
content is graded. Max 50p.
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Assignment
• Individual assignment (=homework)
• Due Thu 4/9 8:00am
• Problems:
– 10.11, 10.15, 10.19, 10.21, 10.22 (10p each)
• Follow format in course book EF chapter 4
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Reminders
• Look for (broken) products to take apart later
– good product will replace one bad assignment grade
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Newton’s Law - Application
• Equilibrium of forces and moments
F
0
x ,i
F
y ,i
M
0
i
i
O ,i
0
i
System:
F
Free Body Diagram:
F
FA
FBY
FBX
If F is 50N, what is FA+ FB, why?
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Whiteboard example
F  3i  4 j
System:
j
i
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Whiteboard example 2
You are pushing a lawn mower with force of 703N. The lawn mower
weighs 30 kg. The mower is not moving in the y-direction. What is
the acceleration of the mower on the grass? The friction coefficient
between the wheels and the grass is 0.60. At what angle should you
push to maximize the acceleration? What is a reasonable angle if you
were to redesign the mower?
F
N
y
45°
F
EGR 102 03/24/2009
mg
x
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Pressure
F
• Pressure: Force acting over an area p 
A

F N
 2  Pa
• SI unit Pascal:  p  
A m

F  lb f
 2  psi
• US customary unit:  p 
A in
• Pressure analogous
to stress s
F
s
A
EGR 102 03/24/2009
F=mg
A1=pr12
F=mg
A2=pr22
© Katja Hölttä-Otto 2009
Pressure in Engineering
• Pressure and stress
• Hydrostatics
• Hydrodynamics
• Hydraulics/Pneumatics
AP
• Aerodynamics
www.howstuffworks.com
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Pressure - Fluids
• Pascal’s law: for fluid at rest,
pressure at a point is the
same in all directions.
• For a fluid at rest, pressure
increases with the depth of
fluid: p  gh
h
kg m
kg m 1
N
 p   g h  3 2 m  2 2  2  Pa
m s
s m
m
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Pressure - Air
• Analogous to fluids
• Atmospheric pressure 101.325 kPa
(14.696 psi)
– Based on the weight of air in the atmosphere above
the surface of the earth divided by the area at the
base of the column
h1
EGR 102 03/24/2009
h2
h
h
© Katja Hölttä-Otto 2009
Measuring Pressure
• Absolute pressure vs gauge pressure
pabsolute  p gauge  patmosphere
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Pressure Application
• Hydraulic systems
F1
F2
p1 
F1
A1
F
p2  2
A2
www.immersivetechnologies.com/Images/machine/cat5130.jpg
F1 F2
A2
p1  p2 

 F2 
F1
A1 A2
A1
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Hydraulic system –
whiteboard example
F2
F1
EGR 102 03/24/2009
r1 = 1 cm
r2 = 3 cm
F1 = 10 N
F2 = ?
© Katja Hölttä-Otto 2009
Example (whiteboard)
How much force (F1) do you need to apply to compress the spring by 5.00cm?
F1
Rigid beam
A1
A2 = 20.0 A1
F1 = ?
k=3000.0 N/m
x=5.00 cm
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
• Mass
–
–
–
–
–
Mass
Density
Mass Flow
Mass Moment of Inertia
Momentum
• Force
–
–
–
–
–
Force basics and applications
Spring forces
Friction forces
Newton’s laws
Pressure
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Hooke’s law - revisited
F  kx
F = applied force (N)
k = spring constant (N/mm, N/m)
x = deformation of the spring (mm, m)
• Applicable in the elastic range of the spring
– Elastic means there is no permanent deformation
after the force is removed
x
EGR 102 03/24/2009
F
© Katja Hölttä-Otto 2009
Hooke’s law cont’d
F  kx
F = applied force
k = a constant
x = deformation
• Applicable in the elastic range of the material
– Elastic means there is no permanent deformation
after the force is removed
Elastic stress is not
typically visible
x
EGR 102 03/24/2009
F
© Katja Hölttä-Otto 2009
Stress-Strain curve
Remember: Hooke’s law valid in “elastic range”
Elastic stress
Stress (Pa)
Ultimate stress
Upper yield stress
Lower yield stress
Fracture stress
Modulus of Elasticity E
L
s E
L
Hooke’s law restated:
s  E
EGR 102 03/24/2009
Strain (mm/mm)
© Katja Hölttä-Otto 2009
Modulus of Elasticity & Shear Modulus
• Material property
• Modulus of elasticity, E,
for linear pulling (or
pushing) of material
• Shear modulus for
shearing or twisting
material
EGR 102 03/24/2009
F
F
F
F
F
F
© Katja Hölttä-Otto 2009
Tensile Strength &
Compressive Strength
• Tensile strength is the stress
when a material is pulled apart
– Yield strength in the elastic region
(commonly used in engineering)
– Ultimate stress in the maximum
stress material can handle before
failure
• Compressive strength is the
stress when a material is
compressed
EGR 102 03/24/2009
F
F
F
F
F
© Katja Hölttä-Otto 2009
Modulus of elasticity of materials
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Stress (Pa)
Stress (Pa)
Stress-Strain curve
Brittle and Ductile material
Strain (mm/mm)
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Strength of materials
• When designing a piece, it is typically
designed to withstand loads to its yield
stress + a factor of safety
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Example (whiteboard)
• A rod is pulled in a tensile testing
machine with force 50.0 kN. The rod has
a diameter of 4.0 cm. The stain is
0.00001 What material could the rod be?
Hooke’s law restated:
s  E
EGR 102 03/24/2009
Modulus of Elasticity E
L
s E
L
© Katja Hölttä-Otto 2009
Replacing an engine - problem
Supporting cable
10°
ring, not a pulley
80°
Engine
EGR 102 03/24/2009
You are replacing an engine in your
car using a rope as shown in figure.
How much force do you need to lift
the engine? What is the tension force
on the supporting cable? Choose a
rope diameter to accommodate the
largest stress. Use a reasonable
factor of safety. The tensile strength
of the rope is 42 MPa.
The engine weighs 321 kg.
To solve, draw a free body
diagram, make the sum of
forces (as vectors or as
components) add to 0
(equilibrium), and solve for
unknown forces. From
forces, calculate the needed
diameter for the worst case.
Mathematical answer for
diameter is 1.0 cm, what is
your engineering answer?
© Katja Hölttä-Otto 2009
Reminders
• Look for (broken) products to take apart later
– good product will replace one bad assignment grade
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Agenda
•
•
•
•
•
•
•
•
•
Force basics and applications
Spring forces
Friction forces
Free body diagrams
Newton’s laws
Pressure
Stress and strain
Moments
Work
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Moment
• Moment is force acting at a distance
M  Fd
M   F d   Nm
F
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Moment & Torque in Engineering
http://eml.ou.edu/Photo/Struct/Traffic%20light2.jpg
www.craneoperator.com
http://content.answers.com/main/content/wp/en/thumb/b
/b4/500px-The_Little_Belt_Bridge_(1935).jpeg
www.mech.uwa.edu.au/DANotes/ge
ars/intro/gearbox.jpeg
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Calculating moments
1. Draw a free body diagram
2. Write the force and moment equilibrium
equations
3. Solve for unknowns
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Moment – Example 1
F
System:
L/2
L
Free Body Diagram:
F
FBx
FA
FBy
1.
2.
EGR 102 03/24/2009
3.
Draw a free body diagram
Write the force and moment
equilibrium equations
Solve for unknowns
© Katja Hölttä-Otto 2009
Moment – Example 2
F
System:
L/2
L
Free Body Diagram:
M
FAx
F
FAy
1.
2.
EGR 102 03/24/2009
3.
Draw a free body diagram
Write the force and moment
equilibrium equations
Solve for unknowns
© Katja Hölttä-Otto 2009
Moment – Example 3
• EF problem 10.25
1.
2.
EGR 102 03/24/2009
3.
Draw a free body diagram
Write the force and moment
equilibrium equations
Solve for unknowns
© Katja Hölttä-Otto 2009
Moment – in class task
L/4
Calculate the
forces and
moments in
terms on given
mi , L, and g.
L/8
L/2
1.
2.
3.
m1g
m2g
L
M
EGR 102 03/24/2009
L
FAx FAy
Draw a free body diagram
Write the force and moment
equilibrium equations
Solve for unknowns
© Katja Hölttä-Otto 2009
Agenda
•
•
•
•
•
•
•
•
•
Force basics and applications
Spring forces
Friction forces
Free body diagrams
Newton’s laws
Pressure
Stress and strain
Moments
Work
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Work
• Work is the force to the direction of the
movement need to move an item
W  Fd
W  F d
F
Fwork
W   F d   Nm  J
F
Fwork
d
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Work in Engineering
mowabb.com/aimages/images/2005/07-15-05.jpg
www.mhinfo.com/mhi_new/images/conve
yor.jpg
EGR 102 03/24/2009
www.otis.com
© Katja Hölttä-Otto 2009
Work - Example
• How much work is required to push a lawn
mower for 100m? How much force do you need
to push the mower? F=200N. Pushing angle is
60°.
F
N
Fy
60°
Fwork
F
EGR 102 03/24/2009
mg
© Katja Hölttä-Otto 2009
Work - Example
• How much work is required to push a lawn
mower for 100m?
F =200i+231j
N
mg
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
Work – In class task
• How much work is required to push w/ force F
the crate 1m down the ramp at constant speed.
Friction force is 800N. The crate weighs 100
kg.
mg
EGR 102 03/24/2009
40
© Katja Hölttä-Otto 2009
Assignment
• Individual assignment (=homework)
• Due Thu 4/16 8:00am
• Problems:
– 10.23, 10.24, 10.27, + the pulley problem on the next slide (10p
for each problem in EF + 20p for the pulley problem)
• Follow format in course book EF chapter 4
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
• A pulley is hanging on 2
cables on the ceiling.
You are pulling a known
mass on a rope as
shown in the figure.
What percentage of the
effort (force) you use
goes to work done to lift
the mass? Which
rope/cable is under
most stress?
40
30
Cable A
Cable B
25
Rope
m
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009
40
TAy
40
TA
TB
TAx
30
30
TBy
TBx
25
G
Fpull_y
Fpull_x
EGR 102 03/24/2009
© Katja Hölttä-Otto 2009