Transcript Chapter 5 - Applications of Newton`s Laws

Chapter 5
Applying Newton’s
Laws
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Copyright © 2012 Pearson Education Inc.
Congratulations!
Copyright © 2012 Pearson Education Inc.
Congratulations!
Copyright © 2012 Pearson Education Inc.
Congratulations!
Copyright © 2012 Pearson Education Inc.
Congratulations!
Copyright © 2012 Pearson Education Inc.
Congratulations!
On the first try….
Thursday!
• Ahmed, Christian, Nick, Fern
• Troy, Ali, Russell, Moses, Andrew
• Dolly, Samuel, Ray, Hye
Friday!
 Hairu, Guohao, Dan, Quinlan
On the second try….
•
•
•
•
•
Robert, Saminder, Jaikar, Kelsi
Joshua, Rex, Steven, Shean, Hesham
Jordan, Camille, Andrew, Bryan
David, Omar, Sugei, Louis, Eduardo
Roger, Sue, Becky, Adolfo, Jasmine
Copyright © 2012 Pearson Education Inc.
Goals for Chapter 5
• Use Newton’s 1st law for bodies in
equilibrium (statics)
• Use Newton’s 2nd law for accelerating
bodies (dynamics)
• Study types of friction & fluid resistance
• Solve circular motion problems
Copyright © 2012 Pearson Education Inc.
Using Newton’s First Law when forces are in equilibrium
• A body is in equilibrium when it is at rest
or moving with constant velocity in an
inertial frame of reference.
• Follow Problem-Solving Strategy 5.1.
Copyright © 2012 Pearson Education Inc.
Copyright © 2012 Pearson Education Inc.
Using Newton’s First Law when forces are in equilibrium
• VISUALIZE (create a coordinate system; decide what is happening?)
• SKETCH FREE-BODY DIAGRAM
•Isolate one point/body/object
•Show all forces in that coordinate system
ON that body (not acting by that body!)
• LABEL all forces clearly, consistently
•Normal forces from surfaces
•Friction forces from surfaces
•Tension Forces from ropes
•Contact forces from other objects
•Weight from gravity
Copyright © 2012 Pearson Education Inc.
Using Newton’s First Law when forces are in equilibrium
• BREAK ALL applied forces into
components based on your coordinate
system.
• Apply Newton’s Laws to like components
ONLY
•SFx = max; SFy = may
Copyright © 2012 Pearson Education Inc.
One-dimensional equilibrium: Tension in a massless rope
• A gymnast hangs from the end of a massless rope.
• Example 5.1 (mg = 50 kg; what is weight & force on rope?)
Copyright © 2012 Pearson Education Inc.
One-dimensional equilibrium: Tension in a massless rope
• A gymnast hangs from the end of a massless rope.
• Example (mg = 50 kg; what is weight & force on rope?)
Copyright © 2012 Pearson Education Inc.
One-dimensional equilibrium: Tension in a massless rope
• A gymnast hangs from the end of a massless rope.
• Example (mg = 50 kg; what is weight & force on rope?)
Tension of Rope on
Gymnast
Copyright © 2012 Pearson Education Inc.
One-dimensional equilibrium: Tension in a rope with mass
• What is the tension in the previous example if the
rope has mass? (Example 5.2 weight of rope = 120 N)
Copyright © 2012 Pearson Education Inc.
Two-dimensional equilibrium
• A car engine hangs from several chains.
• Weight of car engine = w; ignore chain weights
Copyright © 2012 Pearson Education Inc.
A car on an inclined plane
• An car rests on a slanted ramp (car of weight w)
Copyright © 2012 Pearson Education Inc.
A car on an inclined plane
• Coordinate system choice #1:
– (y) parallel to slope &
–
(x) perpendicular to slope
Copyright © 2012 Pearson Education Inc.
A car on an inclined plane
• Coordinate system choice #1:
– (y) parallel to slope &
–
(x) perpendicular to slope
N (in y)
T (in x)
W (in x & y)
Copyright © 2012 Pearson Education Inc.
A car on an inclined plane
• Coordinate system choice #2:
– (y) perpendicular to ground
–
(x) parallel to ground
N
in both x & y!
y
T (in x & y!)
W (in y only)
x
Copyright © 2012 Pearson Education Inc.
A car on an inclined plane
• Coordinate system choice #2:
– (y) perpendicular to ground
–
(x) parallel to ground
N
T
y
Ny
Tx
Tx
Nx
W
x
Copyright © 2012 Pearson Education Inc.
Example 5.5: Bodies connected by a cable and pulley
• Cart connected to bucket by cable passing over pulley.
•Initially, assume pulley is massless and frictionless!
• Pulleys REDIRECT force – they don’t amplify or reduce.
Tension in rope pulls
upwards along slope
Copyright © 2012 Pearson Education Inc.
SAME tension in rope
pulls upwards on bucket
Example 5.5: Bodies connected by a cable and pulley
• Draw separate free-body diagrams for the bucket and the cart.
Copyright © 2012 Pearson Education Inc.
A note on free-body diagrams
• Only the force of
gravity acts on the
falling apple.

• ma does not belong in
a free-body diagram!
It is the SUM of all the
forces you find!
Copyright © 2012 Pearson Education Inc.
Ex 5.6 Straight-line motion with constant force
• Wind exerts a constant horizontal force on the boat.
• 4.0 s after release, v = 6.0 m/s; mass = 200 kg.
Copyright © 2012 Pearson Education Inc.
Ex 5.6 Straight-line motion with constant force
• Wind exerts a constant horizontal force on the boat.
• 4.0 s after release, v = 6.0 m/s; mass = 200 kg.
• Find W, the force of the wind!
Copyright © 2012 Pearson Education Inc.
Example 5.7: Straight-line motion with friction
• For the ice boat in the previous example, a constant
horizontal friction force of 100 N opposes its motion
• What constant force needed by wind to create the same
acceleration (a = + 1.5 m/s/s)?
Copyright © 2012 Pearson Education Inc.
Example 5.7: Straight-line motion with friction
• For the ice boat in the previous example, a constant
horizontal friction force now opposes its motion
(100N); what constant force needed by wind to create
the same acceleration (a = + 1.5 m/s/s)?
Copyright © 2012 Pearson Education Inc.
Example 5.8: Tension in an elevator cable
• Elevator (800 kg) is moving downward @ 10 m/s but
slowing to a stop over 25.0 m.
• What is the tension in the supporting cable?
Copyright © 2012 Pearson Education Inc.
Example 5.8: Tension in an elevator cable
• Elevator (800 kg) is moving downward @ 10 m/s but
slowing to a stop over 25.0 m.
• What is the tension in the supporting cable?
Copyright © 2012 Pearson Education Inc.
Example 5.8: Tension in an elevator cable
• Compare Tension to Weight while elevator slows?
Copyright © 2012 Pearson Education Inc.
Example 5.8: Tension in an elevator cable
• Compare Tension to Weight while elevator slows?
• What if elevator was accelerating upwards at same rate?
Copyright © 2012 Pearson Education Inc.
Example 5.8: Tension in an elevator cable
• What if elevator was accelerating upwards at same rate?
• Same Free Body Diagram! Same result!
Copyright © 2012 Pearson Education Inc.
Ex 5.9 Apparent weight in an accelerating elevator
• A woman inside the elevator of the previous example is standing
on a scale. How will the acceleration of the elevator affect the
scale reading?
Copyright © 2012 Pearson Education Inc.
Ex 5.9 Apparent weight in an accelerating elevator
• A woman inside the elevator of the previous example is standing
on a scale. How will the acceleration of the elevator affect the
scale reading?
Copyright © 2012 Pearson Education Inc.
Ex 5.9 Apparent weight in an accelerating elevator
• What if she was accelerating downward, rather than slowing?
Increasing
speed
down?
Copyright © 2012 Pearson Education Inc.
Acceleration down a hill
• What is the acceleration of a toboggan sliding down a
friction-free slope?
Copyright © 2012 Pearson Education Inc.
Acceleration down a hill
• What is the acceleration of a toboggan sliding down a
friction-free slope? Follow Example 5.10.
Copyright © 2012 Pearson Education Inc.
Two common free-body diagram errors
• The normal force must be perpendicular to the surface.

• There is no separate “ma
force.”
Copyright © 2012 Pearson Education Inc.
Two bodies with the same acceleration
• We can treat the milk carton and tray as separate bodies,
or we can treat them as a single composite body.
Copyright © 2012 Pearson Education Inc.
Two bodies with the same acceleration
• Push a 1.00 kg food tray with constant 9 N force, no
friction. What is acceleration of tray and force of tray on
the carton of milk?
Copyright © 2012 Pearson Education Inc.
Two bodies with the same magnitude of acceleration
• The glider on the air track and the falling weight move in different
directions, but their accelerations have the same magnitude and
relative direction (both increasing, or both decreasing)
Copyright © 2012 Pearson Education Inc.
Two bodies with the same magnitude of acceleration
• What is the tension T, and the acceleration a, of the system?
Copyright © 2012 Pearson Education Inc.
Frictional forces
• When a body rests or
slides on a surface, the
friction force is parallel
to the surface.
• Friction between two
surfaces arises from
interactions between
molecules on the
surfaces.
Copyright © 2012 Pearson Education Inc.
Kinetic and static friction
• Kinetic friction acts when a body slides over a
surface.
• The kinetic friction force is fk = µkn.
• Static friction acts when there is no relative motion
between bodies.
• The static friction force can vary between zero and
its maximum value: fs ≤ µsn.
Copyright © 2012 Pearson Education Inc.
Static friction followed by kinetic friction
•
Before the box slides, static friction acts. But once it starts to slide,
kinetic friction acts.
Copyright © 2012 Pearson Education Inc.
Some approximate coefficients of friction
Copyright © 2012 Pearson Education Inc.
Friction in horizontal motion – example 5.13
• Move a 500-N crate across a floor with friction by
pulling with a force of 230 N.
• Initially, pull harder to get it going; later pull easier
(at 200N once it is going). What are ms? mk?
Copyright © 2012 Pearson Education Inc.
Friction in horizontal motion
• Before the crate moves, static friction acts on it.
Copyright © 2012 Pearson Education Inc.
Friction in horizontal motion
• After it starts to move, kinetic friction acts.
Copyright © 2012 Pearson Education Inc.
Static friction can be less than the maximum
• Static friction only has its maximum value just
before the box “breaks loose” and starts to slide.
Force
Force builds in time to
maximum value, then object
starts moving & slipping
Time
Copyright © 2012 Pearson Education Inc.
Pulling a crate at an angle
• The angle of the pull affects the normal force,
which in turn affects the friction force.
• Follow Example 5.15.
Copyright © 2012 Pearson Education Inc.
Motion on a slope having friction – ex 5.16
• Consider a toboggan
going down a slope at
constant speed. What
is m?
• Now consider same
toboggan on steeper
hill, so it is now
accelerating.
What is a?
Copyright © 2012 Pearson Education Inc.
Fluid resistance and terminal speed
• Fluid resistance on a body
depends on the speed of the
body.
• Resistance can depend upon
v or v2 and upon the shape
moving through the fluid.
• Fresistance = -kv
or - Dv2
• These will result in different
terminal speeds
Copyright © 2012 Pearson Education Inc.
Fluid resistance and terminal speed
• A falling body reaches its
terminal speed when resisting
force equals weight of the
body.
• If F = -kv for a falling body,
vterminal = mg/k
• If F = - Dv2 for a falling body,
vterminal = (mg/D)½
Copyright © 2012 Pearson Education Inc.
Dynamics of circular motion
• If something is in
uniform circular motion,
both its acceleration and
net force on it are
directed toward center of
circle.
• The net force on the
particle is Fnet = mv2/R,
always towards the
center.
Copyright © 2012 Pearson Education Inc.
Using Newton’s First Law when forces are in equilibrium
• IF you see uniform circular
motion…
r
v
Copyright © 2012 Pearson Education Inc.
Using Newton’s First Law when forces are in equilibrium
• IF you see uniform circular
motion…
THEN remember centripetal
force is NOT another force – it is
the SUM of one or more forces
already present!
r
•SFx = mv2/R
NOT
•mv2/R + T – mg = ma
Copyright © 2012 Pearson Education Inc.
v
Avoid using “centrifugal force”
• Figure (a) shows the
correct free-body
diagram for a body in
uniform circular
motion.
• Figure (b) shows a
common error.
• In an inertial frame of
reference, there is no
such thing as
“centrifugal force.”
Copyright © 2012 Pearson Education Inc.
Force in uniform circular motion
• A 25 kg sled on frictionless ice is kept in uniform
circular motion by a 5.00 m rope at 5 rev/minute.
What is the force?
Copyright © 2012 Pearson Education Inc.
Force in uniform circular motion
• A 25 kg sled on frictionless ice is kept in uniform
circular motion by a 5.00 m rope at 5 rev/minute.
What is the force?
Copyright © 2012 Pearson Education Inc.
What if the string breaks?
• If the string breaks, no net force acts on the ball, so it
obeys Newton’s first law and moves in a straight
line.
Copyright © 2012 Pearson Education Inc.
A conical pendulum – Ex 5.20
• A bob at the end of a wire moves in a horizontal
circle with constant speed.
Copyright © 2012 Pearson Education Inc.
A car rounds a flat curve
• A car rounds a flat unbanked curve. What is its
maximum speed?
Copyright © 2012 Pearson Education Inc.
A car rounds a flat curve
• A car rounds a flat unbanked curve. What is its
maximum speed?
Copyright © 2012 Pearson Education Inc.
A car rounds a banked curve
• At what angle should a curve be banked so a car can
make the turn even with no friction?
Copyright © 2012 Pearson Education Inc.
A car rounds a banked curve
• At what angle should a curve be banked so a car can
make the turn even with no friction?
Copyright © 2012 Pearson Education Inc.
Uniform motion in a vertical circle – Ex. 5.23
• A person on a Ferris wheel moves in a vertical circle
at constant speed.
• What are forces on person at top and bottom?
Copyright © 2012 Pearson Education Inc.
Uniform motion in a vertical circle – Ex. 5.23
• A person on a Ferris wheel moves in a vertical circle.
Copyright © 2012 Pearson Education Inc.