Download Ch6 - Force and Motion-II

Goals for Chapter 6
• 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
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
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
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?)
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
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
Ceiling on Rope!
One-dimensional equilibrium: Tension in a rope with mass
• What is the tension in the previous example if the rope
has mass? (Say weight of rope = 120 N)
Two-dimensional equilibrium
• A car engine hangs from several chains.
• Weight of car engine = w; ignore chain weights
A car on an inclined plane
An car rests on a slanted ramp (car of weight w)
A car on an inclined plane
Coordinate system choice #1:
(y) parallel to slope &
(x) perpendicular to slope
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)
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
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
Example: 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
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.
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!
Ex: 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.
Ex: 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!
Example: 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)?
Example: 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)?
Example: 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?
Example: 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?
Example: Tension in an elevator cable
Compare Tension to Weight while elevator slows?
Example 5.8: Tension in an elevator cable
Compare Tension to Weight while elevator slows?
What if elevator was accelerating upwards at same
rate?
Example 5.8: Tension in an elevator cable
What if elevator was accelerating upwards at same
rate?
Same Free Body Diagram! Same result!
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?
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?
Ex 5.9 Apparent weight in an accelerating elevator
What if she was accelerating downward, rather than slowing?
Increasing
speed
down?
Acceleration down a hill
What is the acceleration of a toboggan sliding down
a friction-free slope?
Acceleration down a hill
What is the acceleration of a toboggan sliding down
a friction-free slope?
Two common free-body diagram errors

The normal force must be perpendicular to the surface.
There is no separate “ma force.”
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)
Two bodies with the same magnitude of acceleration
What is the tension T, and the acceleration a, of the system?
6-1 Friction

Friction forces are essential:
o
Picking things up
o
Walking, biking, driving anywhere
o
Writing with a pencil
o
Building with nails, weaving cloth
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction

But overcoming friction forces is also important:
o
o
Efficiency in engines
(20% of the gasoline used in an automobile goes to
counteract friction in the drive train)
o
Roller skates, fans
o
Anything that we want to remain in motion
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction

Three experiments:
o
Slide a book across a counter.
o
The book slows and stops, so there must be an acceleration
o
Direction?
o
parallel to the surface and
o
opposite the direction of motion.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction

Three experiments:
o
o
o
Push a book at a constant speed across the counter.
There must be an equal and opposite force opposing you,
otherwise the book would accelerate.
Again the force is parallel to the surface and
opposite the direction of motion.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction

Three experiments:
o
o
o
o
Push a crate or other heavy object that does not move.
To keep the crate stationary, an equal and opposite force
must oppose you.
If you push harder, the opposing force must also increase to
keep the crate stationary.
Keep pushing harder… Eventually the opposing force will
reach a maximum, and the crate will slide.
© 2014 John Wiley & Sons, Inc. All rights reserved.
• 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.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction

Microscopic picture: surfaces are bumpy

Friction occurs as contact points slide over each other



Two specially prepared metal surfaces can cold-weld together
and become impossible to slide, because there is so much
contact between the surfaces
Greater force normal to the
contact plane increases the
friction because the surfaces
are pressed together and make
more contact
Sliding that is jerky, due to the
ridges on the surface, produces
squeaking/squealing/sound
Figure 6-2
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 TWO types of friction

The static frictional force:
o
The opposing force that prevents an object from moving
o
Can have any magnitude from 0 N up to a maximum
o

Once the maximum is reached, forces are no longer in
equilibrium and the object slides
The kinetic frictional force:
o
The opposing force that acts on an object in motion
o
Has only one value
o
Generally smaller than the maximum static frictional force
© 2014 John Wiley & Sons, Inc. All rights reserved.
• 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.
© 2014 John Wiley & Sons, Inc. All rights reserved.
•
Before the box slides, static friction acts. But once it starts to slide,
kinetic friction acts.
© 2014 John Wiley & Sons, Inc. All rights reserved.
•
Before the box slides, static friction acts. But once it starts to slide,
kinetic friction acts.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 STATIC Friction

The properties of static friction
1. If the body does not move, then the applied force
and frictional force balance along the direction
parallel to the surface: equal in magnitude,
opposite in direction
2. The magnitude of fs has a maximum fs,max given by:
Eq. (6-1)
where μs is the coefficient of static friction. If the
applied force increases past fs,max, sliding begins.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Kinetic Friction

The properties of kinetic friction
3. Once sliding begins, the frictional force decreases
to fk given by:
Eq. (6-2)
where μk is the coefficient of kinetic friction.


Magnitude FN of the normal force measures how
strongly the surfaces are pushed together
The values of the friction coefficients are unitless and
must be determined experimentally
© 2014 John Wiley & Sons, Inc. All rights reserved.
• 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.

Assume that μk does not depend on velocity

Note that these equations are not vector equations
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction
Answer: (a) 0
(b) 5 N
(c) no
(d) yes (e) 8 N
© 2014 John Wiley & Sons, Inc. All rights reserved.
Some approximate coefficients of friction
Friction in horizontal motion
• 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?
Friction in horizontal motion
• Before the crate moves, static friction acts on it.
Friction in horizontal motion
• After it starts to move, kinetic friction acts.
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
Pulling a crate at an angle
• The angle of the pull affects the normal force, which in turn
affects the friction force.
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?
6-1 Friction
Example For a force applied at an angle:
o
STEP 1: COORDINATES
o
STEP 2: COMPONENTS!
o
o
o
Decompose the force into x and y components
STEP 3: No Fruit Salad!
o
Balance the vertical components (FN, Fg, Fy)
o
Balance the horizontal components (f, Fx)
STEP 4: You’ll need Normal to get Friction!
o
Note FN and f are related
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-1 Friction
Example For a force applied at an angle:
o
STEP 1: COORDINATES
o
STEP 2: COMPONENTS!
o
o
o
Decompose the force into x and y components
STEP 3: No Fruit Salad!
o
Balance the vertical components (FN, Fg, Fy)
o
Balance the horizontal components (f, Fx)
STEP 4: You’ll need Normal to get Friction!
o
Note FN and f are related
© 2014 John Wiley & Sons, Inc. All rights reserved.
Example: 6.01 Angled Force to Stationary Block
Force of 12.0 N applied to 8 kg
block at downward angle of 30
degrees;
static friction coefficient is ms =
0.700, and kinetic friction
coefficient is mk = 0.400.
Does it slide?
If so, what is the acceleration?
If not, what is the friction force?
© 2014 John Wiley & Sons, Inc. All rights reserved.
Example: 6.01 Angled Force to Stationary Block
Force of 12.0 N applied to 8 kg
block at UPWARD angle of 30
degrees;
static friction coefficient is ms =
0.700, and kinetic friction
coefficient is mk = 0.400.
Now does it slide?
If so, what is the acceleration?
If not, what is the friction force?
Sliding on ice…
How far does it take for a car moving at a
constant speed of 10 m/s to slide to a stop on a
horizontal road if the coefficient of kinetic friction
is 0.60?
If it is on ice (mk = 0.10?)
If it is on a icy HILL (with slope = 20 degrees)?
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-2 The Drag Force and Terminal Speed


A fluid is anything that can flow (gas or liquid)
When there is relative velocity between fluid and an
object there is a drag force:
o
Opposes the relative motion
o
Points along direction of flow, relative to the body
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6-2 The Drag Force and Terminal Speed

Here we examine the drag force for
o
Air
o
With a body that is not streamlined
o
For motion fast enough that the air becomes turbulent
(breaks into swirls)
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
6-2 The Drag Force and Terminal Speed

For this case, the drag force is:

Where:
o
v is relative velocity
o
ρ is air density (mass/volume)
o
C is experimentally determined drag coefficient
o
A is effective cross-sectional area of body
( area taken perpendicular to the relative velocity)
© 2014 John Wiley & Sons, Inc. All rights reserved.
Drag force
Weight
6-2 The Drag Force and Terminal Speed

For this case, the drag force is:

Where:
o
v is relative velocity (m/s)
o
ρ is air density (kg/m3)
o
C is drag coefficient (#)
o
A is cross-sectional area (m2)
Units: kg/m3 * m2 * (m/s)2 = kg m/s2 = N = Force!
© 2014 John Wiley & Sons, Inc. All rights reserved.
Drag force
Weight
6-2 The Drag Force and Terminal Speed


The drag force from the air opposes a falling object
Once the drag force equals the gravitational force, the
object falls at a constant terminal speed:

Terminal speed can be increased by reducing A

Terminal speed can be decreased by increasing A

Skydivers use this to control descent
© 2014 John Wiley & Sons, Inc. All rights reserved.
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
6-2 The Drag Force and Terminal Speed
Example Speed of a rain drop:
o Spherical drop feels gravitational force F = mg:
o Express in terms of density of water
o Use A = πR2 for the cross-sectional area
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-2 The Drag Force and Terminal Speed
Example Speed of a rain drop:
o So plug in to the terminal velocity equation using the values
provided in the text:
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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.
Using Newton’s First Law when forces are in equilibrium
•IF you see uniform circular
motion…
Radius r
Tension T
v
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!
• SFx = mv2/r
NOT
• mv2/r + T – mg = ma
Radius r
Tension T
v
6-3 Uniform Circular Motion

Centripetal force is not a new kind of force, it is simply
an application of force
Eq. (6-18)

For the puck on a string, the
string tension supplies the
centripetal force necessary
to maintain circular motion
Figure 6-8
© 2014 John Wiley & Sons, Inc. All rights reserved.
Avoid using “centrifugal force”
• Figure (a) shows the correct
free-body diagram for a body
in uniform circular motion.
Avoid using “centrifugal force”
• Figure (b) shows a common
error.
• In an inertial frame of
reference, there is no such
thing as “centrifugal force.”
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?
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?
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.
A conical pendulum
• A bob at the end of a wire moves in a horizontal circle with
constant speed.
A car rounds a flat curve
• A car rounds a flat unbanked curve. What is its maximum
speed?
6-3 Uniform Circular Motion

Recall that circular motion requires a centripetal
acceleration
Examples You are a passenger:
o
For a car, rounding a curve, the car accelerates toward the
center of the curve due to a centripetal force provided by
the inward friction on the tires.
Your inertia makes you want to go straight ahead so you
may feel friction from your seat and may also be pushed
against the side of the car.
These inward forces keep you in uniform circular motion in
the car.
A car rounds a flat curve
• A car rounds a flat unbanked curve. What is its maximum
speed?
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?
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?
Maximum Speeds for Cornering on Banked Curves
Maximum Cornering Speed for
Banked Curves
Radius (m)
30
30
30
30
30
30
30
40
50
60
60
40
Coefficient of
Static Friction
0.8
0.8
0.8
0.8
0.6
0.4
0.2
0.2
0.2
0.2
0.2
0.8
Angle
(degrees)
30
20
10
0
30
30
30
30
30
30
0
5
max speed (m/s)
27
22
18
15
23
19
16
19
21
23
11
19
max speed (mph)
60
48
40
34
51
43
35
41
46
50
24
43
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-3 Uniform Circular Motion
Example Car in a banked circular turn:
Figure 6-11
o
Sum components along the radial direction:
Eq. (6-23)
o
Sum components along the vertical direction:
Eq. (6-24)
o
Divide and replace (sin θ)/(cos θ) with tangent.
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-3 Uniform Circular Motion
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-3 Uniform Circular Motion
Answer: (a) accel downward, FN upward
(b) accel upward, FN upward
(c) the magnitudes must be equal for the motion to be uniform
(d) FN is greater in (b) than in (a)
© 2014 John Wiley & Sons, Inc. All rights reserved.
6-3 Uniform Circular Motion
Example Bicycle going around a vertical loop:
Figure 6-9
o
o
At the top of the loop we have:
Solve for v and plug in our known values,
including FN = 0 for the minimum answer:
© 2014 John Wiley & Sons, Inc. All rights reserved.
Eq. (6-19)