Download Kinetics Key to Success Eight Step Process Friction

Kinetics
Key to Success
Kinetics is the study of forces that cause accelerated motion.
⇒
Newton’s 2nd Law governs:
∑F
= ma
Rectangular Coordinates:
∑ Fx
∑ Fy
= may
Normal and Tangential:
∑F
∑F
„
„
„
„
at = v&
an =
v2
ρ
1. Impending motion known
to exist. Draw FBD with
friction force in opposite
direction of impending motion.
Solve for unknown force or
coefficient of friction.
3)
4)
5)
6)
7)
„
„
n
= m an
t
1)
2)
„
„
= m at
The Free Body Diagram = The Kinetic Diagram
Remember:
^
e
Friction
„
„
t
e^n
^ x
i
Eight Step Process
„
„
y
^
j
= m ax
8)
Decide what needs to be isolated. (may be the hardest part).
Draw the isolated Free Body & Kinetic Diagrams (complete with all
external boundaries) and set them equal to each other.
Choose a Coordinate System (C.S.).
a)
Add all EXTERNALLY APPLIED forces & moments
acting ON the Free Body Diagram:
a1)
Given forces and moments including weight.
a2)
Support reactions (where the body is cut from the
rest of the world).
b)
Add all mass*acceleration terms to the Kinetic Diagram.
Add all necessary dimensions.
Enforce Newton’s 2nd Law:
a)
If necessary, set up any required Kinematic equations.
b)
Solve ALL equations for ALL unknowns.
Check work and answers for units, directions, proper notation,
S.D.,
reasonableness, etc.
Fmax - about to slip
(impending motion)
F
F = μs N
F
=
P
F < Fmax & F < P
45o
No Motion
P
Motion
2. Determine whether body is at rest or is sliding. Friction force
F required to maintain equilibrium is obtained from FBD.
If:
If:
If:
F < μs N
No sliding
F = μs N
Impending motion
F > μs N
Body is sliding. F = μ k N
Solve using F = ma
1
Non-Uniform Circular Motion
v = v eˆt
v2
a = v& eˆt +
a =
a =
ρ
time 2
^
e
t
Motion along a circular path with constant speed
path
r
2
n
⎛ v2 ⎞
v& + ⎜⎜ ⎟⎟
⎝ρ⎠
time 1
^
e
t
2
2
Time rate of change of the direction of the velocity
∑ Ft = m at = mv&
v2
∑ Fn = m an = m
ρ
Time rate of change of the magnitude of the velocity
Note that the acceleration and the
force are inward. They are a
centripetal acceleration and force.
A force acting on an object as it moves through a
fluid (liquid or gas).
‰
It is opposite the direction of motion
„
„
Similar to sliding friction
Linear
FD = bv
a =
„
Quadratic
=
e^n
^
e
an eˆn
∑ Ft = m at = 0
t
Time rate of change of the direction of the velocity
v
∑ Fn = m an = m
2
ρ
Note that the acceleration and the
force are inward. They are a
centripetal acceleration and force.
mg
=
b
vter =
2mg
ρAC D
m1m2
r212
„
Scalar form
F =G
„
Vector form
F12 = − G
Where
„
„
„
„
1
FD = ρAC D v 2
2
ρ
r
eˆn = at eˆt + an eˆn
time 1
‰
vter
v2
eˆn
r
v2
e^n
Newton’s Law of Universal Gravitation
Drag & Terminal Velocity
„
a = v& eˆt +
t
path
v = v eˆt
e^n
a +a
2
t
v = constant
time 2
^
e
Motion along a circular path with constant speed
e^n
eˆn = at eˆt + an eˆn
Uniform Circular Motion
„
„
m1m2
r̂21
r212
G = universal gravitational constant = 6.67 x 10-11 N⋅m2/kg
m1 = mass of body 1
m2 = mass of body 2
r21 = position vector from body 2 to body 1
r21 = magnitude of the position vector from body 2 to body 1
r̂21 = position vector from body 2 to body 1
2
Satellite Motion
„
Kepler’s Laws:
The objective is to “balance” the force of gravity
with the normal (as in direction) acceleration.
Fr = G
mS mE
v2
=
m
S
r2
r
m
⇒ G E = v2
r
„
„
‰
‰
‰
‰
‰
‰
‰
Must be on Engineering paper
Front side ONLY
In YOUR handwriting
Put your name at top
Anything you want to write
You are to turn it in with the exam
NO other books, notes, etc. are allowed.
„
During equal time intervals the radius
vector from the Sun to a planet sweeps
out equal areas.
„
If T is the time that it takes for a planet
to make one full revolution around the
Sun, and if S is half the major axis of
the ellipse (S reduces to the radius of
the planet’s orbit if that orbit is circular),
then:
where C is a constant
T2
= C whose value is the
3
same for all planets.
S
m1mSun
v2
= m1 1
r1
r12
v1 =
2πr1
T1
⇒ C=
4π 2
GmSun
Exam 2
„
Covers Module 2
Will be a combination of short, medium, and long
problems
You can bring one sheet of notes (i.e., a “cheat sheet”)
Planets move in planar elliptical paths
with the Sun at one focus of the ellipse.
ΣFn = G
Exam 2 – Wednesday, Oct. 17
„
„
You must show ALL work on ALL problems
Coordinate system on all problems
-
When needed and not provided
-
FBD = KD on all problems that need them
-
Equations
-
Units
-
-
-
Make sure they are easily read, everything is labeled and the are complete
Provide generic versions of the equations
Final answers must have correct and consistent units
Watch unit conversions and unit consistency
-
Significant Figures (SF)
-
Vector symbols and answers when required
-
-
-
Final answers must have the correct SF
Use over bars or over arrows (hats for unit vectors)
Vector answers must be in either I j notation or magnitude and direction
format. Some questions may specify which format.
etc.
MORE SLIDES TO FOLLOW!!!
3
ConcepTest 5.1a Gravity and Weight I
ConcepTest 5.1b Gravity and Weight II
What can you say
1) Fg is greater on the feather
What can you say
1)
about the force of
2) Fg is greater on the stone
about the acceleration
2) it is greater on the stone
3) Fg is zero on both due to vacuum
of gravity acting on the
3) it is zero on both due to vacuum
4) Fg is equal on both always
stone and the feather?
gravity Fg acting on a
stone and a feather?
it is greater on the feather
4) it is equal on both always
5) it is zero on both always
5) Fg is zero on both always
The acceleration is given by F/m so
here the mass divides out. Since we
The force of gravity (weight) depends
know that the force of gravity (weight)
on the mass of the object!! The stone
is mg, then we end up with acceleration
has more mass, therefore more weight.
g for both objects.
Follow-up: Which one hits the bottom first?
ConcepTest 5.2 On the Moon
An astronaut on Earth kicks a
bowling ball and hurts his
foot. A year later, the same
astronaut kicks a bowling
ConcepTest 5.3a Tension I
1) more
2) less
3) the same
You tie a rope to a tree and you
1) 0 N
pull on the rope with a force of
2) 50 N
100 N. What is the tension in
ball on the Moon with the
3) 100 N
4) 150 N
the rope?
5) 200 N
same force. His foot hurts...
Ouch!
The masses of both the bowling ball
The tension in the rope is the force that the rope
and the astronaut remain the same, so
“feels” across any section of it (or that you would
his foot feels the same resistance and
feel if you replaced a piece of the rope). Since you
hurts the same as before.
are pulling with a force of 100 N, that is the tension
in the rope.
Follow-up: What is different about
the bowling ball on the Moon?
ConcepTest 5.3b Tension II
ConcepTest 5.3c Tension III
Two tug-of-war opponents each
1) 0 N
You and a friend can
pull with a force of 100 N on
2) 50 N
each pull with a force of
3) 100 N
20 N. If you want to rip
4) 150 N
a rope in half, what is
5) 200 N
the best way?
opposite ends of a rope. What
is the tension in the rope?
1) you and your friend each pull on
opposite ends of the rope
2) tie the rope to a tree, and you both
pull from the same end
3) it doesn’t matter -- both of the above
are equivalent
4) get a large dog to bite the rope
This is literally the identical situation to the
Take advantage of the fact that the tree can
previous question. The tension is not 200 N !!
pull with almost any force (until it falls down,
Whether the other end of the rope is pulled by a
that is!). You and your friend should team up
person, or pulled by a tree, the tension in the rope
on one end, and let the tree make the effort on
is still 100 N !!
the other end.
Page 1
1
ConcepTest 5.4 Three Blocks
ConcepTest 5.5 Over the Edge
Three blocks of mass 3m, 2m, and
1) T1 > T2 > T3
In which case does block m experience 1) case 1
m are connected by strings and
2) T1 < T2 < T3
a larger acceleration? In (1) there is a
pulled with constant acceleration a.
10 kg mass hanging from a rope and
3) T1 = T2 = T3
What is the relationship between
the tension in each of the strings?
5) tensions are random
constant downward force of 98 N.
4) depends on value of m
Assume massless ropes.
5) case 2
In (2) the tension is 98 N
T1 pulls the whole set
a
the tension is less than
must be the largest.
98 N because the block
T2 pulls the last two
T3
3m
T2
2m
masses, but T3 only
m
T1
m
m
due to the hand. In (1)
of blocks along, so it
pulls the last mass.
3) both cases are the same
falling. In (2) a hand is providing a
4) all tensions are zero
2) acceleration is zero
a
a
10kg
F = 98 N
is accelerating down.
Only if the block were at
Case (1)
Case (2)
rest would the tension
Follow-up: What is T1 in terms of m and a?
be equal to 98 N.
ConcepTest 5.6a Going Up I
ConcepTest 5.6b Going Up II
A block of mass m rests on the floor of
1) N > mg
A block of mass m rests on the
1) N > mg
an elevator that is moving upward at
2) N = mg
floor of an elevator that is
2) N = mg
3) N < mg (but not zero)
accelerating upward. What is
3) N < mg (but not zero)
constant speed. What is the
relationship between the force due to
the relationship between the
gravity and the normal force on the
4) N = 0
block?
5) depends on the size of the
elevator
4) N = 0
force due to gravity and the
5) depends on the size of the
elevator
normal force on the block?
The block is accelerating upward, so
The block is moving at constant speed, so
v
a>0
mg
so N must be greater than mg in order
just like the block at rest on a table.
m
to give the net upward force!
Follow-up: What is the normal force if
the elevator is in free fall downward?
ConcepTest 5.7 Normal Force
Σ F = N – mg = ma > 0
∴ N > mg
ConcepTest 5.8 On an Incline
1) case 1
Consider two identical blocks,
1) case A
2) case 2
one resting on a flat surface,
2) case B
and the other resting on an
3) it’s the same for both
incline. For which case is the
4) depends on the magnitude of
the force F
normal force greater?
3) both the same (N = mg)
4) both the same (0 < N < mg)
5) both the same (N = 0)
5) depends on the ice surface
In Case A, we know that N = W.
Case 1
In Case 1, the force F is pushing down
In Case B, due to the angle of
(in addition to mg), so the normal force
the incline, N < W. In fact, we
needs to be larger. In Case 2, the force F
can see that N = W cos(θ).
is pulling up, against gravity, so the
m
forces on it are N (up) and mg (down),
on it are N (up) and mg (down), so N = mg,
Below you see two cases: a
physics student pulling or
pushing a sled with a force F
which is applied at an angle θ.
In which case is the normal
force greater?
N
it must have a net upward force. The
it must have no net force on it. The forces
y
N
f
x
Case 2
θ Wy
normal force is lessened.
θ
W
Page 2
2
ConcepTest 5.9 Friction
ConcepTest 5.10 Antilock Brakes
on a frictionless truck bed.
1) the force from the rushing air
pushed it off
When the truck accelerates
2) the force of friction pushed it off
forward, the box slides off
3) no net force acted on the box
the back of the truck
4) truck went into reverse by accident
A box sits in a pickup truck
because:
Antilock brakes keep the
car wheels from locking
and skidding during a
sudden stop. Why does
this help slow the car
down?
1) μk > μs so sliding friction is better
2) μk > μs so static friction is better
3) μs > μk so sliding friction is better
4) μs > μk so static friction is better
5) none of the above
5) none of the above
Generally, the reason that the box in the truck bed would move
Static friction is greater than sliding friction, so
with the truck is due to friction between the box and the bed.
by keeping the wheels from skidding, the static
friction force will help slow the car down more
If there is no friction, there is no force to push the box along,
efficiently than the sliding friction that occurs
and it remains at rest. The truck accelerated away, essentially
during a skid.
leaving the box behind!!
ConcepTest 5.11 Going Sledding
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
ConcepTest 5.12 Will it Budge?
1) pushing her from behind
3) both are equivalent
easiest way to
4) it is impossible to move the sled
accomplish this?
5) tell her to get out and walk
In Case 1, the force F is pushing down
3) moves up
4) moves down
5) the box does not move
maximum of μsN = 40 N. The
force is larger. In Case 2, the force F
Static friction
(μs = 0.4 )
tension in the rope is only 30 N.
1
is pulling up, against gravity, so the
enough to overcome friction.
the frictional force is proportional to
2
the normal force.
Follow-up: What happens if the tension is 35 N? What about 45 N?
ConcepTest 5.13a Sliding Down I
ConcepTest 5.13b Sliding Down II
A mass m is placed on an
inclined plane (μ > 0) and
slides down the plane with
constant speed. If a similar
block (same μ) of mass 2m
were placed on the same
incline, it would:
1) component of the gravity force
parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board
decreased
4) both #1 and #3
5) all of #1, #2 and #3
1) not move at all
2) slide a bit, slow down, then stop
3) accelerate down the incline
4) slide down at constant speed
5) slide up at constant speed
As the angle increases, the component
The component of gravity acting down
of weight parallel to the plane increases
N
f
the plane is double for 2m. However,
and the component perpendicular to the
the normal force (and hence the friction
Normal
plane decreases (and so does the Normal
force) is also double (the same factor!).
force). Since friction depends on Normal
This means the two forces still cancel
Net Force
force, we see that the friction force gets
smaller and the force pulling the box
T
m
So the pulling force is not big
normal force is lessened. Recall that
z
2) moves to the right
The static friction force has a
(in addition to mg), so the normal
A box sits on a flat board.
You lift one end of the
board, making an angle
with the floor. As you
increase the angle, the
box will eventually begin
to slide down. Why?
1) moves to the left
A box of weight 100 N is at
rest on a floor where μs = 0.5.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
2) pulling her from the front
to give a net force of zero.
Weight
Wy
θ
Wx
W
θ
down the plane gets bigger.
Page 3
3
ConcepTest 5.14 Tetherball
In the game of tetherball,
the struck ball whirls
around a pole. In what
direction does the net
force on the ball point?
ConcepTest 5.15a Around the Curve I
1) Toward the top of the pole
You are a passenger in a
car, not wearing a seat
belt. The car makes a
sharp left turn. From
your perspective in the
car, what do you feel is
happening to you?
2) Toward the ground
3) Along the horizontal component of
the tension force
4) Along the vertical component of the
tension force
5) Tangential to the circle
The vertical component of the
T
T
(4) You are thrown to the ceiling
(5) You are thrown to the floor
line. From your perspective in the
car, it feels like you are being
thrown to the right, hitting the
points toward the center of the
passenger door.
W
ConcepTest 5.15b Around the Curve II
During that sharp left turn,
you found yourself hitting
the passenger door. What
is the correct description of
what is actually happening?
(3) You are thrown to the left
to continue moving in a straight
W
provides the centripetal force that
circle.
(2) You feel no particular change
The passenger has the tendency
tension balances the weight. The
horizontal component of tension
(1) You are thrown to the right
ConcepTest 5.15c Around the Curve III
(1) centrifugal force is pushing you
into the door
You drive your dad’s car
too fast around a curve
and the car starts to skid.
What is the correct
description of this
situation?
(2) the door is exerting a leftward
force on you
(3) both of the above
(4) neither of the above
(1) car’s engine is not strong enough to
keep the car from being pushed out
(2) friction between tires and road is not
strong enough to keep car in a circle
(3) car is too heavy to make the turn
(4) a deer caused you to skid
(5) none of the above
The friction force between tires and
The passenger has the tendency
road provides the centripetal force
to continue moving in a straight
that keeps the car moving in a circle.
line. There is a centripetal force,
If this force is too small, the car
provided by the door, that forces
continues in a straight line!
the passenger into a circular path.
Follow-up: What could be done to
the road or car to prevent skidding?
ConcepTest 5.16 Missing Link
ConcepTest 5.17 Ball and String
1) T2 = 1/4 T1
A ping pong ball is shot into a
Two equal-mass rocks tied to strings are
circular tube that is lying flat
whirled in horizontal circles. The radius of
(horizontal) on a tabletop. When
circle 2 is twice that of circle 1. If the period
3) T2 = T1
of motion is the same for both rocks, what
4) T2 = 2 T1
is the tension in cord 2 compared to cord 1?
5) T2 = 4 T1
the ping pong ball leaves the
track, which path will it follow?
2) T2 = 1/2 T1
The centripetal force in this case is given by the
z
Once the ball leaves the tube, there is no longer
tension, so T = mv2/r. For the same period, we find
a force to keep it going in a circle. Therefore, it
that v2 = 2v1 (and this term is squared). However, for
simply continues in a straight line, as Newton’s
the denominator, we see that r2 = 2r1 which gives us
First Law requires!
the relation T2 = 2T1.
Follow-up: What physical force provides the centripetal force?
Page 4
4
ConcepTest 5.18 Barrel of Fun
A rider in a “barrel of fun”
finds herself stuck with
her back to the wall.
Which diagram correctly
shows the forces acting
on her?
ConcepTest 5.19a Going in Circles I
You’re on a Ferris wheel moving in a
vertical circle. When the Ferris wheel is
at rest, the normal force N exerted by
1
3
2
4
5
1) N remains equal to mg
2) N is smaller than mg
your seat is equal to your weight mg.
3) N is larger than mg
How does N change at the top of the
4) None of the above
Ferris wheel when you are in motion?
The normal force of the wall on the
rider provides the centripetal force
You are in circular motion, so there
needed to keep her going around
has to be a centripetal force pointing
in a circle. The downward force of
inward. At the top, the only two
gravity is balanced by the upward
forces are mg (down) and N (up), so
frictional force on her, so she does
N must be smaller than mg.
not slip vertically.
Follow-up: Where is N larger than mg?
Follow-up: What happens if the rotation of the ride slows down?
ConcepTest 5.19b Going in Circles II
ConcepTest 5.19c Going in Circles III
A skier goes over a small round hill
1) Fc = N + mg
You swing a ball at the end of string
1) Fc = T – mg
with radius R. Since she is in circular
2) Fc = mg – N
in a vertical circle. Since the ball is
2) Fc = T + N – mg
motion, there has to be a centripetal
3) Fc = T + N – mg
force. At the top of the hill, what is
4) Fc = N
Fc of the skier equal to?
Fc points toward the center of
the circle, i.e., downward in
this case. The weight vector
points down and the normal
force (exerted by the hill)
points up. The magnitude of
the net force, therefore, is:
Fc = mg – N
in circular motion there has to be a
centripetal force. At the top of the
ball’s path, what is Fc equal to?
4) Fc = T
5) Fc = mg
5) Fc = mg
v
Fc points toward the center of the
circle, i.e. downward in this case. The
mg
3) Fc = T + mg
N
weight vector points down and the
v
mg
T
tension (exerted by the string) also
R
points down. The magnitude of the
net force, therefore, is:
Follow-up: What happens when
the skier goes into a small dip?
Fc = T + mg
R
Follow-up: What is Fc at the bottom of the ball’s path?
Page 5
5