Download Chapter 6 - Force along a line

Chapter 6. Dynamics I: Motion Along a Line
Chapter Goal: To learn how
to solve problems about
motion in a straight line.
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Student Learning Objectives – Ch. 6
• To draw and make effective use of free-body
diagrams.
• To recognize and solve simple equilibrium
problems.
• To distinguish mass, weight, and apparent weight.
• To learn and use simple models of friction.
• To apply the full strategy for force and motion
problems to problems in single-particle dynamics.
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Equilibrium
An object on which the net force is zero is said to
be in equilibrium.
•Static equilibrium: object is at rest.
•Dynamic equilibrium: moving along a straight
line with constant velocity.
Both are identical from a Newtonian perspective
because the net force and the acceleration are zero.
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Problem-Solving Strategy: Equilibrium
Problems
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Problem-Solving Strategy: Equilibrium
Problems
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Example Equilibrium Problem (#31)
A 500 kg piano is being lowered into position by a
crane while 2 people steady it with ropes pulling to
the side. Bob’s rope pulls left, 150 below horizontal,
with 500 N of tension. Ellen’s rope pulls right, 250
below horizontal.
a. What tension must Ellen maintain in her rope to
keep the piano descending at a steady speed?
b. What is the tension in the main cable?
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Example Problem (#31) – Freebody Diagram
T3
T1 is Bob’s side, T2 is
Ellen’s side, T3 is the
main cable. Same
system for the angles
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Problem #31 - Apply Newton’s 1st Law
T3
ΣFy = 0, ΣFx = 0
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Problem #31 – Solve and assess
T2 = 533 N, T3 = 5.25 x 103 N. The cable must
supports the weight of the piano (4900 N) plus the
added downward components of the tension in the
supporting ropes.
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When your car rapidly accelerates forward, you are
pressed into the seat. When your car suddenly comes
to a halt, you lunge forward. These two phenomena
are an example of:
A. Newton’s 1st Law; the sum of the forces upon you is
zero.
B. Newton’s 2nd Law; you experience a net force and
you accelerate
C. Newton’s 3rd Law; you experience a force equal and
opposite to that of the car
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Problem-Solving Strategy for Newton’s 2nd
Law Problems
1. Use the problem-solving strategy outlined for
Newton’s 1st Law problems to draw the free body
diagram and determine known quantities.
2. Use Newton’s Law in component form to find the
values for any individual forces and/or the
acceleration.
3. If necessary, the object’s trajectory (time, velocity,
position, acceleration) can be determined by using
the equations of kinematics.
4. Reverse # 2 and 3 if necessary.
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For which of the following is may =
F1 -F2 cosθ - F3 sin θ
A
C
B
D
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Example Dynamics Problem
A 75-kg snowboarder starts down a 50-m high, 100 slope
on a frictionless board. What is his speed at the
bottom?
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Example Dynamics Problem – Visualize
Freebody Diagram
A 75-kg snowboarder starts
down a 50-m high, 100 slope
on a frictionless board.
What is his speed at the
bottom?
n
FG
Find
v1
a
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Example Dynamics Problem – Newton’s 2nd Law in
component form to solve for acceleration
A 75-kg skier starts down a 50m high, 100 slope on a
frictionless board. What is
his speed at the bottom?
ΣFy = may = 0
ΣFx = max a = 1.7 m/s2
Supports earlier statement that
a = g sinθ
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n
FG
Example Dynamics Problem – Use kinematics
to find speed. Is time important?
A 75-kg snowboarder starts
down a 50-m high, 100 slope
on a frictionless board.
What is his speed at the
bottom?
Note (the slope is 50 m high,
not long!)
v1 = 31.3 m/s. That’s about 60
mph!
Find
v1
a = 1.7 m/s2, from
previous
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Use the graph to answer the question #1
The figure shows a force acting
on a 2.0-kg object moving
along the x-axis. The object
is at rest at the origin at t=0.
What is the acceleration of
the object at t = 2s?
A. 4 m/s2
B. 2 m/s2
C. 8 m/s2
D. 0 m/s2
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Use the graph to answer the question #2
The figure shows a force acting on a
2.0-kg object moving along the xaxis. The object is at rest at the
origin at t=0. What is the velocity
of the object at t = 6s?
A. v = 4m/s
B. v = -1 m/s
C. v = 0 m/s
D. v = 2 m/s
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Mass and Gravity
• Mass is a scalar quantity that describes the
amount of matter in an object.
• Mass is an intrinsic property of an object.
• The force of gravity is an attractive, long-range
“inverse square” force between any two objects.
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The Earth and the Moon
The figure shows the moon (m1)
and the earth (m2). The earth is
approximately 80 times as massive
as the moon. The red arrow shown
is the force that the earth exerts on
the moon (F2on1 ). The moon also
exerts a force on the earth, F1on2,
shown in blue (not to scale!). The
magnitude of this force is:
a.
b.
c.
d.
about 80 smaller than F2on1
somewhat smaller than F2on1
Equal to F2on1
Not related to F2on1.
moon
?

F1on2
earth
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Consider an object of mass m, on or near the surface of a
planet. We can write the gravitational force even more
simply as:
gravitational (weight )force
where the quantity g is defined to be
• M, R represent the mass and radius of the planet.
• The weight force is not an intrinsic property of an object
and does not have a unique value.
• The direction of the gravity vector defines true vertical.
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Apparent Weight
• Apparent weight w, is a contact force (e.g. T, n, or
Fsp), which can be thought of as “what the scale
says”, although there is not always a scale.
• If object and scale are in vertical static or dynamic
equilibrium w = FG = mg.
• If object and scale accelerate vertically, w ≠ mg. It
must be calculated using Newton’s 2nd Law.
• The use of w instead of n or T is optional, as long as
you know which force is the apparent weight.
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An elevator suspended by a cable is
moving upward and slowing to a stop. As
it does, your apparent weight is:
A. less than your true weight, which is mg.
B. equal to your true weight, which is mg.
C. more than your true weight, which is mg.
D. zero.
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Workbook Problem # 18
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Apparent Weight Problem
A 50-kg woman gets in a 1000-kg elevator at rest. The
elevator has a scale in it (I hate when that happens).
As the elevator begins to move, the scale reads 600
N for the first 3 seconds.
a. Can you tell which direction she moved? If so, what
is it?
b. How far has the elevator moved in those 3 s?
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Apparent Weight Problem (a)
•
Calculate the true weight (mg) and
draw a free-body diagram that
correctly shows relative vector lengths
•
In this case the scale reads heavy (I
hate when that happens) so net force
and acceleration are in the positive
direction.
•
Elevator must be going up and
speeding up or going down and
slowing down.
•
Only one is possible due to initial
velocity constraint.
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known:
v0 = 0 m/s
m = 50 kg
n = 600 N
g = 9.8 m/s2
Apparent Weight Problem (b)
How far has the elevator
moved in those 3 s?
y1, t1 = 3s v1
The pictorial representation
shows not enough
information to solve
problem, so we go to
Newton’s 2nd Law analysis
a0
0m
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y0 = t0 = v0 = 0
Apparent Weight Problem (b)
ΣFy = may
using known values:
n – mg = ma
a = 2.2 m/s2
a
known:
m = 50 kg
n = 600 N
g = 9.8 m/s2
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Apparent Weight Problem (b)
How far has the elevator moved in those
3 s?
y1, t1 = 3s v1
Time is important so use position
equation with v0 = 0m/s:
a0 = 2.2 m/s2, determined
from N’s 2nd Law
∆ y = ½ a ∆t12
∆ y = 9.9 m
0m
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y0 = t0 = v0 = 0
Friction
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Kinetic Friction
Experiments show that the
kinetic friction force is nearly
constant and
proportional to the magnitude
of the normal force.
where the proportionality
constant μk is called the
coefficient of kinetic friction
(table in text).
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Static Friction
The box is in static
equilibrium, so the static
friction must exactly balance
the pushing force:
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Static friction
• An object remains at rest as
long as fs < fs max
• The object slips when fs = fs max
• A static friction force fs > fs max
is not physically possible.
• fs max is always >fk for the
same surfaces
where the proportionality constant μs is
called the coefficient of static
friction.
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Comparison of static and kinetic friction
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Rolling Friction
• Rolling friction acts much
like kinetic friction, but
values for ur are much less
than those for uk.
• Rolling friction is a resistive
force. It is not the same as
the static friction that
provides the propulsion
force that move the wheel
forward.
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A model of friction
• “ motion” indicates motion relative to the two surfaces
• the max value static friction, fs max occurs at the very instant
the object begins to move (which often means 1 ns before, for
problem-solving purposes.
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Rank in order, from largest to smallest, the
magnitude of the friction forces in these five
different situations. The box and the floor are
made of the same materials in all situations.
The push force is not necessarily the same.
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Example kinetic friction problem
A 75-kg snowboarder starts down a 50-m high, 100 slope
with μk = 0.06. What is his speed at the bottom?
This is the same problem as before only the slope is no
longer frictionless. Before, the velocity was 31.3 m/s.
How does friction change that?
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Example kinetic friction problem
A 75-kg snowboarder starts
down a 50-m high, 100 slope
μs = 0.12 and μk = 0.06
What is his speed at the
bottom?
n
fk
FG
Find
v1
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Example kinetic friction problem
A 75-kg snowboarder starts
down a 50-m high, 100 slope
on a frictionless board.
What is his speed at the
bottom?
v1 = 25.4 m/s. Compare with
“frictionless” problem.
Friction acts to slow him
down, although not by
much.
Find
v1
a, from previous
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Example “max” problem
A truck is hauling a crate
when it starts up a 10.0˚
hill. The coefficients of
friction are μs = 0.35,
and μk = 0.15,
respectively. What is the
maximum acceleration
the truck can have as he
goes up the hill, without
the crate slipping
backward?
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Example static friction problem
n
fs
θ
Fg
known
θ = 10˚
us = .35
uk = .15
find
amax
When does amax occur?
Find n using Newton’s 2nd law in the y direction.
Find amax using Newton’s 2nd law in the x
direction.
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Example static friction problem
n
fs
θ
Fg
known
θ = 10˚
us = .35
uk = .15
amax =1.68 m/s/s
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find
amax
Keep the picture up
A person is trying to judge whether a picture of mass
1.10 kg is properly positioned by pressing it against a
wall. The pressing force is perpendicular to the wall.
The coefficient of static friction between picture and
wall is 0.660. What is the minimum amount of
pressing force required?
Draw a freebody diagram. In which direction is the
normal force in this problem? Does it have anything
to do with the weight of the picture?
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Keep the picture up Freebody diagram
fbd - picture
Knowns
m = 1.10 kg
μs = 0.660
fs
Fpush
n
Find
Fpush
Fg
Forces which are usually x are y in this problem and vice
versa (with the exception of gravity). Newton’s Laws
still work.
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Keep the picture up Freebody diagram
Knowns
m = 1.10 kg
μs = 0.660
Find
Fpush
fbd - picture
fs
Fpush
n
Fg
Newton’s Law in the x direction
tells us that n = Fpush but nothing
else about the value of either.
Moving right along to the ydirection:
ΣFy = may = 0 = fs – FG or fs = mg. No matter how hard
you press, the picture will not levitate up. Fact. However, if
you don’t push hard enough….
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Keep the picture up - answer
Knowns
m = 1.10 kg
μs = 0.660
Find
Fpush
fbd - picture
fs
Fpush
n
FG
The minimum value of n must be
the value that allows fsmax to be
equal to the weight of the picture:
ΣFy = 0 = fsmax – FG
or μs |n| = mg
n = Fpush = 16.3 N
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