Download ch 5 - Applying Newton`s Laws

Chapter 5
Applying Newton’s
Laws
PowerPoint® Lectures for
University Physics, Twelfth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by James Pazun
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Goals for Chapter 5
• To use and apply Newton’s First Law
• To use and apply Newton’s Second Law
• To study friction and fluid resistance
• To consider forces in circular motion
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Introduction
• It’s not hard to state Newton’s
laws well; how should we apply
them to everyday situations?
• We can start with forces
balanced, nothing moving
(statics).
• Next, we’ll study unbalanced
forces causing motion
(dynamics).
• We’ll end Chapter 5 considering
the role of forces in circular
motion.
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Using Newton’s First Law when forces are in equilibrium
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5.1 using Newton's 1st law: particles in equilibrium
• A body is in equilibrium when it is at rest or moving
with constant velocity in an inertial frame of
reference.
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Example 5.1: tension in a massless rope
• A gymnast with mass mG = 50.0 kg suspends herself
from the lower end of a hanging rope. The upper end of
the rope is attached to the gymnasium ceiling.
a.What is the gymnast’s weight? 490 N
b.What force (magnitude and direction) does the rope exert
on her? 490 N
c. What is the tension at the top of the rope? Assume that
490 N
the mass of the rope itself is negligible.
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Example 5.2: tension in a rope with mass
• Suppose that in example 5.1, the weight of
the rope is not negligible but is 120 N. find the
tension at each end of the rope.
• Lower end of the
rope: 490 N
• Upper end of
the rope: 610 N
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Example 5.3: two-dimensional equilibrium
• A car’s engine with weight w hangs from a chain that is
linked at ring O to two other chains, one fastened to the
ceiling and the other to the wall. Find the tension in each
of the three chains in terms of w, the weights of the ring
and chains are negligible.
Engine:T1 = w
T3 = 1.155w
T2 = 0.577w
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Example 5.4: an inclined plane
• A car of weight w rests on a slanted ramp leading to a
car-transporter trailer. Only a cable running from the
trailer to the car prevents the car from rolling backward
off the ramp. Find the tension in the cable and the force
that the tracks exert on the car’s tires.
T  w sin 
n  w cos 
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Example 5.5 – tension over a frictionless pulley
• Blocks of granite are to be hauled up a 15o slope out of a quarry and
dirt is to be dumped into the quarry to fill up old holes to simplify the
process, you design a system in which a granite block on a cart with
steel wheels (weight w1, including both block and cart) is pulled uphill
on steel rails by a dirt-filled bucket (weight w2, including both dirt and
bucket) dropping vertically into the quarry. How must the weights w1
and w2 be related in order for the system to move constant speed?
Ignore friction in the pulley and wheels and the weight of the cable.
Block and cart: w2=0.26w1
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Test your understanding 5.1
A traffic light of weight w hangs from two
lightweight cables, one on each side of the
light. Each cable hands at a 45o angle from the
horizontal. What is the tension in each cable?
a. w/2
b. w/√2
T
T
45o
45o
c. w
d. w√2
e. 2w
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w
5.2 using Newton's 2nd law: dynamics of particles
Caution: ma doesn’t belong in free-body diagrams.
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Example 5.6 Straight-line motion with a constant force
• An ice boat is at rest on a perfectly frictionless horizontal
surface. A wind is blowing (along the direction of the
runners) so that 4.0 s after the iceboat is release, it attains
a velocity of 6.0 m/s. what constant horizontal force Fw does
the wind exert on the iceboat? The mass of iceboat and
rider is 200 kg.
Fw = 300 N
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Example 5.7 Straight-line motion with friction
• Suppose a constant horizontal friction force with magnitude
100 N opposes the motion of the iceboat in Example 5.6. In
this case, what constant force FW must the wind exert on
the iceboat to cause the same constant x-acceleration
ax=1.5 m/s2?
Fw = 400 N
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Example 5.8 Tension in an elevator cable
• An elevator and its load have a total mass of 800 kg. The elevator
is originally moving downward at 10.0 m/s; it slows to a stop with
constant acceleration in a distance of 25.0 m. Find the tension T in
the supporting cable while the elevator is being brought to rest.
T = 9440 N
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Example 5.9 Apparent weight in an accelerating elevator
• A 50.0 kg woman stands on a bathroom scale while
riding in the elevator in Example 5.8. What is the
reading on the scale?
n = 390 N
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Apparent weight and apparent weightlessness
• When a passenger with mass m rides in an elevator with y-acceleration ay,
a scale shows the passenger’s apparent weight to be
n = m∙(g + ay)
• When the elevator is accelerating upward, ay is positive and n is greater
than the passenger’s weight w = mg. when the elevator is accelerating
downward, ay is negative and n is less than the weight.
• When ay = g, the elevator is in free fall, n = 0 and the passenger seems to
be weightless.
• Similarly, an astronaut orbiting the earth in a spacecraft experiences
apparent weightlessness.
• In each case, the person is not truly weightless because there is still a
gravitational force acting. But the person's sensation in this free-fall
condition are exactly the same as though the person were in outer space
with no gravitational force at all.
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Example 5.10 Acceleration down a hill
• A toboggan loaded with vacationing students (total weight
w) slides down a long, snow-covered slope. The hill slopes
at a constant angle α, and the toboggan is so well waxed
that there is virtually no friction. What is its acceleration?
a = g sinα
n = mg cosα
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Watch for this common error—Figure 5.13
• A good “road sign” is to be sure that the normal
force comes out perpendicular to the surface.
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Example 5.11 Two bodies with same acceleration
• You push a 1.00 kg food tray through the cafeteria line with a
constant 9.0 N force. As the tray moves, it pushes on a 0.50 kg
carton of milk. The tray and carton slide on a horizontal surface
that is so greasy that friction can be neglected. Find the
acceleration of the tray and carton and the horizontal force that the
tray exerts on the carton.
a = 6.0 m/s2
F T on C = 3.0 N
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Double Trouble (a.k.a., Two Body Problems)
• Two body-problems can typically be approached using
one of two basic approaches.
• One approach is the system analysis, the two objects are
considered to be a single object moving (or accelerating)
together as a whole.
• Another approach is the individual object analysis, either
one of the two objects is isolated and considered as a
separate, independent object.
Sometimes, you need to combine both approaches to solve a problem
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Example 5.12 Two bodies with the same magnitude of acceleration
• An air-track glider with mass m1 moving on a level,
frictionless air track in the physics lab. The glider is
connected to a lab weight with mass m2 by a light, flexible,
non stretching string that passes over a small frictionless
pulley. Find the acceleration of each body and the tension
in the string.
a
m2
g
m1  m2
T
m1m2
g
m1  m2
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Test Your Understanding 5.2
• Suppose you hold the glider in Example 5.12 so
that it and the weight are initially at rest. You give
the glider a push to the left and then release it. The
string remains taut as the glider moves to the left,
comes instantaneously to rest, than moves to the
right. At the instant the glider has zero velocity,
what is the tension in the string?
a. Greater than in Example 5.12
b. The same as in Example 5.12
c. Less than in Example 5.12, but greater than zero
d. zero
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5.3 Frictional forces, kinetic and static
The friction and normal forces are
really components of a single
contact force.
• Friction can keep an object
from moving or slow its
motion.
• Microscopic imperfections
cause non ideal motion.
• The direction of friction force is
opposite of the direction of
motion.
• Lubricant are used to reduce
friction.
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Kinetic Friction
• The kind of friction that acts when a body slides over a
surface is called kinetic friction force fk. The magnitude of
the kinetic friction force can be represented by the
equation:
f k  k n
μk is a constant called the coefficient of kinetic friction. μk
depends on the types of surface. Because it is a ratio of
two force magnitudes, μk is a pure number, without units.
The equation is only an approximate representation of a
complex phenomenon. As a box slides over the floor,
bonds between the two surfaces form and break, and the
total number of such bonds varies; hence the kinetic friction
force is not perfectly constant.
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Static Friction
• Friction forces may also act when there is no
relative motion. If you try to slide a box across
the floor, the box may not move at all because
the floor exerts an equal and opposite friction
force on the box. This is called a static friction
force fs.
f s  s n
Both equations of friction are relationships between
magnitudes, not vector relationships.
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Coefficients of friction
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Applied force is proportional until the object moves—Figure 5.19
• Notice the transition between static and kinetic friction.
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Example 5.13 Friction in horizontal motion
• You are trying to move a 500 N crate across a level floor.
To start the crate moving, you have to pull with a 230 N
horizontal force. Once the crate “breaks loose” and starts
to move, you can keep it moving at constant velocity with
only 200 N. What are the coefficients of static and kinetic
friction?
 s  0.46
 k  0.40
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5.14 Static friction can be less than the maximum
• In Example 5.13, what is the friction force if the
crate is at rest on the surface and a horizontal
force of 50 N is applied to it?
f s  50 N
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Example 5.15 Minimizing kinetic friction
• In Example 5.13, suppose you try to move the crate by
tying a rope around it and pulling upward on the rope at
an angle of 30o above the horizontal. How hard do you
have to pull to keep the crate moving with constant
velocity? Is this easier or harder than pulling
horizontally? Assume w = 500 N and µk = 0.40.
T = 188 N
The tension
required is less.
n = 406 N
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Example 5.16 Toboggan ride with friction I
• Let’s go back to Example 5.10. the wax has worn off
and there is now a nonzero coefficient of kinetic friction
µk. The slope has just the right angle to make the
toboggan slide with constant speed. Derive an
expression for the slope angle in terms of w and µk.
  arctan k
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Example 5.17 Toboggan ride with friction II
• The same toboggan with same coefficient of
friction as in example 5.16 accelerates down a
steeper hill. Derive and expression for the
acceleration in terms g, a, µk, and w.
a  g (sin   k cos  )
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Rolling friction
• It’s a lot easier to move a loaded filing cabinet
across a horizontal floor using a cart with wheels
than to slide it.
• We can define a coefficient of rolling friction μr ,
the tractive resistance. Typical values of μr are
0.002 to 0.003 for steel wheels on steel rails and
0.01 to 0.02 for rubber tires on concrete.
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Example 5.18 Motion with rolling friction
• A typical car weighs about 12,000 N. if the
coefficient of rolling friction is µr = 0.015, what
horizontal force is needed to make the car move
with constant speed on a level road? Neglect air
resistance.
f r  180 N
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Fluid Resistance and Terminal speed
•
Fluid resistance is the force that a fluid (a gas or liquid) exerts on a
body moving through. The direction of the fluid resistance force
acting on body is always opposite the direction of the body’s
velocity relative to the fluid.
•
The magnitude of the fluid resistance force usually increase with
speed of the body through the fluid.
•
For very low speed, the magnitude f of the fluid resistance force
can be described using the following equation
f  kv
Fluid resistance at low speed
Where k is a proportionality constant that depends on the shape
and size of the body and the properties of the fluid. The unit of k
is N∙s/m or kg/s
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Air drag
• In motion through air at the speed of a tossed tennis ball or
faster, the resisting force is approximately proportional to v2
rather than v. It is then called air drag or simply drag.
Airplanes, falling raindrops, and bicyclists all experience air
drag. The air drag on a typical car is negligible at low speeds
but comparable to or greater than rolling friction at highway
speeds.
f  Dv 2
Fluid resistance at high speed
The value of D depends on the shape and size of the body
and on the density of the air. The units of the constant D are
N∙s2/m2 or kg/m
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• Because of the effects of fluid resistance, an object falling in a
fluid does not have a constant acceleration.
• To find acceleration at a point of time, we need to use Newton’s
second law.
Terminal velocity – slow moving object
• When the rock first starts to move, vy = 0, the resisting
force is zero, and the initial acceleration is ay = g. As the
speed increases, the air resistance force also increases,
until finally it is equal in magnitude to the weight. At this
time mg – k∙vy = 0, the acceleration becomes zero, and
there is no further increase in speed. The final speed vt,
called the terminal speed is given by mg – k∙vt = 0 or
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Terminal velocity – fast moving object
∑Fy = mg + (-Dvy2) = may
• When the rock first starts to move, vy = 0, the resisting
force is zero, and the initial acceleration is ay = g. As the
speed increases, the air resistance force also increases,
until finally it is equal in magnitude to the weight. At this
time mg – D∙vy2 = 0, the acceleration becomes zero, and
there is no further increase in speed. The final speed vt,
called the terminal speed is given by mg – D∙vy2 = 0, or
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Graphs of the motion of a body falling with fluid
resistance proportional to the speed.
5.19 Determine the terminal speed of a sky diver
• For a human body falling
through air in a spreadeagle position, the numerical
value of the constant D is
about 0.25 km/m. Find the
terminal speed for a
lightweight 50 kg skydiver.
vt = 44 m/s
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Test Your Understanding 5.3
Consider a box that is placed on different surfaces.
a. In which situation(s) is there no friction force acting on the box?
b. In which situations(s) is there a static friction force acting on the box?
c. In which situation(s) is there a kinetic friction force on the box?
i. The box is at rest on a rough horizontal surface.
ii. The box is at rest on a rough tilted surface.
iii. The box is on the rough-surfaced flat bed of a truck, the tuck is moving at
a constant velocity on a straight, level road, and the box remains in the
same place in the middle of the truck bed.
iv. The box is on the rough-surfaced flat bed of a truck, the truck is speeding
up on a straight, level road, and the box remains in the same place in the
middle of the truck bed.
v. The box is on the rough surfaced flat bed of a truck, the truck is climbing
a hill, the box is sliding toward the back of the truck.
a.
b.
c.
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i, iii
ii, iv
v
The dynamics of uniform circular motion
• In uniform circular
motion, both the
acceleration and force
are centripetal.
• Cut the cord
restraining an object in
such motion and
observe the object’s
behavior without the
inward force.
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Equations in Uniform Circular Motion
v2
ac 
r
Fnet
v2
 Fc  mac  m
r
2r
T
v
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4 2 r
ac  2
T
Example 5.20 Force in uniform circular motion
• A sled with a mss of 25.0 kg rests on a horizontal sheet of
essentially frictionless ice. It is attached by a 5.00 m rope to a
post set in the ice. Once given a push, the sled revolves uniformly
in a circle around the post. If the sled make five complete
revolutions every minute, find the force F exerted on it by the
rope.
F  34.4N
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Example 5.21 The conical pendulum
• An inventor proposes to make a pendulum clock using a
pendulum bob with mass m at the end of a thin wire of length L.
instead of swinging back and for, the bob moves in a horizontal
circle with constant speed v, with the wire making a constant
angle β with the vertical direction. This system is called a conical
pendulum because the suspending wire traces out a cone. Find the
tension F in the wire and the period T in terms of β, g and L.
mg
F
cos 
L cos 
T  2
g
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Example 5.22 Rounding a flat curve
• A sports car is rounding a flat, unbanked curve with
radius R. if the coefficient of static friction between tires
and roads is µs, what is the maximum speed vmax at
which the driver can take the curve without sliding?
vmax   s gR
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Example 5.3 Rounding a banked curve
• For a car traveling at a certain speed, it is possible to bank a curve
at just the right angle so that no friction at all is needed to
maintain the car’s turning radius. Then a car can safely round the
curve even on wet ice. Your engineering firm plans to rebuild the
curve in Example 5.22 so that a car moving at speed v can safely
make the turn even with no friction. At what angle β should the
curve be banked?
v2
  arctan
gR
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Banked curves and the flight of airplanes
• Example 5.23 also
apply to an airplane
when it makes a turn in
level flight.
• Lcosβ = mg
• Lsinβ=mv2/R
• tanβ = v2/gR
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Examining a misnomer—Figure 5.30
• People have adopted
the pop-culture use of
“centrifugal force” but
it really results from
reference frames.
• It is fictional and
results from a car
turning while a person
continues in straightline motion (for
example).
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Example 5.24 Uniform circular motion in a vertical circle
• A passenger on a carnival Ferries wheel moves in a vertical circle
of radius R with constant speed v. the seat remains upright during
the motion. Find expressions for the force the seat exerts on the
passenger at the top of the circle and at the bottom.
nTop
v2
 m( g  )
R
v2
nBottom  m( g  )
R
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Test Your Understanding 5.4
• Satellites are held in orbit by the force of our planet’s
gravitational attraction. A satellite in a small-radius orbit moves at
a higher speed than a satellite in an orbit of large radius. Based on
this information, what you can conclude about the earth’s
gravitational attraction for the satellite?
i.
It increases with increasing distance from the earth.
ii. It is the same at all distances from the earth.
iii. It decreases with increasing distance from the earth
iv. This information by itself isn’t enough to answer the question.
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5.5 The fundamental Forces of Nature
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