Glencoe Physics Chapter 7 Forces and Motion in Two Dimensions

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Glencoe Physics Chapter 7Forces and Motion in
Two Dimensions
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Objectives 7.1

• Determine the force that produces equilibrium
  when three forces act on an object
• Analyze the motion of an object on an inclined
  plane with and without friction

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Objectives 7.2

• Recognize that the vertical and horizontal
  motions of a projectile are independent
• Relate the height, time in air (hang time), and
  initial vertical velocity of a projectile using
  its vertical motion , then determine the range
• Explain how the shape of the trajectory of a
  moving object depends upon the frame of reference
  from which it is observed

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Objectives 7.3

• Explain the acceleration of an object moving in a
  circle at a constant speed
• Describe how centripetal acceleration depends
  upon the objects speed and the radius of the
  circle
• Recognize the direction of the force that causes
  centripetal acceleration
• Explain how the rate of circular motion is
  changed by exerting torque on it

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7.1 Forces in Two Dimensions
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• We have already learned about how to find the
  resultant force of two forces acting on an
  object, as well as 3 or more forces.
• What we want to consider now are cases where Net
  Force acting on an object is zero, and the object
  is in Static equilibrium.

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Static Equilibrium

• Condition of an object when net forces equal zero
• Object is motionless

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Hanging sign f.b.d.
Free Body Diagram
Since the sign is not accelerating in any
direction, its in equilibrium. Since its not
moving either, we call it Static Equilibrium.
Thus, red green black 0.
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Equilibrant

• If you remove one force from a static condition,
  the system is no longer in static equilibrium
• The removed force is called the EQUILIBRANT.
• The equilibrant is the force, that if added to
  other forces, will bring the other forces into
  static equilibrium.
• Equilibrant is always equal to, but opposite, the
  sum of all the other force vectors

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What force represents the equilibrant of the
tensions of the strings?
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Example problem

• What is the tension in the ropes holding the mass
  in place?

22.5
T 22.5 m 168 N
168 N
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Solution

• Fnet 0 (system is in equilibrium)
• Fax -Fbx
• Fay Fby mg 0
• FaSin 22.5 FbSin 22.5 168 N
• 2Fa Sin 22.5 168 N
• Fa 220 N

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Weight of the Picture?
Weight is equal to F1y F2y So. 25N 25 N 50
N
What is the weight of this picture?
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Components Scalar Equations
If in Equilibrium..the following would be true
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Sample Problem

• A mother and daughter are outside playing on the
  swings. The mother pulls the daughter and swing
  (total mass 55.0 kg) back so that the swing makes
  an angle of 40.0 with the vertical (50.0 from
  horizontal)
• A. What is the tension in each chain holding the
  swing seat and the daughter?
• B. How hard did the mother have to pull to hold
  the daughter at that position?

A. 703N
B. 452N
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How to budge a stubborn mule
It would be pretty difficult to budge this mule
by pulling directly on his collar. But it would
be relatively easy to budge him by tying the rope
to a tree and then pulling up (or pushing
sideways) in the middle. Why would this work?????
The tension in the rope has two components, one
horizontal and one vertical
FT
Little Force
FTx
Big Force
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Sample Problem

• Is this a case of equilibrium?
• Calculate the magnitude of the net force

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Equilibrium or Motion along an Inclined Plane
Is the sled on the inclined plane in
equilibrium? What are the forces acting on the
sled? Draw a Free Body Diagram of the forces
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Inclined Plane

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Inclined Plane )
Be careful what ? you use in your
calculation..remember you must use the angle, as
measured from (X) (East)
Fgx
Fgy
This calculation will look different than what is
in your book (p. 152)but is easier to use if you
remember the above caution!!!!!
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The next 3 slides show the effect
Fn on block
of increasing the angle.
Fgx
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Fn on block
F gx
Fg on block
CD 5.3 p-22
CD 5.3 p 21
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Fn on block
F gx
Fg on block
CD 5.3 p-22
CD 5.3 p 21
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net force greatest at
acceleration greatest at
Speed increases, acceleration
A
A
A
decreases
B
C
Fe on block
CD 5.3 p 22
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Example p. 153 of your book.

• A 62 kg person on skis is going down a hill
  sloped at 37. The coefficient of kinetic
  friction between the skis and the snow is 0.15.
• A) What is the Horizontal component of the
  skiers weight?
• B) What is the Vertical component of the skiers
  weight?
• C) What is the Normal force acting on the skier?
• D) What is the Frictional force acting on the
  skier?
• E) What is the Net Force acting on the skier?
• F) What is the acceleration of the skier?
• G) How fast is the skier traveling after 5.0s,
  starting from

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7.2 Projectile Motion in Two Dimensions
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Fountain at Explora! Science Museum,
Albuquerque, NM
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What is Projectile Motion?

• Any object moving horizontally, with only the
  force of gravity acting on it, will exhibit
  projectile motion. (not moving straight up or
  down)
• An object displaying projectile motion will
  follow a trajectory, or a curved,
  parabolic-shaped path.

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Projectile Motion
At a given location on the earth and in the
absence of air resistance, all objects fall with
the same uniform acceleration. Thus, two objects
of different masses, dropped from the same
height, will hit the ground at the same time.
An objects horizontal and vertical motion are
independent. Object projected horizontally will
reach the ground in the same time as an object
dropped vertically. No matter how large the
horizontal velocity is, the downward pull of
gravity is always the same.
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Why does a projectile follow a parabolic path?
What happens to the horizontal displacement
during each successive time period? What happens
to the vertical displacement during each
successive time period?
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Horizontal motion.
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Vertical parabolic path
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Horizontal Vs. Vertical Velocities

• We can see from the diagrams that the horizontal
  velocity of a projectile remains constant (with
  no air resistance)
• We can also see that that vertical velocity is
  constantly changing, at a rate of 9.8 m/s2.

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Horizontal Vs. Vertical Velocities

• The horizontal and vertical velocities are also
  independent of each other, that is, they have no
  effect on each other.
• The only thing the horizontal and vertical
  motions have in common is that each motion takes
  exactly the same time to occur.

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Equations of Projectile Motion
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Working Projectile motion problems

• Separate the problem into two problemsvertical
  motion and horizontal motion
• Vertical motion is exactly that of an object
  dropped or thrown straight up or down. Use
  free-fall equations (or accelerated motion
  equations)
• Horizontal motion is constant velocity (Newtons
  1st Law), work this part using constant velocity
  equations.
• Both Vertical and horizontal motion are tied
  together by TIME. (TvTh)

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Sample Problem.

• A rock is thrown horizontally off a tall cliff at
  20.m/s. If the cliff is 70.m high
• A) How far horizontally from the bottom of the
  cliff does the rock land?
• B) What is the magnitude of the rocks total
  velocity at the instant before landing?

Another example in book on page 157
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Monkey Hunter

• A young native hunter in the Amazon is hunting
  monkeys with a bow and arrow. The hunter see a
  monkey sitting on a branch, and since the bow
  does not have a sight, the hunter pulls back the
  bow and points the arrow directly at the monkey.
  As the hunter releases the arrow the monkey
  notices the movement and drops off the branch.
  What happens next?

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Throw at monkey with no gravity
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Throw above monkey with gravity
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Throw at monkey fast with gravity
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Horizontal RangeWhat is the horizontal range of
the projectile?
Firstfind time using vertical motion..
Then, use that time to find range!
?Vy at

• dx vxt
• dx (v0 cos T)t

t ?vy a
Notice nice equation to use to find vertical time
in air!!!
- V0Sin25.0 -V0Sin25.0 -9.8m/s2
dx (v0Cos T) (V0 Sin T/4.9)
2.58s
dx (30.0 m/s Cos 25.0)(30.0 m/s Sin 25.0)/4.9
dx 70.4 m
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Alternate Range equationsame thing.
Range (dx) V02 Sin 2 T g
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Max height hang time depends only on initial
vertical velocity
Each initial velocity vector below has the a
different magnitude (speed) and the projectiles
will have different ranges (green the greatest),
but each object will spend the same time in the
air and reach the same max height. This is
because each vector has the same vertical
component.
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Max Range at 45?
Over level ground at a constant launch speed,
what angle maximizes the range, R ? First
consider some extremes When ? 0, R 0,
since as soon as the object is launched its back
on the ground. When ? 90?, the object goes
straight up and lands right on the launch site.
The best angle is 45?, smack dab between the
extremes.
All launch speeds are the same only the angle
varies.
76?
45?
38?