Download Lecture 8 - UD Physics

Phys 207
Announcements
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Hwk3 submission deadline = 1st thing Monday morning
Quiz 2 average score = 1.7/4 (N=110)
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Today’s Agenda
More discussion of dynamics
ÍRecap
Í The Free Body Diagram
ÍThe tools we have for making & solving problems:
» Ropes & Pulleys (tension)
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Topics
1. Position, speed, acceleration
2. Motion – constant acceleration
3. Scalar vs. Vector
4. Decomposing vectors into components along x, y, z
5. Adding subtracting vectors
6. Decompose 2D motion into x and y components
7. Tangential vs. radial velocity, acceleration
8. Applying 1D kinematics to x or y motion
9. Centripetal acceleration
10.Systems of equations for solving 2D motion
problems
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TA Office Hours
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R: Philip Castro / 7:00 – 9:00 pm / 227 SHL
F: Dan Pajerowski / 1:30 – 3:30 pm / 131B SHL
F: Xing Chen / 5:00 – 7:00 pm / 225 SHL (Commons
Room)
T,R: Nowak / 9:30 – 10:30am / 228 SHL
R: Nowak / 1:30 – 2:30pm / 228 SHL
and by appt.
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Review: Newton's Laws
Law 1: An object subject to no external forces is at rest or
moves with a constant velocity if viewed from an
inertial reference frame.
Law 2: For any object, FNET = ma
Where FNET = Σ F
Law 3: Forces occur in action-reaction pairs, FA ,B = - FB ,A.
Where FA ,B is the force acting on object A due to its
interaction with object B and vice-versa.
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Gravity:
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What is the force of gravity exerted by the earth on a typical
physics student?
ÍTypical student mass m = 55kg
Íg = 9.81 m/s2.
ÍFg = mg = (55 kg)x(9.81 m/s2 )
ÍFg = 540 N = WEIGHT
FE,S = Fg = mg
FS,E = -mg
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Lecture 8, Act 1
Mass vs. Weight
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An astronaut on Earth kicks a bowling ball and hurts his foot.
A year later, the same astronaut kicks a bowling ball on the
moon with the same force.
Ouch!
His foot hurts...
(a)
more
(b)
less
(c)
the same
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Lecture 8, Act 1
Solution
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The masses of both the bowling
ball and the astronaut remain the
same, so his foot will feel the same
resistance and hurt the same as
before.
Ouch!
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Lecture 8, Act 1
Solution
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However the weights of the
bowling ball and the astronaut are
less:
W = mgMoon
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Wow!
That’s light.
gMoon < gEarth
Thus it would be easier for the
astronaut to pick up the bowling
ball on the Moon than on the
Earth.
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The Free Body Diagram
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Newton’s 2nd Law says that for an object F = ma.
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Key phrase here is for an object.
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So before we can apply F = ma to any given object
we isolate the forces acting on this object:
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The Free Body Diagram...
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Consider the following case
ÍWhat are the forces acting on the plank ?
P = plank
F = floor
W = wall
E = earth
FW,P
FP,W
FF,P
FP,F
FE,P
FP,E
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The Free Body Diagram...
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Consider the following case
ÍWhat are the forces acting on the plank ?
Isolate the plank from
the rest of the world.
FW,P
FP,W
FF,P
FE,P
FP,F
FP,E
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The Free Body Diagram...
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The forces acting on the plank should reveal themselves...
FW,P
FF,P
FE,P
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Aside...
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In this example the plank is not moving...
ÍIt is certainly not accelerating!
ÍSo FNET = ma becomes FNET = 0
FW,P
FW,P + FF,P + FE,P = 0
FF,P
FE,P
ÍThis is the basic idea behind statics, which we will
discuss in a few weeks.
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Example
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Example dynamics problem:
A box of mass m = 2 kg slides on a horizontal frictionless
floor. A force Fx = 10 N pushes on it in the x direction.
What is the acceleration of the box?
y
F = Fx i
a=?
m
x
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Example...
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Draw a picture showing all of the forces
y
FF,B
F
x
FB,F
FE,B
FB,E
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Example...
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Draw a picture showing all of the forces.
Isolate the forces acting on the block.
y
FF,B
F
x
FB,F
FE,B = mg
FB,E
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Example...
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z
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Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
y
FF,B
F
x
mg
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Example...
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Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
Solve Newton’s equations for each component.
y
Í FX = maX
Í FF,B - mg = maY
FF,B
x
F
mg
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Example...
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FX = maX
Í So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
FF,B - mg = maY
Í But aY = 0
Í So FF,B = mg.
N
y
FX
x
mg
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The vertical component of the force
of the floor on the object (FF,B ) is
often called the Normal Force (N).
Since aY = 0 , N = mg in this case.
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Example Recap
N = mg
y
FX
aX = FX / m
mg
x
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Lecture 8, Act 2
Normal Force
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A block of mass m rests on the floor of an elevator that is
accelerating upward. What is the relationship between
the force due to gravity and the normal force on the block?
(a) N > mg
(b) N = mg
a
(c) N < mg
m
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Lecture 8, Act 2
Solution
All forces are acting in the y direction,
so use:
N
Ftotal = ma
a
m
N - mg = ma
mg
N = ma + mg
therefore N > mg
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Tools: Ropes & Strings
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Can be used to pull from a distance.
Tension (T) at a certain position in a rope is the magnitude
of the force acting across a cross-section of the rope at that
position.
ÍThe force you would feel if you cut the rope and
grabbed the ends.
ÍAn action-reaction pair.
T
cut
T
T
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Tools: Ropes & Strings...
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Consider a horizontal segment of rope having mass m:
ÍDraw a free-body diagram (ignore gravity).
m
T1
a
T2
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Using Newton’s 2nd law (in x direction):
FNET = T2 - T1 = ma
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So if m = 0 (i.e. the rope is light) then T1 = T2
x
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Tools: Ropes & Strings...
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An ideal (massless) rope has constant tension along the
rope.
T
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T
If a rope has mass, the tension can vary along the rope
Í For example, a heavy rope
hanging from the ceiling...
T = Tg
T=0
We will deal mostly with ideal massless ropes.
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Tools: Ropes & Strings...
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The direction of the force provided by a rope is along the
direction of the rope:
T
Since ay = 0 (box not moving),
m
T = mg
mg
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Lecture 8, Act 3
Force and acceleration
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A fish is being yanked upward out of the water using a fishing
line that breaks when the tension reaches 180 N. The string
snaps when the acceleration of the fish is observed to be is
12.2 m/s2. What is the mass of the fish?
snap !
(a) 14.8 kg
(b) 18.4 kg
(c)
a = 12.2 m/s2
8.2 kg
m=?
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Lecture 8, Act 3
Solution:
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Draw a Free Body Diagram!!
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Use Newton’s 2nd law
in the upward direction:
a = 12.2 m/s2
T
m=?
FTOT = ma
T - mg = ma
mg
T = ma + mg = m(g+a)
m=
T
g +a
m =
(9 . 8
180 N
+ 12 . 2 ) m s
2
= 8 . 2 kg
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Tools: Pegs & Pulleys
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Used to change the direction of forces
ÍAn ideal massless pulley or ideal smooth peg will
change the direction of an applied force without altering
the magnitude:
F1
| F1 | = | F2 |
ideal peg
or pulley
F2
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Tools: Pegs & Pulleys
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Used to change the direction of forces
ÍAn ideal massless pulley or ideal smooth peg will
change the direction of an applied force without altering
the magnitude:
FS,W = mg
T
T = mg
m
mg
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Lecture 8, Act 4
Force and acceleration
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A block weighing 4 lbs is hung from a rope attached to a
scale. The scale is then attached to a wall and reads 4 lbs.
What will the scale read when it is instead attached to
another block weighing 4 lbs?
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m
m
m
(2)
(1)
(a)
0 lbs.
(b) 4 lbs.
(c)
8 lbs.
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Lecture 8, Act 4
Solution:
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Draw a Free Body Diagram of one
of the blocks!!
Use Newton’s 2nd Law
in the y direction:
T
m
T = mg
a = 0 since the blocks are
stationary
mg
FTOT = 0
T - mg = 0
T = mg = 4 lbs.
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Lecture 8, Act 4
Solution:
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The scale reads the tension in the rope, which is T = 4 lbs in
both cases!
T
T
T
T
m
T
T
T
m
m
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