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Physics 111: Lecture 5, Pg 1
Physics 111: Lecture 5
Today’s Agenda
More discussion of dynamics
Recap
The Free Body Diagram
The tools we have for making & solving problems:
» Ropes & Pulleys (tension)
» Hooke’s Law (springs)
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Physics 111: Lecture 5, Pg 2
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 = 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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Physics 111: Lecture 5, Pg 3
Gravity:
What is the force of gravity exerted by the earth on a typical physics student?
Typical student mass m = 55kg
g = 9.81 m/s2.
Fg = mg = (55 kg)x(9.81 m/s2 )
Fg = 540 N = WEIGHT
FE,S = -mg
FS,E = Fg = mg
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Physics 111: Lecture 5, Pg 4
Lecture 5, Act 1 Mass vs. Weight
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. His foot hurts...
(a) more
(b) less
(c) the same
Ouch!
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Physics 111: Lecture 5, Pg 5
Lecture 5, Act 1 Solution
Ouch!
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.
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Physics 111: Lecture 5, Pg 6
Lecture 5, Act 1 Solution
Wow!
That’s light. However the weights of the
bowling ball and the astronaut are less:
Thus it would be easier for the astronaut to pick up the bowling ball on the Moon than on the Earth.
W = mgMoon gMoon < gEarth
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Physics 111: Lecture 5, Pg 7
The Free Body Diagram
Newton’s 2nd Law says that for an object F = ma.
Key phrase here is for an object.
So before we can apply F = ma to any given object we isolate the forces acting on this object:
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Physics 111: Lecture 5, Pg 8
The Free Body Diagram...
Consider the following case
What are the forces acting on the plank ?
P = plank
F = floor
W = wall
E = earth FW,P
FP,W
FP,F FP,E
FF,P FE,P
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Physics 111: Lecture 5, Pg 9
The Free Body Diagram...
Consider the following case
What are the forces acting on the plank ?
Isolate the plank from
the rest of the world.
FW,P
FP,W
FP,F FP,E
FF,P FE,P
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Physics 111: Lecture 5, Pg 10
The Free Body Diagram...
The forces acting on the plank should reveal themselves...
FP,W
FP,F FP,E
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Physics 111: Lecture 5, Pg 11
Aside...
In this example the plank is not moving...
It is certainly not accelerating!
So FNET = ma becomes FNET = 0
This is the basic idea behind statics, which we will discuss in a few weeks.
FP,W + FP,F + FP,E = 0
FP,W
FP,F FP,E
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Physics 111: Lecture 5, Pg 12
Example
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?
F = Fx i a = ?
m
y
x
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Physics 111: Lecture 5, Pg 13
Example...
Draw a picture showing all of the forces
F FB,F
FF,B FB,E
FE,B
y
x
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Physics 111: Lecture 5, Pg 14
Example...
Draw a picture showing all of the forces.
Isolate the forces acting on the block.
F FB,F
FF,B FB,E =
mg
FE,B
y
x
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Physics 111: Lecture 5, Pg 15
Example...
Draw a picture showing all of the forces.
Isolate the forces acting on the block.
Draw a free body diagram.
F FB,F
mg
y
x
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Physics 111: Lecture 5, Pg 16
Example...
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.
FX = maX
FB,F - mg = maY
F FB,F
mg
y
x
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Physics 111: Lecture 5, Pg 17
Example...
FX = maX
So aX = FX / m = (10 N)/(2 kg) = 5 m/s2.
FB,F - mg = maY
But aY = 0
So FB,F = mg.
The vertical component of the force of the floor on the object (FB,F ) is often called the Normal Force (N).
Since aY = 0 , N = mg in this case.
FX
N
mg
y
x
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Physics 111: Lecture 5, Pg 18
Example Recap
FX
N = mg
mg
aX = FX / m y
x
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Physics 111: Lecture 5, Pg 19
Lecture 5, Act 2 Normal Force
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?
m
(a) N > mg
(b) N = mg
(c) N < mg
a
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Physics 111: Lecture 5, Pg 20
Lecture 5, Act 2 Solution
m
N
mg
All forces are acting in the y direction, so use:
Ftotal = ma
N - mg = ma
N = ma + mg
therefore N > mg
a
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Physics 111: Lecture 5, Pg 21
Tools: Ropes & Strings
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.
The force you would feel if you cut the rope and grabbed the ends.
An action-reaction pair.
cut
T
T
T
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Physics 111: Lecture 5, Pg 22
Tools: Ropes & Strings...
Consider a horizontal segment of rope having mass m:
Draw a free-body diagram (ignore gravity).
Using Newton’s 2nd law (in x direction): FNET = T2 - T1 = ma
So if m = 0 (i.e. the rope is light) then T1 =T2
T1 T2
m
a x
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Physics 111: Lecture 5, Pg 23
Tools: Ropes & Strings...
An ideal (massless) rope has constant tension along the rope.
If a rope has mass, the tension can vary along the rope
For example, a heavy rope hanging from the ceiling...
We will deal mostly with ideal massless ropes.
T = Tg
T = 0
T T
2 skateboards
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Physics 111: Lecture 5, Pg 24
Tools: Ropes & Strings...
The direction of the force provided by a rope is along the direction of the rope:
mg
T
m
Since ay = 0 (box not moving),
T = mg
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Physics 111: Lecture 5, Pg 25
Lecture 5, Act 3 Force and acceleration
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?
m = ? a = 12.2 m/s2
snap ! (a) 14.8 kg
(b) 18.4 kg
(c) 8.2 kg
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Physics 111: Lecture 5, Pg 26
Lecture 5, Act 3 Solution:
Draw a Free Body Diagram!!
T
mg
m = ? a = 12.2 m/s2
Use Newton’s 2nd law in the upward direction:
FTOT = ma
T - mg = ma
T = ma + mg = m(g+a)
mT
g a
kg28
sm21289
N180m
2.
..
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Physics 111: Lecture 5, Pg 27
Tools: Pegs & Pulleys
Used to change the direction of forces
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
F1
ideal peg
or pulley
F2
| F1 | = | F2 |
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Physics 111: Lecture 5, Pg 28
Tools: Pegs & Pulleys
Used to change the direction of forces
An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:
mg
T
m T = mg
FW,S = mg
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Physics 111: Lecture 5, Pg 29
Springs
Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.
relaxed position
FX = 0
x
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Physics 111: Lecture 5, Pg 30
Springs...
Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.
relaxed position
FX = -kx > 0
x x 0
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Physics 111: Lecture 5, Pg 31
Springs...
Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
FX = -k x Where x is the displacement from the relaxed position and k is the constant of proportionality.
FX = - kx < 0
x x > 0
relaxed position
Horizontal
springs
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Physics 111: Lecture 5, Pg 32
Scales:
Springs can be calibrated to tell us the applied force.
We can calibrate scales to read Newtons, or...
Fishing scales usually read weight in kg or lbs.
0 2 4 6 8
1 lb = 4.45 N
Spring/string
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Physics 111: Lecture 5, Pg 33
m m m
(a) 0 lbs. (b) 4 lbs. (c) 8 lbs.
(1) (2)
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Lecture 5, Act 4 Force and acceleration
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?
Scale
on a
skate
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Physics 111: Lecture 5, Pg 34
Lecture 5, Act 4 Solution:
Draw a Free Body Diagram of one of the blocks!!
Use Newton’s 2nd Law in the y direction:
FTOT = 0
T - mg = 0
T = mg = 4 lbs.
mg
T
m T = mg
a = 0 since the blocks are stationary
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Physics 111: Lecture 5, Pg 35
Lecture 5, Act 4 Solution:
The scale reads the tension in the rope, which is T = 4 lbs in both cases!
m m m
T T T T
T T T
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Physics 111: Lecture 5, Pg 36
Recap of today’s lecture..
More discussion of dynamics.
Recap (Text: 4-1 to 4-6)
The Free Body Diagram (Text: 4-5)
The tools we have for making & solving problems:
» Ropes & Pulleys (tension) (Text: 4-5,4-7)
» Hooke’s Law (springs). (Text: 4-4, ex. 4-5)
Look at Textbook problems Chapter 4: # 51, 53, 57, 66