physics 211: lecture 5, pg 1 physics 101 l more discussion of dynamics recap newton's laws free...
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Physics 211: Lecture 5, Pg 1
Physics Physics 101101
More discussion of dynamics
Recap Newton's Laws
The Free Body DiagramFree Body Diagram
The tools we have for making & solving problems:Ropes & Pulleys (tension)Hooke’s Law (springs)
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Physics 211: Lecture 5, Pg 2
Review: Newton's LawsReview: 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, FFNET = maa
Where FFNET = FF
Law 3: Forces occur in action-reactionaction-reaction pairs, FFA ,B = - FFB ,A.
Where FFA ,B is the force acting on object A due to its interaction with object B and vice-versa.
m is “mass” of object
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Physics 211: Lecture 5, Pg 3
Gravity:Gravity:Mass vs WeightMass vs Weight
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
FFE,S = -= -mg g
FFS,E = F = Fg = = mg g
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Physics 211: Lecture 5, Pg 4
Mass vs. WeightMass vs. Weight An astronaut on Earth kicks a bowling ball straight ahead
and hurts his foot. A year later, the same astronaut kicks a bowling ball on the moon in the same manner with the same force.
His foot hurts...
(a) more
(b) less (c) the same
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Physics 211: Lecture 5, Pg 5
SolutionSolution 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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Physics 211: Lecture 5, Pg 6
SolutionSolution 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 211: Lecture 5, Pg 7
The Free Body DiagramThe Free Body Diagram
Newton’s 2nd Law says that for an object FF = maa.
Key phrase here is for an objectfor an object.. Object has mass and experiences forcesObject has mass and experiences forces
So before we can apply FF = maa to any given object we isolate the forces acting on this object:
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Physics 211: Lecture 5, Pg 8
The Free Body Diagram...The Free Body Diagram...
Consider the following case as an example of this…. What are the forces acting on the plank ? Other forces act on F, W and E. focus on plank
P = plank
F = floor
W = wall
E = earthFFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,P
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Physics 211: Lecture 5, Pg 9
The Free Body Diagram...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.
FFW,P
FFP,W
FFP,F FFP,E
FFF,P
FFE,P
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Physics 211: Lecture 5, Pg 10
The Free Body Diagram...The Free Body Diagram...
The forces acting on the plank should reveal themselves...
FFP,W
FFP,F FFP,E
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Physics 211: Lecture 5, Pg 11
Aside...Aside...
In this example the plank is not moving... It is certainly not accelerating! So FFNET = maa becomes FFNET = 0
This is the basic idea behind statics, which we will discuss in a few weeks.
FFP,W + FFP,F + FFP,E = 0
FFP,W
FFP,F FFP,E
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Physics 211: Lecture 5, Pg 12
ExampleExample
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 xx direction. What is the acceleration of the box?
FF = Fx ii aa = ?
m
y y
x x
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Physics 211: Lecture 5, Pg 13
Example...Example...
Draw a picture showing all of the forces
FFFFB,F
FFF,BFFB,E
FFE,B
y y
x x
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Physics 211: Lecture 5, Pg 14
Example...Example...
Draw a picture showing all of the forces. Isolate the forces acting on the block.
FFFFB,F
FFF,BFFB,E = mgg
FFE,B
y y
x x
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Physics 211: Lecture 5, Pg 15
Example...Example... Draw a picture showing all of the forces. Isolate the forces acting on the block. Draw a free body diagram.
FFFFB,F
mgg
y y
x x
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Physics 211: Lecture 5, Pg 16
Example...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
FFFFB,F
mgg
y y
x x
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Physics 211: Lecture 5, Pg 17
Example...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 forceof the floor on the object (FB,F ) isoften called the Normal Force Normal Force (N).
Since aY = 0 , N = mg in this case.
FX
N
mg
y y
x x
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Physics 211: Lecture 5, Pg 18
Example RecapExample Recap
FX
N = mg
mg
aX = FX / m y y
x x
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Physics 211: Lecture 5, Pg 19
Normal ForceNormal 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)(a) N > mgN > mg
(b)(b) N = mgN = mg
(c)(c) N < mgN < mg
a
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Physics 211: Lecture 5, Pg 20
SolutionSolution
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 211: Lecture 5, Pg 21
Tools: Ropes & StringsTools: Ropes & Strings
Can be used to pull from a distance. TensionTension (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
TT
T
Tension doesn’t have a direction. When you hook up a wire to an object the direction is determined by geometry of the hook up.
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Physics 211: Lecture 5, Pg 22
Tools: Ropes & Strings...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 xx direction): FNET = T2 - T1 = ma
So if m = 0 (i.e. the rope is light) then T1 =T2 T is constant anywhere in a rope or string
as long as its massless
T1 T2
m
a x x
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Physics 211: Lecture 5, Pg 23
Tools: Ropes & Strings...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
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Physics 211: Lecture 5, Pg 24
Tools: Ropes & Strings...Tools: Ropes & Strings...
What is force acting on box by the rope in the picture below? (always assume rope is massless unless told different)
mg
T
m
Since ay = 0 (box not moving),
T = mg
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Physics 211: Lecture 5, Pg 25
Force and accelerationForce 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 211: Lecture 5, Pg 26
Solution:Solution: Draw a Free Body Diagram!!
T
mg
m = ?a = 12.2 m/s2
Use Newton’s 2nd lawin 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 211: Lecture 5, Pg 27
Tools: Pegs & PulleysTools: 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:
FF1 ideal peg
or pulley
FF2
| FF1 | = | FF2 |
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Physics 211: Lecture 5, Pg 28
Tools: Pegs & PulleysTools: 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 211: Lecture 5, Pg 29
SpringsSprings
Hooke’s Law: 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 211: Lecture 5, Pg 30
Springs...Springs...
Hooke’s Law: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
xx 0
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Physics 211: Lecture 5, Pg 31
Springs...Springs...
Hooke’s Law: 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
xx > 0
relaxed position
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Physics 211: Lecture 5, Pg 32
Scales: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.
02468
1 lb = 4.45 N
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Physics 211: Lecture 5, Pg 33
m m m
(a)(a) 0 lbs. (b)(b) 4 lbs. (c)(c) 8 lbs.
(1) (2)
?
Force and accelerationForce 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?
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Physics 211: Lecture 5, Pg 34
Solution:Solution: Draw a Free Body Diagram of one
of the blocks!!
Use Newton’s 2nd Lawin 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 211: Lecture 5, Pg 35
Solution:Solution:
The scale reads the tension in the rope, which is T = 4 lbs in both cases!
m m m
T T T T
TTT
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Physics 211: Lecture 5, Pg 36
Recap of today’s lecture..Recap of today’s lecture..
More discussion of dynamics
Recap The Free Body DiagramFree Body DiagramThe tools we have for making & solving problems:
» Ropes & Pulleys (tension)
» Hooke’s Law (springs).