7 problems of newton law
TRANSCRIPT
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Physics 111: Lecture 5, Pg 1
Physics 111: Lecture 5Physics 111: Lecture 5
Today’s AgendaToday’s Agenda
More discussion of dynamics
Recap
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 111: 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.
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Physics 111: Lecture 5, Pg 3
Gravity:Gravity:
What is the force of gravity exerted by the earth on a typical physics student?
Typical student mass m = 55kgg = 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 111: Lecture 5, Pg 4
Lecture 5, Lecture 5, Act 1Act 1Mass vs. WeightMass 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, Lecture 5, Act 1Act 1SolutionSolution
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, Lecture 5, Act 1Act 1SolutionSolution
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 DiagramThe Free Body Diagram
Newton’s 2nd Law says that for an object FF = maa.
Key phrase here is for an objectfor an object..
So before we can apply FF = maa 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...The Free Body Diagram...
Consider the following caseWhat are the forces acting on the 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 111: Lecture 5, Pg 9
The Free Body Diagram...The Free Body Diagram...
Consider the following caseWhat 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 111: 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 111: 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 111: 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 111: 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 111: 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 111: 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 111: 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 111: Lecture 5, Pg 17
Example...Example... FX = maX
So aX = FX / m = (10 N)/(2 kg) = 5 m/s 2.
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 111: Lecture 5, Pg 18
Example RecapExample Recap
FX
N = mg
mg
aX = FX / m y y
x x
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Physics 111: Lecture 5, Pg 19
Lecture 5, Lecture 5, Act 2Act 2Normal 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 111: Lecture 5, Pg 20
Lecture 5, Lecture 5, Act 2Act 2SolutionSolution
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 & 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
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Physics 111: 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
T1 T2
m
a x x
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Physics 111: 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
2 skateboards
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Physics 111: Lecture 5, Pg 24
Tools: Ropes & Strings...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, Lecture 5, Act 3Act 3Force 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 111: Lecture 5, Pg 26
Lecture 5, Lecture 5, Act 3Act 3Solution: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 111: 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 111: 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 111: 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 111: 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 111: 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
Horizontalsprings
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Physics 111: 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
Spring/string
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Physics 111: Lecture 5, Pg 33
m m m
(a)(a) 0 lbs. (b)(b) 4 lbs. (c)(c) 8 lbs.
(1) (2)
?
Lecture 5, Lecture 5, Act 4Act 4Force 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?
Scale on a skate
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Physics 111: Lecture 5, Pg 34
Lecture 5, Lecture 5, Act 4Act 4Solution: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 111: Lecture 5, Pg 35
Lecture 5, Lecture 5, Act 4Act 4Solution: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 111: Lecture 5, Pg 36
Problem: AccelerometerProblem: Accelerometer
A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road, and has an acceleration a in the x direction. The string makes an angle with respect to the vertical (y) axis. Solve for in terms of a and g.
a
i i
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Physics 111: Lecture 5, Pg 37
Accelerometer...Accelerometer...
Draw a free body diagram for the mass: What are all of the forces acting?
m
TT (string tension)
mgg (gravitational force)
i i
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Physics 111: Lecture 5, Pg 38
Accelerometer...Accelerometer...
Using components Using components (recommended):
ii: FX = TX = T sin = ma
jj: FY = TY mg
= T cos mg = 0 TT
mgg
mmaa
jj
ii
TX
TY
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Physics 111: Lecture 5, Pg 39
Accelerometer...Accelerometer...
Using components Using components :
ii: T sin = ma
jj: T cos - mg = 0
Eliminate T :
mgg
m
maaT sin = ma
T cos = mgtan
a
g
TX
TY
jj
ii
TT
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Physics 111: Lecture 5, Pg 40
Accelerometer...Accelerometer...
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):
Find the total vector force FFNET:
TT
mgg
FFTOT
m
TT (string tension)
mgg (gravitational force)
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Physics 111: Lecture 5, Pg 41
Accelerometer...Accelerometer...
Alternative solution using vectorsAlternative solution using vectors (elegant but not as (elegant but not as systematic):systematic):
Find the total vector force FFNET: Recall that FFNET = ma:
So
maa
tan ma
mg
a
g
TT
mgg
tan a
g
m
TT (string tension)
mgg (gravitational force)
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Physics 111: Lecture 5, Pg 42
Accelerometer...Accelerometer...
Let’s put in some numbers:
Say the car goes from 0 to 60 mph in 10 seconds: 60 mph = 60 x 0.45 m/s = 27 m/s.Acceleration a = Δv/Δt = 2.7 m/s2. So a/g = 2.7 / 9.8 = 0.28 .
= arctan (a/g) = 15.6 deg
tan a
g
a
Cart w/ accelerometer
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Physics 111: Lecture 5, Pg 43
Problem: Inclined planeProblem: Inclined plane
A block of mass m slides down a frictionless ramp that makes angle with respect to the horizontal. What is its acceleration a ?
ma
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Physics 111: Lecture 5, Pg 44
Inclined plane...Inclined plane...
Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only.
ma
ii
jj
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Physics 111: Lecture 5, Pg 45
Inclined plane...Inclined plane...
Consider x and y components separately: ii: mg sin =ma. a = g sin
jj: N - mg cos = 0. N = mg cos
mgg
NN
mg sin
mg cos
maa
ii
jj
Incline
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Physics 111: Lecture 5, Pg 46
Inclined plane...Inclined plane...
Alternative solution using vectors:
m
mgN
a = g sin ii
N = mg cos jj
ii
jj
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Physics 111: Lecture 5, Pg 47
Angles of an Inclined planeAngles of an Inclined plane
ma = mg sin
mgN
The triangles are similar, so the angles are the same!
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Physics 111: Lecture 5, Pg 48
Lecture 6, Lecture 6, Act 2Act 2Forces and MotionForces and Motion
A block of mass M = 5.1 kg is supported on a frictionless ramp by a spring having constant k = 125 N/m. When the ramp is horizontal the equilibrium position of the mass is at x = 0. When the angle of the ramp is changed to 30o what is the new equilibrium position of the block x1?
(a) x1 = 20cm (b) x1 = 25cm (c) x1 = 30cm
x = 0
M
k
x 1 = ?
Mk
= 30o
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Physics 111: Lecture 5, Pg 49
Lecture 6, Lecture 6, Act 2Act 2SolutionSolution
x 1
Mk
x
y
Choose the x-axis to be along downward direction of ramp.
Mg
FBD: The total force on the block is zero since it’s at rest.
N
Fx,g = Mg sin
Force of gravity on block is Fx,g = Mg sin Consider x-direction:
Force of spring on block is Fx,s = -kx1
F x,s = -kx 1
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Physics 111: Lecture 5, Pg 50
Lecture 6, Lecture 6, Act 2Act 2SolutionSolution
x 1
Mk
x
y
Since the total force in the x-direction must be 0:
F x,g = Mg sin
F x,s = -kx 1
Mg sinkx1κ
sin Μgx
1
θ
m20mN125
50sm189kg15x
2
1.
...
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Physics 111: Lecture 5, Pg 51
Problem: Two Blocks Problem: Two Blocks
Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface. If a force of magnitude F is applied to the box of mass m1, what is the force on the block of mass m2?
mm11 mm22
F
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Physics 111: Lecture 5, Pg 52
Problem: Two BlocksProblem: Two Blocks
Realize that F = (mm11+ mm22) a : Draw FBD of block mm22 and apply FNET = ma:
F2,1F2,1 = mm22 a
F / (mm11+ mm22) = a
m22,1
m2m1
FF
Substitute for a :
mm22
(m1 + m2)m2F2,1 F
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Physics 111: Lecture 5, Pg 53
Problem: Tension and AnglesProblem: Tension and Angles
A box is suspended from the ceiling by two ropes making an angle with the horizontal. What is the tension in each rope?
mm
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Physics 111: Lecture 5, Pg 54
Problem: Tension and AnglesProblem: Tension and Angles
Draw a FBD:
T1 T2
mg
Since the box isn’t going anywhere, Fx,NET = 0 and Fy,NET = 0
T1sin T2sin
T2cos T1cos
jj
ii
Fx,NET = T1cos - T2cos = 0 T1 = T2
2 sin mg
T1 = T2 =Fy,NET = T1sin + T2sin - mg = 0
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Physics 111: Lecture 5, Pg 55
Problem: Motion in a CircleProblem: Motion in a Circle
A boy ties a rock of mass m to the end of a string and twirls it in the vertical plane. The distance from his hand to the rock is R. The speed of the rock at the top of its trajectory is v.What is the tension T in the string at the top of the rock’s
trajectory?
R
v
TT
Tetherball
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Physics 111: Lecture 5, Pg 56
Motion in a Circle...Motion in a Circle...
Draw a Free Body Diagram (pick y-direction to be down): We will use FFNET = maa (surprise) First find FFNET in y direction:
FFNET = mg +T
TTmgg
y y
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Physics 111: Lecture 5, Pg 57
Motion in a Circle...Motion in a Circle...
FFNET = mg +T
Acceleration in y direction:
maa = mv2 / R
mg + T = mv2 / R
T = mv2 / R - mg
R
TT
v
mgg
y y
F = ma
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Physics 111: Lecture 5, Pg 58
Motion in a Circle...Motion in a Circle... What is the minimum speed of the mass at the top of the trajectory such that the string does not go limp?
i.e. find v such that T = 0.
mv2 / R = mg + T
v2 / R = g
Notice that this doesnot depend on m.
R
mgg
v
T= 0
v Rg
Bucket
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Physics 111: Lecture 5, Pg 59
Lecture 6Lecture 6, , Act 3Act 3Motion in a CircleMotion in a Circle
A skier of mass m goes over a mogul having a radius of curvature R. How fast can she go without leaving the ground?
R
mgg NN
vv
(a) (b) (c) Rgv =mRgv =mRg
v =
Trackw/ bump
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Physics 111: Lecture 5, Pg 60
Lecture 6Lecture 6, , Act 3Act 3SolutionSolution
mv2 / R = mg - N
For N = 0:
R
v
mgg NN
v Rg
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Physics 111: Lecture 5, Pg 61
Recap of Today’s lecture:Recap of Today’s lecture:
The Free Body DiagramFree Body Diagram
The tools we have for making & solving problems:
» Ropes & Pulleys (tension)
» Hooke’s Law (springs)
Accelerometer Inclined plane Motion in a circle