physics 111 - valparaiso university 111 lecture 11 • mid-term survey results • ch 5: newton’s...
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Physics 111
Title page
Thursday,September 30, 2004
Physics 111 Lecture 11
• Mid-term survey results
• Ch 5: Newton’s 3rd Law
• Ch 6: TensionExamples
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Help this week:Wednesday, 8 - 9 pm in NSC 118/119
Sunday, 6:30 - 8 pm in CCLIR 468
Help sessions
AnnouncementsThursSept
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Don’t forget to read over the labwrite-up and be ready for the quiz.
labs
AnnouncementsThursSept
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Response rate: 23 out of 33
Several students provided little to nofeedback on the free-response page
I need to hear from you if you havesuggestions or comments!
Help sessions
AnnouncementsThursSept
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Most questions had mean of 3.0
Least favorite activities:PhysletsMastering Physics
Favorite activities:Interactive Exercises
Average time on class: ~7 hours/week
Help sessions
AnnouncementsThursSept
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What can you do?Lots of recommended problems…including Mastering PhysicsOne exam problem will be MPOne exam problem will be Walker
Help sessions
AnnouncementsThursSept
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What I will do in the future:
• Make sure we cover what you need beforeMastering Physics assignment is due.
• However, Physlets will cover material beforewe cover it in class.
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We’ve looked at a lot of problems with but asingle “system” of interest.
Interacting Systems
Ch 5: Newton’s LawsThursSept
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Let’s now formalize the way to handle problemswith multiple systems.
In the process, we’ll encounter Newton’s 3rd Law.
Example of me pushing on the wall.
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Just because neither I nor the wall seem to beaccelerating, however, does NOT mean thatthere are no forces acting. In fact, there areseveral relevant forces involved in this process.
I can huff and puff andpush on the wall, but itdoesn’t seem to beaccelerating. Andneither am I!!!
Big Bad Wolf
Ch 5: Newton’s LawsThursSept
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Let’s just look at theforces acting on me.
A STATICSystem: all
the forces arein balance.
Nothingaccelerates!
All forces on hero - picture
Ch 5: Newton’s LawsThursSept
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N = normal forceof wall onme
N = normalforce of flooron me
f = frictional forceof floor on me
W = force ofEarth’s gravityon me
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Let’s just look at theforces acting on me.
friction
weight
Normal (floor)
All forces on hero -FBD
Normal (wall)GOODGOOD
free-bodyfree-bodydiagramdiagram
Ch 5: Newton’s LawsThursSept
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Let’s now look at theforces acting on the wall.
A STATICSystem: all
the forces arein balance.
Nothingaccelerates!
All forces on wall - picture
Ch 5: Newton’s LawsThursSept
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C = contact forceof wall onme
f = frictional forceof floor on wall
N = normalforce of flooron wall
W = force ofEarth’s gravityon wall
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All forces on wall -FBD
friction of flooron wall
Contact Forceof the mepushingon the wall.
weightof wall
Normal forceof floor on wall
GOODGOODfree-bodyfree-bodydiagramdiagram
Ch 5: Newton’s LawsThursSept
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Let’s now look atthe forces actingon the wall.
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Let’s just look at theforces acting on the floor.
All forces on floor - picture
Ch 5: Newton’s LawsThursSept
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N = normal forceof the Earth onthe floor
W = force ofEarth’s gravityon the floor
fmf = frictional forceof me on floor
Cwf = contactforce of wallon floor
Cmf = contact force of me onfloor
fwf = frictional forceof wall on floor
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Let’s just look atthe forces acting
on the floor.
Contactof wall onthe floor
Friction (me)
Normal (Earth)
All forces on floor - FBD
Contactof me onthe floor
Friction (wall)
weightof floor
GOODGOODfree-bodyfree-bodydiagramdiagram
Ch 5: Newton’s LawsThursSept
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Finally, let’s look at the forces actingon the Earth in this problem.
EARTH
All forces on Earth - picture
Ch 5: Newton’s LawsThursSept
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C = Contactof floor onthe Earth
Fghero = GravitationalForce of Meon the Earth
Fgf = GravitationalForce of flooron the Earth
Fgw = GravitationalForce of wallon the Earth
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Finally, let’s look at the forces actingon the Earth in this problem.
GravitationalForce of Meon the Earth
EARTH
Contactof floor onthe Earth
GravitationalForce of wallon the Earth
GravitationalForce of flooron the Earth
All forces on Earth - FBD
Ch 5: Newton’s LawsThursSept
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GOODGOODfree-bodyfree-bodydiagramdiagram
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If one object exerts a force on a second object,the second object necessarily exerts an equalbut oppositely directed force on the first.
We’re talking aboutTWO DIFFERENTFORCES HERE!!!
Law 3
Ch 5: Newton’s LawsThursSept
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NormalForce
Force of thewall pushingon me.
ContactForce
Force of theme pushingon the wall.
Interaction Pair -- Newton’s Third Lawalways act on DIFFERENT objects
Super Hero: normal & contact
NOT a free-body diagram
Ch 5: Newton’s LawsThursSept
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NormalForce
Force of thewall pushingon me.
ContactForce
Force of theme pushingon the wall.
Interaction Pair -- Newton’s Third Lawalways appear in DIFFERENT FBDs.
Super Hero: normal & contact
NOT a free-body diagram
Ch 5: Newton’s LawsThursSept
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Several notes:
• The NORMAL force acts perpendicular tosome surface. It is NOT NECESSARILYequal to mg!
• Each and every object in a problem has itsown free body diagram! Draw each oneseparately.
• Third Law Force pairs act on DIFFERENTOBJECTS. They NEVER act on the same object!
Summary notes
Ch 5: Newton’s LawsThursSept
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King Henry and the Earth bothpossess “gravitational massgravitational mass”and exert equal but oppositelydirected forces on one another.
F
Earth on Henry
F
Henry on Earth
W1: Henry & Earth
1)
2)
3)
F
Earth on Henry=F
Henry on Earth
F
Earth on Henry<F
Henry on Earth
Worksheet Problem #1
Ch 5: Newton’s LawsThursSept
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F
Earth on Henry>F
Henry on Earth
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I’m going to jump off a chair.
Demo: my gravity on Earth
Ch 5: Newton’s LawsThursSept
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Watch as the Earth rushesup to meet me!
Do you want to seethat again?
What’s going on here?
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I need two student volunteers
Demo: students and 3rd Law
Ch 4: Newton’s LawsThursSept
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Exp. #1: Each of you take onescale. Push on each otherthrough the scales and call outthe readings.
Exp. #2: One of you sit in the rolling chair. Theother push through the scale. Both call outreadings on the scales.
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Our avant-guarde socialite pulls on the ropethat’s wrapped around the tree. Nothing happens.
Tension
Ch 6: Applying Newton’s LawsTuesSept
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skip
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What must be true about the forces acting...
here here
here
tension along rope
Ch 6: Applying Newton’s LawsTuesSept
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Let’s examine this piece more carefully...
Tension Tension
The forces balance -- The rope does NOT accelerate.
piece of rope
Ch 6: Applying Newton’s LawsTuesSept
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In fact, no matter which little segment of therope I examine in this case, the tension forcesbalance in either direction, and the rope remainsstationary.
Okay, let’s look at tension in a rope that resultsin the acceleration of an object...
balance
Ch 6: Applying Newton’s LawsTuesSept
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Remember this one?
What exactly is it that causesthe green block to accelerate?
block & ice
Ch 6: Applying Newton’s LawsTuesSept
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Force Meter
Frictionless pond of ice
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Let’s look at the free-bodydiagram for the green block.What forces are acting onthe green block?
Weight Normalforce
Contact
The contact force of therope on the block resultsin the block accelerating.
block accelerates
Ch 6: Applying Newton’s LawsTuesSept
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What is the acceleration ofthe green block?
Weight Normalforce
Contact
block accelerates
Ch 6: Applying Newton’s LawsTuesSept
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a
block=F
contact
mblock
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What if we look at a pieceof the rope in this case?
Some mass mr
a =F
net
mr
=T
2−T
1
mr
Some mass mbRemember, thewhole systemis accelerating
at the same rate, a.
a
Tension 1 Tension 2
rope accelerates?
Ch 6: Applying Newton’s LawsThursSept
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Some mass mb
And clearly the block’s acceleration will bedictated by the magnitude of the contact force at theend of the rope (which in magnitude is equal to thetension in the rope at that end of the rope)connected to the block...
a =F
net
mb
=F
contact
mb
=T
end
mb
Contact
block again
Ch 6: Applying Newton’s LawsThursSept
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Back to the ropefor a minute...
TensionrightTensionleft
T
left>T
right
unequal tensions?
Ch 6: Applying Newton’s LawsThursSept
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Frictionless pond of ice
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However, what happensif mr = 0?
F
net=T
2−T
1= m
r
a = 0
a = 0
T
2=T
1
Tension 1 Tension 2
Massless string
Ch 6: Applying Newton’s LawsThursSept
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T
left=T
right
equal tensions
Ch 6: Applying Newton’s LawsThursSept
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TensionrightTensionleft
Frictionless pond of ice
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Worksheet Problem #2
CQ2 Tension
Ch 6: Applying Newton’s LawsThursSept
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Worksheet Problem #2
CQ2 – Tension the ?
In the 17th century, Otto von Güricke, a physicist in Mag-deburg, fitted two hollow bronze hemispheres together andremoved the air from the resulting sphere with a pump.Two 8-horse teams could not pull the halves apart even though thehemispheres fell apart when air was readmitted. Suppose vonGüricke had tied both teams of horses to one side and bolted theother side to a heavy tree trunk. In this case, the tension on thehemispheres would be
PI, Mazur (1997)
≥≥
1. twice2. exactly the same as3. half what it was before.
Ch 6: Applying Newton’s LawsThursSept
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A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.
W1: Worker & Crate
38o111 N
y
x
Problem Sheet #1
Ch 6: Applying Newton’s LawsThursSept
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a
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A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.
W1: Worker & Crate
FBD
Ch 6: Applying Newton’s LawsThursSept
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Ffloor
n
θ
T
W
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A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.
W1: Worker & Crate
Ch 6: Applying Newton’s LawsThursSept
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Knowns:θ= 380 , ay = 0Tx = (450 N) cos θ = 354.6 NTy = (450 N) sin θ = 277.0 NFfloor = -111 N xm = 310 kgg = 9.81 m/s2
Unknowns:ax, n
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W1: Worker & Crate
(a)
Fx,net = Tx + Ffloor
Fx,net = 354.6 N – 111 N
Fx,net = m ax = (310 kg) ax
Fx,net = 254.6 N
ax= Fx,net / m
ax= 254.6 N/ 310 kg
ax= 0.821 m/s2
Ch 6: Applying Newton’s LawsThursSept
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A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.
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W1: Worker & Crate
(b)
Fy,net = Ty + W + n = 0
0 = |T| sin 38o – mg + n
- n = 277 N – (310 kg) (9.81 m/s2)
Fy,net = m ay = (310 kg) (0 m/s2) = 0
n = 2760 N y
A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.
Ch 6: Applying Newton’s LawsThursSept
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A worker sits in a bosun’s chair that is supported by amassless rope that runs over a massless, frictionlesspulley and back down to the man’s hand. Thecombined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed? (b) How would theforce be different if it were exerted by a second manon the ground instead of the man in the chair?
This looks like a pretty complexproblem…And it can be tricky…so, let’s be careful and useNewton’s Laws explicitly foreach moving object.
? Worker, cart, pulley (W3)
Problem Sheet #2
Ch 6: Applying Newton’s LawsThursSept
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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
First, what are the forces acting on the man?
Normal force ofchair on man.
n
Force ofgravityon man
Wman
Tension ofrope on chair. T
? Soln – F on man
Ch 6: Applying Newton’s LawsThursSept
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FBD: man
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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
Next, what are the forces acting on the chair?
Force of gravityon chair
W
chair
Tension ofrope on chair. T
Contact force ofman on chair.
n '
? Soln – F on chair
Ch 6: Applying Newton’s LawsThursSept
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FBD: chair
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A worker sits in a bosun’s chair that is supported by amassless rope that runs over a massless, frictionlesspulley and back down to the man’s hand. Thecombined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed? (b) How would theforce be different if it were exerted by a second manon the ground instead of the man in the chair?
? Worker, cart, pulley (W3)
Ch 6: Applying Newton’s LawsThursSept
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Knowns:mman+chair = 95.0 kgg = 9.81m/s2
a = 0n = - n’
+y
Unknowns:
Tn
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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
What’s Newton’s Law say about the forces on the man?
F
net= ma
But the acceleration of the man is ZERO!(i.e., constant speed means NO acceleration.)
F
net ,man=T + n −
W
man= ma = m(0) = 0
? Soln – 2N on man
Ch 6: Applying Newton’s LawsThursSept
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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
F
net ,chair=T − n '−
W
chair= ma = m(0) = 0
? Soln – 2N on chair
Ch 6: Applying Newton’s LawsThursSept
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But the acceleration of the chair is ZERO!(i.e., constant speed means NO acceleration.)
What’s Newton’s Law say about the forces on the chair?
F
net= ma
![Page 49: Physics 111 - Valparaiso University 111 Lecture 11 • Mid-term survey results • Ch 5: Newton’s 3rd Law • Ch 6: Tension Examples Help this week: Wednesday, 8 - 9 pm in NSC 118/119](https://reader030.vdocuments.net/reader030/viewer/2022011800/5ad9e96a7f8b9a53618bdcef/html5/thumbnails/49.jpg)
The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
F
net ,cart=T − n '−
W
cart= 0
F
net ,man=T + n −
W
man= 0 Add these two
together.
2T + n − n '−
W
man−
Wcart
= 0
3rd Law PairSame magnitude
2T = m
man
g + m
cart
g
? Soln - combine
Ch 6: Applying Newton’s LawsThursSept
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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?
T = (m
man+ m
cart)g / 2
T = =( . )( . ) /950 9 81 2 466kg Nm
s2
? Soln - tension
Ch 6: Applying Newton’s LawsThursSept
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2T = m
man
g + m
cart
g
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(b) How would the force be different if it were exertedby a second man on the ground instead of the man inthe chair?
Now the man on the groundmust exert a force on the ropesuch that the tension in therope balances the force ofgravity on the man+chair.
T m m gman cart= + =( ) 932N
? Soln – ground
Ch 6: Applying Newton’s LawsThursSept
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