Physics 111
Thursday,
September 30, 2004
• Mid-term survey results
• Ch 5:
Newton’s 3rd Law
• Ch 6:
Tension
Examples
Thurs
Sept
.30.
Announcements
Help this week:
Wednesday, 8 - 9 pm in NSC 118/119
Sunday, 6:30 - 8 pm in CCLIR 468
Phys
111
Thurs
Sept
.30.
Announcements
Don’t forget to read over the lab
write-up and be ready for the quiz.
Phys
111
Thurs
Sept
.30.
Announcements
Response rate: 23 out of 33
Several students provided little to no
feedback on the free-response page
I need to hear from you if you have
suggestions or comments!
Phys
111
Thurs
Sept
.30.
Announcements
Most questions had mean of 3.0
Least favorite activities:
Physlets
Mastering Physics
Favorite activities:
Interactive Exercises
Average time on class: ~7 hours/week
Phys
111
Thurs
Sept
.30.
Announcements
What can you do?
Lots of recommended problems
…including Mastering Physics
One exam problem will be MP
One exam problem will be Walker
What I will do in the future:
• Make sure we cover what you need before
Mastering Physics assignment is due.
• However, Physlets will cover material before
we cover it in class.
Phys
111
Thurs
Sept
.30.
Ch 5: Newton’s Laws
We’ve looked at a lot of problems with but a
single “system” of interest.
Let’s now formalize the way to handle problems
with multiple systems.
In the process, we’ll encounter Newton’s 3rd Law.
Example of me pushing on the wall.
Phys
111
Thurs
Sept
.30.
Ch 5: Newton’s Laws
Phys
111
I can huff and puff and
push on the wall, but it
doesn’t seem to be
accelerating. And
neither am I!!!
Just because neither I nor the wall seem to be
accelerating, however, does NOT mean that
there are no forces acting. In fact, there are
several relevant forces involved in this process.
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
Let’s just look at the
forces acting on me.
N = normal force
of wall on
me
f = frictional force
of floor on me
A STATIC
System: all
the forces are
in balance.
Nothing
accelerates!
N = normal
force of floor
on me
W = force of
Earth’s gravity
on me
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
Let’s just look at the
forces acting on me.
Normal (floor)
GOOD
free-body
diagram
Normal (wall)
friction
weight
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
Let’s now look at the
forces acting on the wall.
C = contact force
of wall on
me
f = frictional force
of floor on wall
A STATIC
System: all
the forces are
in balance.
Nothing
accelerates!
N = normal
force of floor
on wall
W = force of
Earth’s gravity
on wall
Thurs
Sept
Ch 5: Newton’s Laws
.30.
Normal force
of floor on wall
friction of floor
on wall
GOOD
free-body
diagram
weight
of wall
Phys
111
Let’s now look at
the forces acting
on the wall.
Contact Force
of the me
pushing
on the wall.
Thurs
Sept
.30.
Ch 5: Newton’s Laws
Phys
111
Let’s just look at the
forces acting on the floor.
Cmf = contact force of me on
floor
Cwf = contact
force of wall
on floor
N = normal force
of the Earth on
the floor
fmf = frictional force
of me on floor
fwf = frictional force
of wall on floor
W = force of
Earth’s gravity
on the floor
Thurs
Sept
Phys
111
Ch 5: Newton’s Laws
.30.
Let’s just look at
the forces acting
on the floor.
GOOD
free-body
diagram
Friction (wall)
Friction (me)
weight
of floor
Contact
of me on
the floor
Contact
of wall on
the floor
Normal (Earth)
Thurs
Sept
Phys
111
Ch 5: Newton’s Laws
.30.
Finally, let’s look at the forces acting
on the Earth in this problem.
C = Contact
of floor on
the Earth
Fgw = Gravitational
Force of wall
on the Earth
Fgf = Gravitational
Force of floor
on the Earth
EARTH
Fghero = Gravitational
Force of Me
on the Earth
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
GOOD
free-body
diagram
Finally, let’s look at the forces acting
on the Earth in this problem.
Contact
of floor on
the Earth
Gravitational
Force of wall
on the Earth
Gravitational
Force of floor
on the Earth
Gravitational
Force of Me
on the Earth
EARTH
Thurs
Sept
.30.
Ch 5: Newton’s Laws
Phys
111
If one object exerts a force on a second object,
the second object necessarily exerts an equal
but oppositely directed force on the first.
We’re talking about
TWO DIFFERENT
FORCES HERE!!!
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
NOT a free-body diagram
Normal
Force
Force of the
wall pushing
on me.
Contact
Force
Force of the
me pushing
on the wall.
Interaction Pair -- Newton’s Third Law
always act on DIFFERENT objects
Thurs
Sept
.30.
Phys
111
Ch 5: Newton’s Laws
NOT a free-body diagram
Normal
Force
Force of the
wall pushing
on me.
Contact
Force
Force of the
me pushing
on the wall.
Interaction Pair -- Newton’s Third Law
always appear in DIFFERENT FBDs.
Thurs
Sept
.30.
Ch 5: Newton’s Laws
Phys
111
Several notes:
• The NORMAL force acts perpendicular to
some surface. It is NOT NECESSARILY
equal to mg!
• Each and every object in a problem has its
own free body diagram! Draw each one
separately.
• Third Law Force pairs act on DIFFERENT
OBJECTS. They NEVER act on the same object!
Thurs
Sept
.30.
Ch 5: Newton’s Laws
King Henry and the Earth both
possess “gravitational mass”
mass
and exert equal but oppositely
directed forces on one another.
Worksheet Problem #1


1) FEarth on Henry > FHenry on Earth


2) FEarth on Henry = FHenry on Earth


3) FEarth on Henry < FHenry on Earth

FEarth on Henry

FHenry on Earth
Phys
111
Thurs
Sept
.30.
Ch 5: Newton’s Laws
I’m going to jump off a chair.
Watch as the Earth rushes
up to meet me!
Do you want to see
that again?
What’s going on here?
Phys
111
Thurs
Sept
.30.
Ch 4: Newton’s Laws
I need two student volunteers
Exp. #1: Each of you take one
scale. Push on each other
through the scales and call out
the readings.
Exp. #2: One of you sit in the rolling chair. The
other push through the scale. Both call out
readings on the scales.
Phys
111
Tues
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
skip
Our avant-guarde socialite pulls on the rope
that’s wrapped around the tree. Nothing happens.
Tues
Sept
Ch 6: Applying Newton’s Laws
.30.
here
here
here
What must be true about the forces acting...
Phys
111
Tues
Sept
.30.
Ch 6: Applying Newton’s Laws
Tension
Phys
111
Tension
Let’s examine this piece more carefully...
The forces balance -- The rope does NOT accelerate.
Tues
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
In fact, no matter which little segment of the
rope I examine in this case, the tension forces
balance in either direction, and the rope remains
stationary.
Okay, let’s look at tension in a rope that results
in the acceleration of an object...
Tues
Sept
.30.
Ch 6: Applying Newton’s Laws
Remember this one?
What exactly is it that causes
the green block to accelerate?
Force Meter
Frictionless pond of ice
Phys
111
Tues
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
Let’s look at the free-body
diagram for the green block.
What forces are acting on
the green block?
Contact
The contact force of the
rope on the block results
in the block accelerating.
Weight
Normal
force
Tues
Sept
.30.
Phys
111
Ch 6: Applying Newton’s Laws
What is the acceleration of
the green block?

ablock

Fcontact
=
mblock
Contact
Weight
Normal
force
Thurs
Sept
Phys
111
Ch 6: Applying Newton’s Laws
.30.
What if we look at a piece
of the rope in this case?
a
Remember, the
whole system
is accelerating
at the same rate, a.

 
 Fnet T2 − T1
a=
=
mr
mr
Some mass mb
Tension 1
Tension 2
Some mass mr
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
Contact
Some mass mb
And clearly the block’s acceleration will be
dictated by the magnitude of the contact force at the
end of the rope (which in magnitude is equal to the
tension in the rope at that end of the rope)
connected to the block...



 Fnet Fcontact Tend
a=
=
=
mb
mb
mb
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Back to the rope
for a minute...
Tensionleft
Tensionright
Frictionless pond of ice


Tleft > T right
Phys
111
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
However, what happens
if mr = 0?
Tension 1
Tension 2

 


Fnet = T2 − T1 = mr a = 0 a = 0


T2 = T1
Phys
111
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Tensionleft


Tleft = Tright
Tensionright
Frictionless pond of ice
Phys
111
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
Worksheet Problem #2
≥
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
Worksheet Problem #2
In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and
removed the air from the resulting sphere with a pump.Two 8horse teams could not pull the halves apart even though the
hemispheres fell apart when air was readmitted. Suppose von
Güricke had tied both teams of horses to one side and bolted the
other side to a heavy tree trunk. In this case, the tension on the
hemispheres would be
1. twice
2. exactly the same as
3. half
what it was before.
PI, Mazur (1997)
Thurs
Sept
Ch 6: Applying Newton’s Laws
.30.
Phys
111
A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force
of 450 N on a rope that is inclined at 38o to the
horizontal, and the floor exerts a horizontal force of
111 N that opposes the motion. (a) Calculate the
acceleration of the crate if its mass is 310 kg.
(b) Calculate the normal force of the floor on the crate.
Problem Sheet #1
y
x
111 N
a
38o
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force
of 450 N on a rope that is inclined at 38o to the
horizontal, and the floor exerts a horizontal force of
111 N that opposes the motion. (a) Calculate the
acceleration of the crate if its mass is 310 kg.
(b) Calculate the normal force of the floor on the crate.
FBD
n
T
θ
Ffloor
W
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force
of 450 N on a rope that is inclined at 38o to the
horizontal, and the floor exerts a horizontal force of
111 N that opposes the motion. (a) Calculate the
acceleration of the crate if its mass is 310 kg.
(b) Calculate the normal force of the floor on the crate.
Knowns:
θ= 380 , ay = 0
Tx = (450 N) cos θ = 354.6 N
Ty = (450 N) sin θ = 277.0 N
Ffloor = -111 N x
m = 310 kg
g = 9.81 m/s2
Unknowns:
ax, n
Thurs
Sept
Ch 6: Applying Newton’s Laws
.30.
Phys
111
A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force
of 450 N on a rope that is inclined at 38o to the
horizontal, and the floor exerts a horizontal force of
111 N that opposes the motion. (a) Calculate the
acceleration of the crate if its mass is 310 kg.
(b) Calculate the normal force of the floor on the crate.
(a) F
x,net
= m ax = (310 kg) ax
Fx,net = Tx + Ffloor
Fx,net = 354.6 N – 111 N
Fx,net = 254.6 N
ax= Fx,net / m
ax= 254.6 N/ 310 kg
ax= 0.821 m/s2
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
A worker drags a crate across a factory floor by pulling
on a rope tied to the crate. The worker exerts a force
of 450 N on a rope that is inclined at 38o to the
horizontal, and the floor exerts a horizontal force of
111 N that opposes the motion. (a) Calculate the
acceleration of the crate if its mass is 310 kg.
(b) Calculate the normal force of the floor on the crate.
(b)
Fy,net = m ay = (310 kg) (0 m/s2) = 0
Fy,net = Ty + W + n = 0
0 = |T| sin 38o – mg + n
- n = 277 N – (310 kg) (9.81 m/s2)
n = 2760 N y
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
A worker sits in a bosun’s chair that is supported by a
massless rope that runs over a massless, frictionless
pulley and back down to the man’s hand. The
combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed? (b) How would the
force be different if it were exerted by a second man
on the ground instead of the man in the chair?
This looks like a pretty complex
problem…And it can be tricky…
so, let’s be careful and use
Newton’s Laws explicitly for
each moving object.
Problem Sheet #2
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?
First, what are the forces acting on the man?
 Tension of
T rope on chair.
FBD: man
Force of
gravity
on man

Wman

n
Normal force of
chair on man.
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?
Next, what are the forces acting on the chair?
FBD: chair
Force of gravity 
Wchair
on chair
 Tension of
T rope on chair.
 Contact force of
n ' man on chair.
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
A worker sits in a bosun’s chair that is supported by a
massless rope that runs over a massless, frictionless
pulley and back down to the man’s hand. The
combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed? (b) How would the
force be different if it were exerted by a second man
on the ground instead of the man in the chair?
+y
Knowns:
Unknowns:
mman+chair = 95.0 kg
g = 9.81m/s2
a=0
n = - n’
T
n
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?
What’s Newton’s Law say about the forces on the man?


Fnet = ma
But the acceleration of the man is ZERO!
(i.e., constant speed means NO acceleration.)

  

Fnet, man = T + n − Wman = ma = m(0) = 0
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?
What’s Newton’s Law say about the forces on the chair?


Fnet = ma
But the acceleration of the chair is ZERO!
(i.e., constant speed means NO acceleration.)

  

Fnet,chair = T − n '− Wchair = ma = m(0) = 0
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?

  
Fnet, man = T + n − Wman = 0

  
Fnet,cart = T − n '− Wcart = 0
Add these two
together.
   

2T + n − n '− Wman − Wcart = 0
3rd Law Pair
Same magnitude



2 T = mman g + mcart g
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
The combined mass of the man and the chair is 95.0 kg.
(a) With what force must the man pull on the rope for
him to rise at a constant speed?



2 T = mman g + mcart g


T = (mman + mcart ) g / 2

m
T = (95.0kg)(9.81 2 ) / 2 = 466N
s
Thurs
Sept
.30.
Ch 6: Applying Newton’s Laws
Phys
111
(b) How would the force be different if it were exerted
by a second man on the ground instead of the man in
the chair?
Now the man on the ground
must exert a force on the rope
such that the tension in the
rope balances the force of
gravity on the man+chair.


T = (mman + mcart ) g = 932N