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Physics 207 – Lecture 7
Lecture 7
Goals:
Solve 1D and 2D problems with forces in equilibrium
and non-equilibrium (i.e., acceleration) using Newton’ 1st
and 2nd laws.
Distinguish static and kinetic coefficients of friction
Differentiate between Newton’s 1st, 2nd and 3rd Laws
Assignment: HW4, (Chapters 6 & 7 (1st part)
due 2/17, 9 am, Wednesday)
For Thursday, Read Chapter 7
1st Exam Thursday, Feb. 17 from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 1
Identifying Forces: Non-contact
All objects having mass exhibit a mutually attractive force
(i.e., gravity) that is distance dependent
At the Earth’s surface this variation is small so little “g” (the
associated acceleration) is typically set to 9.80 or 10. m/s2
FB,G
Physics 207: Lecture 7, Pg 2
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Physics 207 – Lecture 7
Contact (i.e., normal) Forces
Certain forces act to keep an object in place.
These have what ever force needed to balance
all others (until a breaking point).
FB,T
Physics 207: Lecture 7, Pg 3
No net force No acceleration
r r
r
∑ F = Fnet = ma = 0
∑ Fx = 0
∑ Fy = 0
y
(Force vectors are not always
drawn at contact points)
FB,T Normal force is always ⊥ to a surface
∑ Fy = −mg + N = 0
N = mg
FB,G
Physics 207: Lecture 7, Pg 4
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Physics 207 – Lecture 7
No net force No acceleration
r r
r
∑ F = Fnet = ma = 0
If zero velocity then “static equilibrium”
If non-zero velocity then “dynamic equilibrium”
This label, static vs. dynamic, is observer dependent
Forces are vectors
r r
r r r r
∑ F ≡ Fnet = ma = F1 + F2 + F3 + K
Physics 207: Lecture 7, Pg 5
A special contact force: Friction
What does it do?
It opposes motion (velocity, actual or that which
would occur if friction were absent!)
How do we characterize this in terms we have learned?
Friction results in a force in a direction opposite to
the direction of motion (actual or, if static, then
“inferred”)!
j
N
FAPPLIED
fFRICTION
ma
i
mg
g
Physics 207: Lecture 7, Pg 6
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Physics 207 – Lecture 7
Friction...
Friction is caused by the “microscopic” interactions between
the two surfaces:
Physics 207: Lecture 7, Pg 7
Friction...
Force of friction acts to oppose motion:
Parallel to a surface
Perpendicular to a Normal force.
j
N
F
ma
fF
i
mg
g
Physics 207: Lecture 7, Pg 8
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Physics 207 – Lecture 7
Static Friction with a bicycle wheel
You are pedaling hard
and the bicycle is
speeding up.
What is the direction of
the frictional force?
You are breaking and
the bicycle is slowing
down
What is the direction of
the frictional force?
Physics 207: Lecture 7, Pg 9
Important notes
Many contact forces are conditional and, more
importantly, they are not necessarily constant
We have a general notion of forces is from
everyday life.
In physics the definition must be precise.
A force is an action which causes a body to
accelerate.
(Newton’s Second Law)
On a microscopic level, all forces are non-contact
Physics 207: Lecture 7, Pg 10
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Physics 207 – Lecture 7
Pushing and Pulling Forces
A rope is an example of something
that can pull
You arm is an example of an object
that can push or push
Physics 207: Lecture 7, Pg 11
Examples of Contact Forces:
A spring can push
Physics 207: Lecture 7, Pg 12
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Physics 207 – Lecture 7
A spring can pull
Physics 207: Lecture 7, Pg 13
Ropes provide tension (a pull)
In physics we often use a “massless” rope with opposing
tensions of equal magnitude
Physics 207: Lecture 7, Pg 14
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Physics 207 – Lecture 7
Forces at different angles
Case1: Downward angled force with friction
Case 2: Upwards angled force with friction
Cases 3,4: Up against the wall
Questions: Does it slide?
What happens to the normal force?
What happens to the frictional force?
Cases 3, 4
Case 2
Case 1
F
N
N
ff
F
F
N
ff
ff
mg
mg
mg
Physics 207: Lecture 7, Pg 15
Free Body Diagram
A heavy sign is hung between two poles by a rope at
each corner extending to the poles.
Eat at Bucky’s
A hanging sign is an example of static equilibrium
(depends on observer)
What are the forces on the sign and how are they
related if the sign is stationary (or moving with
constant velocity) in an inertial reference frame ?
Physics 207: Lecture 7, Pg 16
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Physics 207 – Lecture 7
Free Body Diagram
Step one: Define the system
T2
T1
θ2
θ1
Eat at Bucky’s
T2
T1
θ1
mg
θ2
mg
Step two: Sketch in force vectors
nd
Step three: Apply Newton’s 2 Law
(Resolve vectors into appropriate components)
Physics 207: Lecture 7, Pg 17
Free Body Diagram
T1
T2
θ2
θ1
Eat at Bucky’s
mg
Vertical :
y-direction
Horizontal :
x-direction
0 = -mg + T1 sinθ1 + T2 sinθ2
0 = -T1 cosθ1 + T2 cosθ2
Physics 207: Lecture 7, Pg 18
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Physics 207 – Lecture 7
Exercise, Newton’s 2nd Law
P is the upward force being exerted on the crate by the person
C is the contact or normal force on the crate by the floor, and
W is the weight (force of the earth on the crate).
!" #$%&!'(' )$& $$! & $'$ #$' '
&#*$+ $ & $ ($#'! ' &#, !" *!'*$''* , & & & $
#%&$- ./&$0 #$ *( ' ('&1$ 2 3 ! ' !$"%&1$4
A.
B.
C.
D.
P+C<W
P+C>W
P=C
P+C=W
Physics 207: Lecture 7, Pg 19
Mass
We have an idea of what mass is from everyday life.
In physics:
Mass (in Phys 207) is a quantity that specifies
how much inertia an object has
(i.e. a scalar that relates force to acceleration)
(Newton’s Second Law)
Mass is an inherent property of an object.
Mass and weight are different quantities; weight is
usually the magnitude of a gravitational (non-contact)
force.
“Pound” (lb) is a definition of weight (i.e., a force), not
a mass!
Physics 207: Lecture 7, Pg 20
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Physics 207 – Lecture 7
Inertia and Mass
The tendency of an object to resist any attempt to
change its velocity is called Inertia
Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its
velocity (acceleration)
r r
a ∝ Fnet
If mass is constant then
If force constant r
| a |∝
1
m
|a|
Mass is an inherent property of an object
m
Mass is independent of the object’s surroundings
Mass is independent of the method used to measure it
Mass is a scalar quantity
The SI unit of mass is kg
Physics 207: Lecture 7, Pg 21
Exercise
Newton’s 2nd Law
An object is moving to the right, and experiencing
a net force that is directed to the right. The
magnitude of the force is decreasing with time
(read this text carefully).
The speed of the object is
A.
B.
C.
D.
increasing
decreasing
constant in time
Not enough information to decide
Physics 207: Lecture 7, Pg 22
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Physics 207 – Lecture 7
Exercise
Newton’s 2nd Law
A 10 kg mass undergoes motion along a line with a
velocities as given in the figure below. In regards to
the stated letters for each region, in which is the
magnitude of the force on the mass at its greatest?
A.
B.
C.
D.
E.
B
C
D
F
G
Physics 207: Lecture 7, Pg 23
Home Exercise
Newton’s 2nd Law
A constant force is exerted on a cart that is initially at rest on
an air table. The force acts for a short period of time and
gives the cart a certain final speed s.
Force
Cart
Air Track
In a second trial, we apply a force only half as large.
To reach the same final speed, how long must the same force be
applied (recall acceleration is proportional to force if mass fixed)?
A.
B.
C.
D.
4 x as long
2 x as long
1/2 as long
1/4 as long
Physics 207: Lecture 7, Pg 24
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Physics 207 – Lecture 7
Home Exercise Newton’s 2nd Law
Solution
Force
Cart
Air Track
F = ma
Since F2 = 1/2 F1
a2 = 1/2 a1
We know that under constant acceleration, v = a ∆t
So,
a2 ∆t2 = a1 ∆t1 we want equal final velocities
1/2 a1 / ∆t2 = a1 / ∆t1
∆t = 2 ∆t
2
1
(B) 2 x as long
Physics 207: Lecture 7, Pg 25
Home Exercise
Newton’s 2nd Law
A force of 2 Newtons acts on a cart that is initially at rest
on an air track with no air and pushed for 1 second.
Because there is friction (no air), the cart stops
immediately after I finish pushing.
It has traveled a distance, D.
Force
Cart
Air Track
Next, the force of 2 Newtons acts again
but is applied for 2 seconds.
A.
B.
The new distance the cart moves relative
to D is:
C.
D.
8 x as far
4 x as far
2 x as far
1/4 x as far
Physics 207: Lecture 7, Pg 26
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Physics 207 – Lecture 7
Home Exercise
Solution
Force
Cart
Air Track
We know that under constant acceleration,
∆x = a (∆t)2 /2
(when v0=0)
Here ∆t2=2∆t1,
F2 = F1 ⇒ a2 = a1
1
a ∆ t 22 (
2 ∆ t1 )2
∆x2 2
=
=
=4
∆ x1 1 a ∆ t 2
∆ t12
1
2
(B) 4 x as long
Physics 207: Lecture 7, Pg 27
Home Exercise: Physics in an Elevator
Gravity and Normal Forces
A woman in an elevator is accelerating upwards
The normal force exerted by the elevator on the woman is,
(A) greater than
(B) the same as
(C) less than
the force due to gravity acting on the woman
(REMEMBER: Draw a FREE BODY DIAGRAM)
Physics 207: Lecture 7, Pg 28
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Physics 207 – Lecture 7
Moving forces around
Massless strings: Translate forces and reverse their
direction but do not change their magnitude
(we really need Newton’s 3rd of action/reaction to justify)
string
T1
-T1
Massless, frictionless pulleys: Reorient force direction
but do not change their magnitude
T2
T1
-T1
| T1 | = | -T1 | = | T2 | = | T2 |
-T2
Physics 207: Lecture 7, Pg 29
Scale Problem
You are given a 1.0 kg mass and you hang it
directly on a fish scale and it reads 10 N (g is
10 m/s2).
10 N
1.0 kg
Now you use this mass in a second
experiment in which the 1.0 kg mass hangs
from a massless string passing over a
massless, frictionless pulley and is anchored
to the floor. The pulley is attached to the fish
scale.
What force does the fish scale now read?
?
1.0 kg
Physics 207: Lecture 7, Pg 30
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Physics 207 – Lecture 7
Scale Problem
Step 1: Identify the system(s).
In this case it is probably best to treat each
object as a distinct element and draw three
force body diagrams.
One around the scale
One around the massless pulley (even
though massless we can treat is as an
“object”)
One around the hanging mass
Step 2: Draw the three FBGs. (Because this
is a now a one-dimensional problem we
need only consider forces in the y-direction.)
?
1.0 kg
Physics 207: Lecture 7, Pg 31
Static and Kinetic Friction
Friction exists between objects and its behavior has been
modeled.
At Static Equilibrium: A block, mass m, with a horizontal force F
applied,
Direction: A force vector ⊥ to the normal force vector N and
the vector is opposite to the direction of acceleration if µ were
0.
Magnitude: f is proportional to the applied forces such that
fs ≤ µs N
µs called the “coefficient of static friction”
Physics 207: Lecture 7, Pg 32
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Physics 207 – Lecture 7
Case study ... big F
Dynamics:
x-axis i :
y-axis j :
max = F − µKN
may = 0 = N – mg or N = mg
F − µKmg = m ax
so
fk
v
j
N
F
i
ma
ax
fk
µK mg
mg
g
Physics 207: Lecture 7, Pg 33
Case study ... little F
Dynamics:
x-axis i :
y-axis j :
so
max = F − µKN
may = 0 = N – mg or N = mg
F − µKmg = m ax
fk
v
j
N
F
i
ma
ax
fk
µK mg
mg
g
Physics 207: Lecture 7, Pg 34
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Physics 207 – Lecture 7
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
Σ Fx = 0 = -F + fs
fs = F
FBD
Σ Fy = 0 = - N + mg N = mg
N
As F increases so does fs
F
m
fs
1
mg
Physics 207: Lecture 7, Pg 35
Static friction, at maximum (just before slipping)
Equilibrium: A block, mass m, with a horizontal force F applied,
Direction: A force vector ⊥ to the normal force vector N and the
vector is opposite to the direction of acceleration if µ were 0.
Magnitude: fS is proportional to the magnitude of N
fs = µs N
N
F
m
mg
Physics 207: Lecture 7, Pg 36
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fs
Physics 207 – Lecture 7
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
Σ Fx = 0 = -F + fk
FBD
fk = F
v
Σ Fy = 0 = - N + mg N = mg
N
F
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
m
fk
1
mg
fk = µk N
Physics 207: Lecture 7, Pg 37
Sliding Friction: Quantitatively
Direction: A force vector ⊥ to the normal force vector N and
the vector is opposite to the velocity.
Magnitude: fk is proportional to the magnitude of N
fk = µk N
( = µK mg
g in the previous example)
The constant µk is called the “coefficient of kinetic friction”
Logic dictates that
µS > µK
for any system
Physics 207: Lecture 7, Pg 38
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Physics 207 – Lecture 7
Coefficients of Friction
Material on Material
µs = static friction
µk = kinetic friction
steel / steel
0.6
0.4
add grease to steel
0.1
0.05
metal / ice
0.022
0.02
brake lining / iron
0.4
0.3
tire / dry pavement
0.9
0.8
tire / wet pavement
0.8
0.7
Physics 207: Lecture 7, Pg 39
An experiment
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find µS
N
T
Static equilibrium: Set
m2
m2 and add mass to
m1 to reach the
breaking point.
Requires two FBDs
Mass 1
Σ Fy = 0 =
T – m1g
fS
T
m1
m2g
m1g
Mass 2
Σ Fx = 0 = -T + fs = -T + µS N
Σ Fy = 0 = N – m2g
T = m1g = µS m2g µS = m1/m2
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Physics 207: Lecture 7, Pg 40
Physics 207 – Lecture 7
A 2nd experiment
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find µK.
T
Dynamic equilibrium:
Set m2 and adjust m1
to find place when
T
a = 0 and v ≠ 0
fk
m1
Requires two FBDs
Mass 1
Σ Fy = 0 =
T – m1g
N
m2
m2g
m1g
Mass 2
Σ Fx = 0 = -T + ff = -T + µk N
Σ Fy = 0 = N – m2g
T = m1g = µk m2g µk = m1/m2
Physics 207: Lecture 7, Pg 41
An experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
N
Design an experiment to find µK.
T
Non-equilibrium: Set m2
T
and adjust m1 to find
regime where a ≠ 0
m1
Requires two FBDs
m1g
Mass 1
Σ Fy = m1a =
fk
m2
m2g
Mass 2
T – m1g
Σ Fx = m2a = -T + fk
Σ Fy = 0 = N – m2g
= -T + µk N
T = m1g + m1a = µk m2g – m2a µk = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 7, Pg 42
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Physics 207 – Lecture 7
Sample Problem
You have been hired to measure the coefficients of friction
for the newly discovered substance jelloium. Today you will
measure the coefficient of kinetic friction for jelloium sliding
on steel. To do so, you pull a 200 g chunk of jelloium across
a horizontal steel table with a constant string tension of 1.00
N. A motion detector records the motion and displays the
graph shown.
What is the value of µk for jelloium on steel?
Physics 207: Lecture 7, Pg 43
Sample Problem
Σ Fx =ma = F - ff = F - µk N = F - µk mg
Σ Fy = 0 = N – mg
µk = (F - ma) / mg & x = ½ a t 2 0.80 m = ½ a 4 s 2
a = 0.40 m/s2
µk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46
Physics 207: Lecture 7, Pg 44
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Physics 207 – Lecture 7
Recap
Assignment:
HW4, (due 2/17, 9 am, Wednesday)
Read Chapter 7
1st Exam Wednesday, Feb. 17 from 7:15-8:45 PM
Physics 207: Lecture 7, Pg 45
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