Lecture 8

Goals:
 Solve 1D & 2D problems introducing forces
with/without friction
 Utilize Newton’s 1st & 2nd Laws
 Begin to use Newton’s 3rd Law in problem solving
Physics 201: Lecture 9, Pg 1
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, FNET =  F = ma
Law 3:
Forces occur in pairs: FA , B = - FB , A
(For every action there is an equal and opposite reaction.)
Read: Force on A by B
Physics 201: Lecture 9, Pg 2
Exercise Newton’s 3rd Law


Two blocks are being pushed by a massless finger on a
horizontal frictionless floor.
How many action-reaction force pairs are present in this
exercise?
a
A.
B.
C.
D.
b
2
4
6
Something else
Physics 201: Lecture 9, Pg 3
Newton’s Law
The acceleration of an object is directly proportional to the
net external force acting upon it.
The constant of proportionality is the mass.

This is a vector expression

F  Fx î  Fy ĵ  Fz k̂
FNETx  ma x
FNETy  ma y
FNETz  ma z
Physics 201: Lecture 9, Pg 4
Analyzing Forces: 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 201: Lecture 9, Pg 5
Free Body Diagram
Step one: Define the system
T2
T1
q2
q1
Eat at Bucky’s
mg
T2
T1
q1
q2
mg
Step two: Sketch in force vectors
Step three: Apply Newton’s 1st & 2nd Laws
(Resolve vectors into appropriate components)
Physics 201: Lecture 9, Pg 6
Free Body Diagram
T1
T2
q2
q1
Eat at Bucky’s
mg
Vertical :
y-direction
Horizontal :
x-direction
0 = -mg + T1 sinq1 + T2 sinq2
0 = -T1 cosq1 + T2 cosq2
Physics 201: Lecture 9, Pg 7
Pushing and Pulling Forces


String or ropes are examples of
things that can pull
You arm is an example of an object
that can push or push
Physics 201: Lecture 9, Pg 8
Scale Problem

You are given a 5.0 kg mass and you hang it
directly on a fish scale and it reads 50 N (g is
10 m/s2).
50 N
5.0 kg


Now you use this mass in a second
experiment in which the 5.0 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?
?
5.0 kg
Physics 201: Lecture 9, Pg 9
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.)
?
5.0 kg
Physics 201: Lecture 9, Pg 10
Scale Problem
3:
T”
T’
1:
2:
T
?
W

-T ’
 Fy = 0 in all cases
1.0 kg
-T
-T
?
-mg
1: 0 = -2T + T ’
2: 0 = T – mg  T = mg
3: 0 = T” – W – T ’ (not useful here)
 Substituting 2 into 1 yields T ’ = 2mg = 100 N
(We start with 50 N but end with 100 N)
5.0 kg
Physics 201: Lecture 9, Pg 11
Newton’s 2nd Law, Forces are conditional
A woman is straining to lift a large crate, without success. It
is too heavy. We denote the forces on the crate as follows:
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).
Which of following relationships between these forces is
true, while the person is trying unsuccessfully to lift the
crate? (Note: force up is positive & down is negative)
A.
B.
C.
D.
P+C<W
P+C>W
P=C
P+C=W
Physics 201: Lecture 9, Pg 12
Another experiment
A block is connected by a horizontal massless string. There
is a known mass 2.0 kg and you apply a constant force of
10 N. What is the acceleration of the block if the table top
is frictionless?
N
FBD
 Fx = max = -T
 Fy = 0 = N - mg
ax = -T/m = 5 m/s2
m
T
1
mg
Physics 201: Lecture 9, Pg 13
Version 2
A block is connected by a horizontal massless string. There
is a known mass 2.0 kg and you apply a constant
horizontal force of 10 N. The surface is frictionless and it
is inclined 30° from horizontal.
What is the acceleration of the block down the slide?
 Fx = max = Tx+Wx
 Fy = 0 = N + Ty+Wy
max = -T cos 30°+W sin 30°
ax = -T/m 3½/2 - g / 2
ax = -5(1.732)/2- 5 = -7.2 m/s2
What is the apparent weight of the block on
the slide?
N = Ty+Wy = T sin 30° -W cos 30°
N = -10(0.5) + 20(0.866) = 12.3 N
FBD
T
30°
30° 30°
mg = W
Physics 201: Lecture 9, Pg 14
Version 3
A block is connected by a horizontal massless string. There
is a known mass 2.0 kg and you apply a constant
horizontal force of 34.6 N. The surface is frictionless and
it is inclined 30° from horizontal.
What is the acceleration of the block down the slide?
 Fx = max = Tx+Wx
 Fy = 0 = N + Ty+Wy
max = -T cos 30°+W sin 30°
ax = -T/m 3½/2 - g / 2
ax = -17.3(1.732)/ 2- 5 = -35 m/s2
What is the apparent weight of the block on
the slide?
N = Ty+Wy = T sin 30° -W cos 30°
N = -34.6(0.5) + 20(0.866) = 0.0 N !!!
FBD
T
30°
30° 30°
mg = W
Physics 201: Lecture 9, Pg 15
Another experiment: A modified Atwood’s machine
Two blocks, m1 & m2, are connected by a massless frictionless
string/pulley on the table as shown. The table surface is
frictionless and little g acts downward.
What is the acceleration of the horizontal block?
N
T
Requires two FBDs and
m2
Newton’s 3rd Law.
Mass 1
 Fy = m1a1y= T – m1g
Mass 2
 Fx = m2a2x = -T
 Fy = 0 = N – m2g
T
m1
m2g
m1g
Correlated motion:
|a1y| = | a2x| ≡ a
If m1 moves up moves m2 right
Physics 201: Lecture 9, Pg 16
Another experiment: A modified Atwood’s machine
Two blocks, m1 & m2, are connected by a massless frictionless
string/pulley on the table as shown. The table surface is
frictionless and little g acts downward.
What is the acceleration of the horizontal block.
N
T
Requires two FBDs and
m2
Newton’s 3rd Law.
T
Mass 1
A.  Fy = m1 a= T – m1g
m1
m2g
m1g
Subbing in T from B into A
Mass 2
B.  Fx = m2 a = -T
 Fy = 0 = N – m2g
m1 a= - m2 a – m1g
m1 a + m2 a = m1g
a = m1g / (m1 + m2 )
Physics 201: Lecture 9, Pg 17
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
ma
fFRICTION
i
mg
Physics 201: Lecture 9, Pg 18
If no acceleration



No net force
So frictional force just equals applied force
Key point: It is conditional!
N
FAPPLIED
fFRICTION
j
i
mg
Physics 201: Lecture 9, Pg 19
Friction...

Friction is caused by the “microscopic” interactions between
the two surfaces:
Physics 201: Lecture 9, Pg 20
Friction: Static friction
Static equilibrium: A block with a horizontal force F applied,
 Fx = 0 = -F + fs
 fs = F
FBD
 Fy = 0 = - N + mg  N = mg
As F increases so does fs
N
F
m
fs
1
mg
Physics 201: Lecture 9, Pg 21
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 m were 0.
Magnitude: fS is proportional to the magnitude of N
fs = ms N
N
F
m
mg
Physics 201: Lecture 9, Pg 22
fs
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
 Fx = 0 = -F + fk
 fk = F
 Fy = 0 = - N + mg  N = mg
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
FBD
v
N
F
m
fk
1
mg
fk = mk N
Physics 201: Lecture 9, Pg 23
Model of Static Friction (simple case)
Magnitude:
f is proportional to the applied forces such that
fs ≤ ms N
ms called the “coefficient of static friction”
Direction:
Opposite to the direction of system acceleration if m were 0
Physics 201: Lecture 9, Pg 24
Sliding Friction

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 = mk N
( = mK mg in the previous example)

The constant mk is called the “coefficient of kinetic friction”

Logic dictates that
mS > mK
for any system
Physics 201: Lecture 9, Pg 25
Coefficients of Friction
Material on Material
ms = static friction
mk = 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 201: Lecture 9, Pg 26
Other Forces are Conditional


Notice what happens if we change the direction of the applied
force
The normal force can increase or decrease
F
j
N
i
Let a=0
fF
mg
Physics 201: Lecture 9, Pg 27
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 mS
N
T
Static equilibrium: Set
m2
m2 and add mass to
m1 to reach the
breaking point.
Requires two FBDs
Mass 1
 Fy = 0 =
T – m1g
fS
T
m1
m2g
m1g
Mass 2
 Fx = 0 = -T + fs = -T + mS N
 Fy = 0 = N – m2g
T = m1g = mS m2g  mS = m1/m2
Physics 201: Lecture 9, Pg 28
Static Friction with a bicycle wheel (not simple)
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 201: Lecture 9, Pg 29
For Thursday
 Read
Chapter 6
Physics 201: Lecture 9, Pg 30