Ishmam Zaman-Static Friction
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Date
Jan 9, 2024
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Experiment
6
Phys410
Friction
Fall
2023
Introduction
The
force
of
friction
is
not
a
fundamental
force
of
nature
since
it
can
be
explained,
at
least
in
principle,
in
terms
of
the
electromagnetic
interaction
between
atoms.
However,
in
many
instances
the
frictional
force
can
be
described
by
simple empirical
equations.
In
this
experiment
you
will
investigate
the
force
of
friction
between
a
wooden
blocksand
the
top
of
your
lab
table.
The
following
empirical
statements
are
sometimes
made
about
friction.
You
should
check
each
of
these
statements
to
see
if
they
apply,
or
nearly
apply,
in
your
experiment.
*The
maximum
static
friction
force
is
proportional
to
the
normal
force:
Jsmax
=
ps
N
*For
a
given
normal
force,
the
frictional
force
is
roughly
independent
of
the
area
of
contact
between
the
surfaces.
Setup
You
have
been
provided
with
a
wood
block.
Note
that
portions
of
the
block's
surface
have been
removed
to
provide
four
different
possible
contact
areas.
1.
Choose
a
small
flat
region
of
the
table
that
is
nearly
2
meters
from
the
edge
of
the
table
where
your
clamp
is
located.
Make
sure
this
region
is
clean.
2.
Set
up
your
equipment
as
suggested
by
the
following
sketch.
the
table
by
using
a
long
string
(measure
its
height
above
the
table
at
two
widely
separated
points).
Method
Maximum
Static
Friction
Versus
Normal
Force
for
a
Fixed
Area
of
Contact
4.
Start
with
a
block
without
any
added
weights.
Note
that
you
can
vary
the
force
of
contact
between
the
block
and
the lab
table
(normal
force,
N)
by
placing
weights
on
top
of
your
block
(don't
forget
to
weigh
your
block
too).
@
5.
Measure
the
maximum
force
that
static
friction
can
provide,
fmax,
for
a
given
normal
force
by
adding
weights
to
the
end
of
the
string
that
hangs
over
the
pulley
until
the
block
just
starts
to
slide.
Move
the
block
slightly
to
ensure
that
it
has not
"stuck”
to
the
table—
but
make
sure
that
the
block
is
absolutely
stationary
when
you
release
it.
Repeat
with
at
least
10
normal
forces
and
complete
the
attached
data
table.
6.
Make
a
graph
of
fsmax On
the
vertical
axis
and
N
on
the
horizontal
axis.
Maximum
Static
Friction
Versus
Surface
Area
of
Contact
for
a
Fixed
Normal
Force
7.
On
your
data
sheet,
make
a
sketch
for
each
face
of
your
block
complete
with
dimensions—use
a
caliper
to
measure
your
block.
8.
Set
up
your
equipment
as
before.
Place
a
fixed
weight
of
intermediate
size
on
your
block
to
produce
a
fixed
normal
force.
9.
Determine
the
maximum
static
friction
for
each
face
without
changing
the
normal
force.
Maximum
Tilt
angle
and
the
Coefficient
of
Static
Friction
10.
Estimate
the
static
friction
between
your
block
and
the
top
of
your
textbook
as
follows:
+Place
the
block
on
top
of
your
textbook
and
slowly
raise
one
end
of
the
textbook
until
the
block
begins
to
slide.
«You
can
calculate
the
coefficient
of
static
friction
between
your
book
and
the
block
from
the
following
equation
Ms
=
tan(emax),
where
Omax
is
the
maximum
angle
that
the
book
makes
with
respect
to
the
horizontal
before
the
block
begins
to
slide.
Questions
1.
What
is
the
slope
and
intercept
of
your
fomax
versus
N
graph?
Based
on
this
graph,
does
it
appear
that
the
maximum
frictional
force
proportional
to
the
normal
force?
Explain.
2.
Based
on
your
fsmax
versus
N
graph,
what
is
the
coefficient
of
static
friction,
15,
between
your
block
and
the
top
of
the
table?
What
are
the
dimensions
of
p5?
3.
Do
your
data
prove
that
the
frictional
force
is
not
proportional
to
the
area
of
contact!?
Do
you
feel
that
your
data
are
good
enough
to
show
that
the
frictional
force
has
no
significant
dependence
on
area
(within
the
limitations
of
your
measurements)?
Please
explain.
IHint:
If
y
is
proportional
to
,
theny
will
double
when
you
double
z.
Does
the
maximum
static
friction
appear
to
double
when
the
area
of
contact
is
doubled?
Partners
I(ré/flf;z,q.}
W
Friction
Data
Sheet
Name
MMW\
3
Maximum
Static
Friction
Versus
Normal
Force
for
a
Fixed
Area
of
Contact
M
(grams)
m
(grams)
|
N
=
Mg
(newtons)
|
fsmax
=
Mg
(newtons)
Wyl
(09|
2.Yly/-
0.837
VLTI
15030
339472
.22G
vYE.ugl
1760
4
(.7
M72
1.
71§
s
e
13004
T,10477
7.2730
L.
ua200.0
9]
[.13%/L
2.
7Y
Y64
|
220,091
F.31477
3.
(30
846,44
|
Y00.0
9
§.
L%
1,41
196y
9
[
vy7.
04
4.27Y477
Y.
LLoL
Y6
14|
446.04
1025
y7L
y.
451
\I1b.y
41
54¢.04
123472
G.753
Maximum
Static
Friction
Versus
Surface
Area
of
Contact
for
a
Fixed
Normal
Force
vy
L
,
l{
m
(grams)
|
fomax
(newtons)
|
A
(contact
area
in
cm?
from
sketch)
[
17%.0
9|
ITIS
70,088
W
172.081
1017
36.%64
s
135.00
|
1.727
17.
7b
Y[
102.09
109916
7.
\4b
Maximum
Tilt
angle
and
the
Coefficient
of
Static
Friction
Bmax
=
,]
li
(;
!
i
/Ls=tan(0max)=’___D'22‘é1
Y44
o
O
/
Z
o
‘W
W
@
0
V5
i
|
-
§
"
t_{
A
§
%
5
iN
‘\
\f(
;
2‘
"‘is
1
:k—i\\
E
7
1|
1
1%
om,
)
~
LA
M
G0
tm
0
Sketch
the
dimensions
of
the
contact
suléf'gcg
for
ea%}i‘%ace
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the
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A
01
em
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is a mass hanging by a spring under the influence of gravity. The force due to gravity, Fg, is acting
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On the right, at the beginning of an experiment, the mass is pushed upward (positive-y direction) by an amount y₁.
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DO
C.D
Frontly
у
Your tasks:
No Deflection
m
k
Fg = mg
Initial Condition
y
m
k
Write down an expression for the total energy If as the sum
Write down an expression for the total energy H
Fg = mg
Figure 3: System schematic for Problem 4.
Yi
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