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Dynamics
Fundamental Forces
>
relative
strength,
strongestbeing assigned
to 1,
smaller numbers
represented
by
fractions
>effective
range
how far can
it reach?
>
force
carrier/exchange
particle
"communicates"force
between
objects
Strong
Nuclear
force
a
strongest
force in nature (strength
of
- SNFL
>responsible
for
binding
the nucleus
>limited
range
sets a
limitto max nuclear
size,
this to the
length
of the
PT
> ifit didn't exist, we
wouldn'thave
any
elements above
Hydrogen
Electromagnetic
force
emEMae
>
binds matter
together
>
originally
thought
to be 2
distinctforces:
James
electrostatic:force of
attraction/repulsion
between charges
Maxwell,
magnetic:
force of
attraction/repulsion
between
magnets/magnetic
material
1864
these I
forces
were "flavors"of the same
force
> force of at
> infinite
range
Weak Nuclear
Force
>binds
neutrons
together
I
responsible
for
Beta
decay
a
necessary
for
stellar fusion
through
transmutation
of
neutrons
I
strength
of 10
>
range
of10-18m
Gravitational
Force
a)
a
i
>
weakestof fundamental forces
r
>
force attraction between
all
objects
with mass
I
force of 6
x 10-
> infinite
range
Newton's
Laws
> 1687
Newton
publishes
Principia
Mathematica
> he
lays
out 3 laws
relating
force,
mass
acceleration
First Law
>
inertial
law
a
=
directly
related
to,
&
inertia
resistance to
acceleration
proportional
to
&
inertia
a mass
>
unbalanced/net
force causes acceleration
>
if no fret,
no
i)
if F
=
0 then
TF
= 0
ii)
it
F;
to then Ff
=
V;
Third Law
S
action-reaction
law
FAB-FBA
GERD
est
Second Law
-|
> relates E, m,
a
mathematically
act
f
=
=
&
experiments:
i)
holding
me
constant,
determine how
a
relates
to
E force,
f
IN=11gm/s
->
f
f
Units: Newtons,
N
ii)
holding
constant,
determine
how
a
relates to m
x Fact:
ME
M
If
= ma (rector
sums,
use this)
kinds of
Friction
Fts
FA
>static
④
>
must
be overcome
in order to
move
from rest
no
movement.
Fnet
=
0
·present
only
when there
isn't relative motion
between the surfaces
FA
Ffk
> Kinetic >
weaker than static
↑netis
to the R
·present
only
when
surfaces are
moving
woteachother
(skidding) a is to the
R
vis
Lor
Friction
Strength
>magnitude
ofthe
friction force is
proportional
to
·
how hard the 2
surfaces are
pressed
together
(normal force,
FN)
the
materials from which the surfaces are made coefficientoffriction, M)
> other attributes that
may
I
may
not
have an
effect
in class lab
Coefficients of Friction
>static:
MS
>
Kinetic:
MK
>
both
depend
on the materials
in contact
·small
for
steel
on
ice/scrambled
egg
on
frying
pan
large
for rubber on
concrete/cardboard
box
on
carpeting
The
bigger
the
coefficient
friction,
the
bigger
the
frictional force
Static Friction Force
Varies
>
Its,
max is
a constantin a
given
problem,
butF, varies
>
Its matches
Fg
until
Ffexceeds fts,
max
Static Friction
Force Kinetic
Friction
Force
FAs
IMsFN-
normal
FAk
=
Mk
FN
-normal
↑
bett.
force
dest.
force
static
Kinetic
of
S
frictional of KE
frictional
force
force
I
once
objectbudges,
forget
about
Ms
I
use
MK
insted
>
Fox
is a constant
if materials don't
change
>no
max
fok
Values
<
typically,
0 if
Mk>
1,
it
would
be easier to lift an
object
carry
it than
to slide itacross the
floor
>
dimensionless (M's have
no units
->
Ff
=
Mfr)
FBD
then 2nd Law
(long
packet)
~
i
gmNx 00kg
on
the
FN EF
=
ma
1 -
-->c
Fr
>FA
FA-Ff
=
ma
buff
=
Mfr
=
Mw
=
rmg
M
=
V
W
FA-
Mmg
=
ma
never use
negatives
when
calculating
force
will
always
(long packet)
EF
=
ma
be
t
Ff
-w
=
Ff
=
m
I
t
Ff
=
w
ma
negecaonebfadirection
an
e
Ff
=
mg
ma Ff
=
mg
Ff
=
m(g
=
m(g
e:
A
paper
bag
holds
7.0kg
of
groceries
Veggies
at bottom of
bag
are
getting
it
wet and F of
bag
reduces to
50.0N
Determine a of
Dag
so it doesn't
rip
T
EF=ma
fg
F+=
ma
+od
=gE
ez down
FN
determine
a
N
EF=ma
exu
=
F-csfT
FT
F
ma
m
=
10.0kg
I
F
=
45.0N
FT
W
F+
Mmg
=
ma
Eg
[x+
Nts
=
M+g
=
mng
FT
=
Mmg=a
FN
T
=
8
= 350
.......
F
c.
c
...- > a
-> f
Fix
Ff
=
max
A.gEximer-manErEsMEFgoeEO
Ewoo-wImg-Fon
ente
350a
=
2.96m/
Inclined
Planes
>
inclined
planes/ramps
are defined
by
the
angle
the bottom
makes the horizontal
I
instead of X
and
y
reference
frames, use
11 and
I
Fr
Fg+=EglosO
Ema
wo
friction
de
gy=fgsino
a
=
En
al=
nO=gsin
O
:Kinematics
of
the
inclined
plane
(no
friction)
determine
angle
of
ramp
if a
box fired
up ramp
at
5.0m/s reaches
a distance
of
3.5m
vf
=
0m/s Vi
=
5m/
Vi=5.0m/s
Vf
=
0m/s
dist a
=
gsinO
vig
wine
e
d
=
3.5m
With
friction
:
situational
depending
stee,
if box
goes
up/down
accel is
always
down
Il
direction:
EF
=
ma, I
direction:
2f
= mal
=
0
F
Fgll-ff
=
ma, Fg
=
Fn
date
f,"
gosof
"
g(sinO-MCOSO)
=
a
EF
= ma
date;
a
t
fgu+ff=ma,
mgsinO+MmgcOsO=
ma,
g(sinO
McosO)
=
a,
Methods for
solving systems
m,
a,
-az
pagenor an
7 FA
M
=
0
①
FA
m,a
=
ma
determine
a
given
fA,m,
me
FA
=
m,a
m,a
FA
a
=
m,
↑FN-
a
Ff
M
↑
Fra
m,
t
mz
F1 <-> --
I
FT
j
FA
- E
W
N
⑤
FA-
Mmg-
m,a
lmg
=
ma
2F
=
m
EF
=
m
FA
Mm.g
-Mmzg
=
m,a
ma
FT
Ff,
=
m,a
but
Ff
=
lmg
② A-fz-z= mama
MMS
a
FT
um,g
=
m,a
-man
F
E
/
EoIm
n
o
-9-m,g
=
a
I
W
⑤
2F
=
m
57w
FT
Ff,
=
m,a
Wz
FT
=
ma
but
Ff
=
lmg
mrg
=
ma
FT
um,g
=
m,a
F
ma
Mm,g
Method 2
Modified FBS
>turn
system
into
single
object
aligns
(more
parts
of world
as
needed) (no
preferred
direction)
>replace
string
w/
massless
rod
>draw
modified
FBD
w/
mats
nodes
---> o
For
th
determine
as
a function of
M. M. Me,
g
new
FBS
aF
=
ma
FN
-- ->a
we
=
(m,
m2)
a
F,
so
we
meg-hm,g -a
m,
me
W,
Besame
te
:
Ff,
Pro
wSF
re
~
I
W
static
Equilibrium
&how
this:
>
the
dynamics
ofnot
accelerating
·
if0,
=
02
F+
=
F
2
conditions:
if
both
conditions met,
object
is static
the more
vertical
cable
supports
greatest
Translational
equilibrium
may
not
be static
yet
vertical load
27
=
0
7
=
T
0,
Or
- rotational equilibrium
·
&
FBD
&
·
FT,
=
1
z
IT
=
0 (tav)
FT
↑ FT
7
determine F.,
Fz
component
method
·
⑦
=
350
X.EFX
=
0
!
m
=
75k
FT,xTFT2X
=
0
FT,X
FT2x
FT,COSO,
=
FTLCSO
whole vestor
solution
FTF
SO
EF
=
0
->
Fozzy
2D
oren
Use
if I of
F+
1
F
2
w
=
0
y
.zfy
=
0 the
tensions
FT,y
Fzzy
=
w
are
given
Wsin
=
FT
=
422 N
FTisin8.
FTzsinOz
=
mg
WCS
=
F
1
=
602N
①x②
use ifno tensions
are
given
EcosO2.
ainDitEsinOmg
EIuCostannAIsinn
e
FT
=
421.58N
F
=
602N
Translational
Equilibrium
[F
=
0
Equilibriant
(fta)
->
Fea
=- Ef
> the
I force
that
brings
an
object
into
translational
equilibrium
-its
"
F,
fz
F
tip
to
tail
method
F
fz
fz
=
0
S
Fea
=
(f,
fz)
Fr -
Fr
Fr
Fa
=
>
--
why
do
powerlines
droop?
7
⑧
->
↓
W
as
f
4
Od
what must
I
be
so O =0? X:
Efx
=
Y:
SFy
=
0
if
0
=
0,
Frx
=
Fxx
2
fy:
W
cannot evaluate on.
Fr
=
2a
0
=
0
F
wsO=
frcosO
2
fsinO
=
L
Fr
=
20
Em
droop
because
they
must
t
Fr
limsin
=
0
0-
calculus
Rotational
Equilibrium
conditions
ET
=
0
or
ETcw
=
ETccu
massless beam,
determine
be 0.
=0.20m
m, mz
m
=
500g
M 1
m=
=
200g
1- ,-
lz
= 7
FBD
TW
=
TW,
FN
Wylz
=
w,b,
wet
e-
my-lz
=
m,g.l,
I
d
=
1
g50.20m
=
50 l
Wz
W,
Uniform
beam
uniform
in
construction
and in
composition
↑
same
I'll
L
X-r
'IIIII
the
geometric
centre
Equivalence
of
Fulcrum
Equilibrium
Stcw=
EICCW
placing
the fulcrum
atthe location of
a force eliminates
it
from
the
calc.
example:
A4.0m
long
uniform
beam
w/
a mass of
25kg
is
held
up
at each end
by
2
supports.
A
10kg
mass is located
as shown. Find
the
upward
force
from
each
support
FNZ
10K
I
FNI
FBD
N
B
↓
↓
F
11
=
2m M
23
=
4m
not
enough
info to use translational
equil.
place
fulcum under
one of the
unknowns
-> Fw,
to
start
ETc
=
Excw
Find
Fr,
with TEQ
IwB+TWm=
IFNz
Fri
FNz
=Wis
Wm
WB.
e, +Wm:lz
=
FNz:
FN
=
147 N
⑲lit
Mm
gilt
re
B
=
beam
01/12/
FBD
(include
form, Fwy)
FBr(*,r's,
F1)
ofwall
↑
↓
solve for Ff(M)
3.5m 4) solve
for
Fux,
fry
(18X
=
=
y
=
↓
- IFwl
Fwx
Fwy
=
Ew
&7.5m
i
Om
long
uniform beam
4
8549 w/mass
of
35kg
↑FT
temp
need O
Wall *
Beam
8
=
tan
not....
r
en
Glam
↓
Wall
Load
3.5m
0
=
Tension
o
V
2.5m
Fwy
WL
Star =ETcew
↑
pz
Yup
Tw
=
4FT
Wb.l.
w(
lz
=
F+
1.lz
↓
mpge,
+mglz
=
fsin8lz
WB
V
FT
=
e.+mcglc
sinGlz
WL
802-85.9.
23
F
=
1975N
X.5FX
= 0
Y
=zfy
=
0
FwX
=
f+x
FTy
=
Fwy+WBTWL
FwX
=
FTCs
Fwy
=
Fig
Wp
W,
Fwx
=
197520s(54.46)
Twy
=
Esino-mpg-mg
Fux
=
1148N
Fury
=
1975sin
35.9.8-85.9.
Fwy
=
43/N
5)Ew=
Fux-Ery
Iwl=AuEr
Fwx
1482
4312
it
in
Ifwl= 1226
N
Tipping
Point
when
a
load is
at the
"tipping
point", only
the
support
dosestto the end
provides any
upward
force FN
Al,
--
re-
light
fre
as
the
'I'
we
stres
neve
"
WB
surface
The Ladder
Problem
empty (nobody
on
it),
uniform
IFNz!=1ffl
a
FBD
↑
Fr
Iw
=
1 Ful
-b
=
0
A
always
at
bottom
a
i
FN,
I
W
M
= 0
