according to stationary observer
DESCRIPTION
accel a. R is reaction force = reading on scales. R. mg. Spring scales. Measured weight in an accelerating Reference Frame. According to stationary observer. F = ma Taking “up” as +ve R - mg = ma R = m(g + a). If a = 0 R = mg normal weight - PowerPoint PPT PresentationTRANSCRIPT
According to stationary observer
R
mgF = ma
Taking “up” as +ve
R - mg = ma
R = m(g + a)If a = 0 R = mg normal weight
If a is +ve R = m(g + a) weight increase
If a is -ve R = m(g - a) weight decrease
R is reaction force
= reading on scales
Measured weight in an accelerating Reference Frame
accel a
Spring scales
According to traveller
F = ma
R - mg = ma
BUT in his ref. frame a = 0!
so R = mg!!
How come he still sees R changing when lift accelerates?
Didn’t we say the laws of physics do not depend on the frame of reference?
R
mg
R is reaction force
= reading on scales
Only if it is an inertial frame of reference! The accelerating lift is NOT!
Forces are Vectors so Directions are Important
Force #1
Force #2
Force #1Force #2
Total Force
Total Force = 0
mgmg
Why doesn’t Mick Doohan fall over?Why doesn’t Mick Doohan fall over?
Friction provides the central force
Friction provides the central force
In the rest reference frameIn the rest reference frame
What is Friction•Surfaces between two materials are not even
•Microscopically the force is atomic
Smooth surfaces have high friction
•Causes wear between surfaces
Bits break off
•Lubrication separates the surfaces
The Source of Friction between two surfaces
Static and Sliding (Dynamic) Friction
• Static frictional force: when nothing is sliding
• Sliding frictional force: when surfaces are sliding
• Static frictional forces always greater than sliding ones
• Lubrication provides microscopic rollers between surfaces
fF
mg
Static Friction
As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero does not move
F is now greater than f
and slipping begins
If no force F
No friction force fSurface with friction
As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly.
fF
f depends on surface properties.
Combine these properties into a coefficient of friction
f N
is usually < 1
Static f < or = s N
Surface with friction
Kinetic f = k N
f
F
f < fmax (= kN )
Static friction
Kinetic friction
Coefficient of Kinetic friction < Coefficient of Static friction
Slipping begins (fmax = sN )fmax
mg
Ff
At crit
F = f
mg sin crit = f = S N
Independent of m, or g.
Property of surfaces only
S = tan crit
mg sin
mg cos
= S mg coscrit
crit
crit
cosmgsinmg
thus S =
Coefficient of Friction
Material on Material
s = static friction
k = 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
F1
F2
Making the most of Friction
A F1 > F2
B F1 = F2
C F1 < F2
mg
N N
mg
fcrit1
fcrit2
fcrit = S N
Friction force does not depend on area!
fcrit = S mg
So why do Petrol Heads use fat tyres?
To reduce wear?
Tyres get hot and sticky which effectively increases .
The wider the tyre the greater the effect?
To reduce wear?
Tyres get hot and sticky which effectively increases .
The wider the tyre the greater the effect?
The tru
th!
Frictio
n is not a
s sim
ple as p
eople think
The tru
th!
Frictio
n is not a
s sim
ple as p
eople think tribophysics
Force of Tyre on road
Force of road on Tyre
acceleration
What force drives the car?
Driving Torque
Braking force
Friction road/tyres
v
d
f
v2 =vo2 + 2a(x-xo)
0 = vo2 + 2ad a
vd
2
20
F = ma
= smgMax value of a is when f is max.
Stopping Distance depends on friction
amax = - sg
-fmax = mamax -smg = mamax
vo
N
mg
fmax = sN
Thus since a
vd
2
20
maxmin a
vd
2
20
gv
ds
min
2
20
dmin depends on v2!! Take care!!
If v0 = 90 kph (24 m s-1) and = 0.6 ==> d = 50 m!!
r
v Fcent
mg
N
rmv
F2
cent
Fcent is provided by friction.
If no slipping the limit is when
Fcent = fs(limit)= sN = smg
grμv
rmv
mgμ
s
2
s
So that
Does not depend on m
So for a given s (tyre quality) and given r there is a maximum vel. for safety.
If s halves, safe v drops to 70%….take care!
Taking a curve on Flat surface
Lateral Acceleration of 4.5 g
The lateral acceleration experienced by a Formula-1 driver on a GP circuit can be as high as 4.5 g
This is equivalent to that experienced by a jet-fighter pilot in fast-turn manoeuvres.
Albert Park GP circuitCentral force provided Central force provided by friction.by friction.Central force provided Central force provided by friction.by friction.
mg
N
= v= v22/Rg/Rg
= 4.3= 4.3
= v= v22/Rg/Rg
= 4.3= 4.3
mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg
R = 70 m mv2/R
V=
55 m
s-1
for racing tyres is ~ 1 (not 4!).
How can the car stay on the road?
for racing tyres is ~ 1 (not 4!).
How can the car stay on the road?
Soft rubber
Grooved tread
Are these just for show, or advertising?
200 km/h
Another version of Newton #2
amF p= mv =momentum
F is a measure of how much momentum is transferred in time t
t
pF
dt
vdm
dt
)vm(d
dt
pd
Momentum p transferred over a time t gives a force:-
Distance travelled in 1 sec @ velocity v Distance travelled in 1 sec @ velocity v
Volume of air hitting each spoiler (area A) in 1 secVolume of air hitting each spoiler (area A) in 1 sec
Area A m2
mass of air (density ) hitting each spoiler in 1 secmass of air (density ) hitting each spoiler in 1 sec
Momentum of air hitting each spoiler in 1 secMomentum of air hitting each spoiler in 1 sec
If deflected by 900, mom change in 1 secIf deflected by 900, mom change in 1 sec
Newton says this is the resulting forceNewton says this is the resulting force
@ 200 kph v = 55 m s-1
A ~ 0.5 m2
~ 1 kg m-3
F ~ 3 x 104 N
~ 3 Tonne!
= v m
= v x A m3
= x v x A kg
= x v2 x A kg m s-1
mvmv22/R = /R = N = N = mgmgmvmv22/R = /R = N = N = mgmg
mvmv22/R = /R = N = N = (m + 3000) g(m + 3000) gmvmv22/R = /R = N = N = (m + 3000) g(m + 3000) g
Big Consequences!
(USGS)
VISCOUS DRAG FORCE
Assumptions
low viscosity (like air)
turbulent flow
What is it?
like fluid friction
a force opposing motion as fluid flows past object
What does the drag force depend on?D D velocity (v velocity (v22))
D D effective area (A) effective area (A)
D D fluid density ( fluid density (
D D A vA v22
D= ½ C A v2
D D velocity (v velocity (v22))
D D effective area (A) effective area (A)
D D fluid density ( fluid density (
D D A vA v22
D= ½ C A v2
C is the Drag coefficient.
It incorporates specifics like
shape, surface texture etc.
v
Fluid of density
V m
Volume hitting object in 1 sec. =AV
Mass hitting object in 1 sec. = AV
momentum (p) transferred to object in 1 sec. = ( AV)V
Force on object = const AV2
t
pF
Area A
In 1 sec a length of V metres hits the object
V
mg
mg
D
mg
D
V
V=0
F = mg - D
F = mg -1/2CAv2
D increases as v2
until F=0
i.e. mg= 1/2CAv2
AC
mg2v
AC
mg2v
term
term2
0mgAv1/2Cdt
dvm 2
F = mg –DD
mg
ma = mg -D
D- mgdt
dvm
2/1Ac
m2
)]e1(Ac
gm2[v
t
2/1]Ac
gm2[v
2/1Ac
m2
)]e1(Ac
gm2[v
t
When entertainment defies reality
D= ½ CAv2
Assume C = 1
v = 700 km h-1
Calculate:
Drag force on presidents wife
Compare with weight force
Could they slide down the wire?
D= ½ CAv2
Assume C = 1
v = 700 km h-1
Calculate:
The angle of the cable relative to horizontal.
Compare this with the angle in the film (~30o)
In working out this problem you will prove the expression for the viscous drag force
2AvC2
1F