Friction
Kinetic friction:
f k  k N
fk- friction force
N - normal force
k
- coefficient of kinetic friction
Static friction:
s
f s  s N
- coefficient of static friction
k   s
Example: When you push a book against a wall, the static friction
between the wall and the book can prevent it from falling.
If you press harder, the friction force will be:
A. Larger than before
fS,book,wall
B. The same
C. Smaller than before.
Fbook,hand
Nbook,wall
Wbook,Earth
For the book not to fall down, fS = W
Pushing harder (increasing Fbook,hand) increases Nbook,wall and
therefore fS,MAX increases, but not the actual value of fS that we had,
which needs to continue to be exactly W.
Example: A 49 kg rock climber is climbing a “chimney” between two
rock slabs. The coefficient of static friction between her shoes and the
rock is 1.2; between her back and the rock it is 0.80. She has reduced
her push against the rock until her back and her shoes are on the
verge of slipping. What is her push against the rock?
fS feet
fs feet
fs back
N
fS back
N
N
N
y
x
W
x:
N  N  max  0
y:
f S ,back  f S , feet  mg  may  0
If the climber dos not slip or move up, the
value of the sum of frictional forces is fixed:
w
f S ,back  f S , feet  mg
fS,back  fS,feet  mg
with fS  SN
 If N is large enough, both forces will be less than SN
As N decreases, the ma ximum static friction gets closer
to the actual value of friction.
Eventually, N is just enough so that fS  SN . If the
climber pushes less than this, the friction will not be
enough to compensate her weight.
S,backN  S,feetN  mg  0
mg
(49 kg)(9.8 m/s2 )
N 

 240 N
( S,back  S,feet )
1.2+0.80
Answer B
Example: Trying to move a trunk
friction
Fby you
ffSk by
by floor
floor
μsN
μkN
Fby you
Static friction
Kinetic friction
Example: Trying to move a trunk
For “small” forces, the trunk
does not move. So there must Fby you
be a friction force fS = Fby you
fS by floor
You increase the applied force, but
the trunk still does not move:
static friction is increasing too.
Fby you
fS by floor
You increase the applied force even
more. Eventually, the trunk moves.
Static friction cannot be larger
than a certain value.
And then friction
becomes kinetic!
Fby you
fSk by
by floor
floor
Example: Box on incline with friction
N  m g cos  0
N
m gsin   f  m a
f
y
Wy 
1)Static: a  0
Wx
W
f s   s N   s m g cos

x
W  mg
Wx  m gsin 
Wy  m g cos
f s  m g sin 
m g sin    s m g cos
t an   s
 s  t an critical
2) Kinetic
f k   k N   k m g cos
m gsin    k m g cos  m a
a g (sin   k cos )
Example: A hand keeps a 20-kg box from sliding down a frictionless
incline. The plane of the incline makes an angle θ = 30° with the
horizontal. What is the magnitude of the force exerted by hand?
mgsinθ – F = m ax = 0
F = mg sinθ
N – mgcosθ = m ay = 0
N = mg cosθ
y
F = (20 kg)(9.8 m/s2)sin(30°)
= 98 N
N
FB,hand
x
mgcosθ
mgsinθ
Directions:
θ
mg
•Draw the free-body diagram
θ
• Choose axes (draw them!)
• Use Newton’s 2nd law in the
x and y-directions.
Centripetal force
F = ma
m v2
 2 
F
 m 2 r  m r
r
T 
Example: A stone of mass m sits at the bottom of a bucket. A string is
attached to the bucket and the whole thing is made to move in circles.
What is the minimum speed that the bucket needs to have at the highest
point of the trajectory in order to keep the stone inside the bucket?
F = ma
m v2
F
r
m v2
N  mg
r
N min  0
m v2
N
 mg
r
vmin  gr
Examples (centripetal force)
(1)
N
mg
(2)
N
F = ma
m v2
F
r
mv
N  mg
r
m v2
N
 mg
r
m v2
N  mg
r
m v2
N
 mg
r
2
mg
(3)
N
2
mg
mv
m g N 
r
m v2
N  m g
r
Example: Little Jacob (15 kg) sits on the edge of a merry-go-round of
radius 1.0 m while big sister makes it turn… faster and faster. How fast
can the system go before Jacob takes off if the coefficient of static
friction between Jacob’s pants and the merry-go-round is 0.5?
N
fs
mg
Static friction provides the needed radial acceleration:
Maximum speed  Maximum static friction:
MAX
fS  mr2
2
S mg  mr MAX
S g
(0.5)(9.8 m/s2 )
1


 2.2 rad/s  turn/s
r
1m
3
Example: A car of mass m with constant speed v drives through a
curve of radius R. What is the minimum value of the coefficient of
static friction between the tires and the road for the car not to slip?
v2
fs  m
R
Fnet  ma
fs
v2
m  S mg
R
fs  S N  S mg
v2
S 
gR