Ch 6
Forces
Question: 9
Problems: 3, 5, 13, 23, 29, 31, 37, 41,
45, 47, 55, 79
Friction
When is friction present in ordinary life?
- car brakes
- driving around a turn
- walking
- rubbing your hands together to make warmth
- belay/rappel device in rock climbing
- moving parts in an engine
- sled pulled across ground
Friction
• Friction is due to surfaces not being perfectly
smooth.
• Friction force ALWAYS opposes the direction of
motion.
• The result of friction forces is energy (heat)
• Two types of friction:
– Static friction force keeps the object at rest, and is
needed to be overcome so the object can begin
motion.
– Kinetic friction occurs when one object is sliding
against another surface.
Even if two surfaces look smooth, they may
not be when viewed on the atomic level.
Points in contact stick and hold the surfaces together.
Only some of the points on the surfaces come into
contact. These points produce static fiction.
If the static friction is overcome, the high points tear
and a continuous re-forming and breaking of contact
points is produced when the objects slide.
1) Push horizontally on a heavy box. The box does not move.
Acceleration is zero. The net force is zero. The frictional
force is counteracting your force.
2) Push harder. The box still does not move. The frictional
force must have increased.
3) Eventually you push hard enough the box begins to slide. At
that point you pushed with a force just greater than the
maximum friction force.
The friction force has an upper limit.
This upper limit is determined by two values.
The coefficient of static friction.
The normal force between surfaces.
Ffs, max
Magnitude of
friction force
sliding
stationary
Breakaway (sliding occurs)
How to calculate frictional force.
Frictional force, Ff , depends on the material of
the two surfaces involves. Given by coefficient
of friction ( ).
There are two coefficients: static ( s) and kinetic ( k)
Ff also depends on the normal force
Static friction force
Ff
sF N
Kinetic friction force Ff = kFN
no sliding
sliding
Static friction
• Ff
sF N
• The static friction force can vary from 0 to sFN.
• sFN is known as the maximum static friction force.
• This is the force needed to be overcome to start
sliding an object across the surface.
When working problems involving friction, you
need to determine the direction of the friction
force.
M1
M2
Depending on the masses of M1 and M2 and the
angle q, the friction force may be pointing up or
down the incline.
Determine which direction M1 would go if there was
no friction. Friction will be in the other direction.
Example
What coefficient of friction is needed to keep
a block from sliding down an incline?
The static friction force balances
out the parallel component of
The weight.
FN
FP
w
FN
Fs balances out FP
From earlier we know FP = mg sin
When the box is about to slide down:
FN
Ff = sFN = smg cos
Therefore:
smg cos = mg sin
Solve for s gives:
s = tan
Fs
FP
w
FN
This tells us the static coefficient, s, needed to
keep an object from sliding down an incline
depends only on the incline.
In this example you are holding a book still, up
against the wall. How hard do you have to
push to keep the book from sliding?
Free body diagram
for the book.
Ff
FPush
FN
W
If the book is not accelerating, the sum of the
forces is zero.
FN and Fpush counteract each other in the
Ff
horizontal direction.
FPush
FN
W
Weight and friction force balance each other
out in the vertical direction. The friction is
what is holding up the book.
Let the book have mass 5 kg and s = .4.
W = (5kg)g = 49 N
For W = Ff, Ff = sFN = 0.4FN = 0.4 Fpush
49 N = 0.4 Fpush
Fpush= 122.5 N
Kinetic Friction Problem
You are pushing a block up an incline. The angle of
inclination is 30 degrees. The mass of the block is
20 kg. The coefficients of friction are s = 0.3 and
k = 0.2. What force must be applied to the box to
keep the speed constant?
FN
Free body diagram
Fapplied
Ff
W
Since block is sliding across surface use
k
= 0.2.
Since velocity is constant, a = 0.
Fnet = ma = 0
Free body diagram
FN
+
Fapplied
We are only concerned with
motion along the incline.
Ff
Incline
W
Fapplied, Ff, and W have components along the incline.
Writing 2nd Law along incline we get:
Fapplied - Ff - mg sin 30 = ma = 0
Fapplied = Ff + mg sin 30 = kmg(cos 30)+mg(sin 30)
Fapplied = 0.2(20kg)g(cos 30) + (20kg)g(sin 30) = 131 N
Pulling with force (F) on box at an angle
2nd Law equations give us:
X-direction: F cos = m ax
Free Body Diagram
FN
W
F
Y-direction: F sin + FN – W = m ay = 0
FN = W – F sin
Normal force is reduced
So the friction force is less than if the rope
is horizontal.
Drag force and terminal speed
A fluid is anything that can flow, gases or liquids.
When there is a relative velocity between an
object and the surround fluid, there will be a drag
force.
examples: air resistance, water resistance
Will deal with cases where the object is blunt, the
fluid is air, and the motion is fast enough so that
the air becomes turbulent behind the object.
Drag force depends on the following values:
drag coefficient
C
density of air
cross-sectional area
A
velocity
v
D = ½ C Av2
The drag force will always be in the direction
opposite of the motion.
D = ½ C Av2
The drag force depends on the velocity. At slow speeds, it
is small, at high speeds becomes very large.
As the velocity of an object increases, so does the drag
force. Eventually the drag force can equal the accelerating
force. When they balance out, the velocity becomes
constant.
This is defined as the terminal velocity.
D
D
D
D
Fg
Fg
Fg
Fg
Fg
When ½ C Av2 – Fg = 0, v will be the terminal
velocity, vt.
vt
2 Fg
C A
Some terminal speeds in air
Object
Terminal speed (m/s)
95% Distance (m)
Shot put
145
2500
Sky diver
60
430
baseball
42
210
Tennis ball
31
115
Basketball
20
47
Ping-pong ball
9
10
Raindrop (r=1.5mm)
7
6
parachutist
5
3
95% distance is the distance needed to fall to reach 95% of the terminal velocity.
Uniform circular motion
a
v2
R
centripetal acceleration
Centripetal force – force that produces the circular motion.
examples: dive car around turn, the centripetal force is the
friction between wheels and road
orbiting satellite, the centripetal force is the gravitational pull
of the Earth
swing a ball in a circle, the centripetal force is the tensioon in
the rope
Applying Newton’s 2nd Law to circular motion,
2
v
we get: F m
R
This force is directed towards the center of the circle.
When working with Newton’s 2nd law and circular
motion, replace the acceleration in F = ma with that
of the centripetal acceleration.
Obtain the sign by looking at what direction the
centripetal force(acceleration) points.
Swing a ball, of mass = 1 kg, on a 0.5 meter long string
in a vertical circle with a constant speed of 2 m/s.
Find the tension in the string when the ball is at the
top and bottom of the circle.
Look at free body diagrams
2m
T
a=v2/R
a=v2/R
Fg
Bottom
Fg T
Top
T
a=v2/R
When the ball is at the bottom of the
circle, Newton’s 2nd law gives us:
Fg
a=v2/R
Fg T
T
mg
T
mg
v2
m
R
v2
m
R
When the ball is at the top of the
circle, Newton’s 2nd law gives:
T
T
mg
v2
m
R
v2
m
R
mg
Sample problem 6-8
Rider leans against wall of rotating cylinder. Floor falls
away, how fast does cylinder have to turn, to keep
rider from falling (what is the velocity of the rider).
radius = 2.1 m
s =0.4
Ff
N
a=v2/R
Fg
For the rider not to fall down, the friction must equal
the weight.
Ff – mg = may ay = 0
Ff = mg
s N = mg
N = mg/
s
In the horizontal plane, the normal force provides
the centripetal acceleration.
N = m v2/R
mg
v
NR
m
s
m
R
gR
s
(9.8 m s 2 )(2.1m)
0.4
7.2m / s
Problems: 8, 20, 30, 44