A crate is located at the center of a flatbed truck. The
truck accelerates toward the east, and the crate
moves with it, not sliding on the bed of the truck.
What is the direction of the friction force exerted
by the bed of the truck on the crate.
A) to the west
B) to the east
C) there is no friction because the crate is not sliding
An applied force F attempts to slide a block
along a surface
– If the block does not move,
fs = F
fs increases when F increases until ….
– fs has a maximum value: fs, max
fs, max = µs N
– If F exceeds fs, then the object moves
and kinetic friction applies. fk
µs : coefficient of static friction
µk : coefficient of kinetic friction
N : normal force
The key concept is that if you see an object
accelerate, then there *had* to be an unbalanced
force in that direction!
Chapter 06: Circular Motion
Hurricane Frances, September 2004
Chapter 6: Uniform circular motion
• For uniform circular motion, the centripetal
acceleration
a = v2/R
which is caused by a force called centripetal force F
F = ma = m(v2/R)
– direction of F: points radially inward
– Centripedal force is not a new kind of force
Question: What are the centripetal forces exerted on
the following object which are in uniform circular
motion?
A) A ball attached to the end of a string
tension on the string
B) A satellite moving around the Earth
Gravitational Force
C) A car driving in a circle
friction
D) A car driving in a circle on a tilted track
friction and gravitational force
When you ride in a Ferris wheel at constant speed,
what are the directions of your acceleration a
and the normal force N when you are:
(a) at the top of the wheel
N
mg
N + mg = mv2/r =>N = mv2/r − mg
(b) at the bottom of the wheel
N − mg = mv2/r =>N = mg + mv2/r
N
mg
(c) at which point, is the normal force the largest in
magnitude?
Definitely at the bottom!
tension on the string causes the object to move in a circle
2
v
v
aC =
R
Example: (Sample problem 6-8 in the book) In
the figure, a person is riding the Rotor.
Suppose that the coefficient of static friction
µs between the rider’s clothing and the
canvas is 0.4 and the cylinder’s radius R is
2.1m.
(A) What minimum speed v must the cylinder
and rider have if the rider is not to fall when
the floor drops.
2
v
v
v
N = ma C = m
R
v2
µSm
= mg
R
Rg
v=
= 7.17 m / s
µS
2
v
v
aC =
R
fS = µSN = mg
fS
N
mg
Car rounding a flat curve
Friction is causing the car to move in a circle.
Car rounding a banked curve
Friction and mgsinθ (the Normal component)
are causing the car to move in a circle.
for the example in the book
2
v
v
f S + FN = ma C = m
R
0
v2
FNr = mg sin θ = m
R
FNy = mg cos θ
v2
tan θ =
gR
Problem Chapter 06-47
v = 480 km/h and θ = 40o. What is the radius?
Fl sinθ
Fl cosθ θ
lift
FC = maC = mv2/R
r: Fl sinθ = mv2/R
y: Fl cosθ = mg
mg
Problem 6-47 continued
Daily Quiz, September 13, 2004
R
The “Corkscrew” ride at Cedar Point has an inverted
section. Assume that the radius is 5 m. What is the
minimum speed that must be maintained at the top to keep
the car on the track? For this problem, let g = 10 m/s2
Daily Quiz, September 13, 2004
0)
1)
2)
3)
9)
0 m/s
5 m/s
7 m/s
49 m/s
none of the above
R
The “Corkscrew” ride at Cedar Point has an inverted
section. Assume that the radius is 5m. What is the
minimum speed that must be maintained at the top to keep
the car on the track? For this problem, let g = 9.8 m/s2
Daily Quiz, September 13, 2004
vminimum = ?
0) 0 m/s 1) 5 m/s
9) none of the above
R
2) 7 m/s
3) 49 m/s
Daily Quiz, September 13, 2004
v min imum = gR = (9.8)(5) = 49 = 7.0m / s
R
0) 0 m/s 1) 5 m/s
9) none of the above
2) 7 m/s
3) 49 m/s
Problem Chapter 06-51
d = L = 1.70m and Tupper = 35N.
Find:
A) Tlower
the ball
B) net force on the ball
D) direction of Fnet,string
C) speed of
Problem 6-51 continued
Problem 6-51 continued
Problem 6-51 continued
Problem Chapter 06-52
m = 0.040 kg, L = 0.90m, circumference = 0.94 m.
Find:
A) String tension
B) period of motion
Problem 6-52 continued