Midterm Exam #1
§ First midterm is next Wednesday, September 23 (80 minutes!!!)
Ø Units 1-5: 1D kinematics – Forces and FBDs (todays lecture)
Ø Things to bring: Pencil(s), calculator UofU ID card
Ø Bring your U of U ID card: you will not receive a grade unless you
show it upon turning in your exam
Ø No makeups!!! (Unless you are unavailable due to University/military
business or documented medical emergency – come talk to me!)
Ø Cheating: You will receive a zero and referred to the University
administration – may result in an “E” for the course and expulsion
§ You must contact Prof. Gerton by Wednesday, September 14 if you
have a conflict wit the exam times/days!!!
§ Exam Location: SFEBB 1110 (large lecture hall on first floor of
Spencer Fox Eccles Business Building)
Mechanics Lecture 6, Slide 1
Classical Mechanics
Lecture 6: Friction
Today's Concept:
Fric%on Mechanics Lecture 6, Slide 2
Time spent on Prelecture #6
Average for quiz 1 = 8.4/15 (class average = 11.6)
Average for quiz 2 = 8.9/15 (class average = 10.4)
Mechanics Lecture 6, Slide 3
Unit 6 Learning Objectives
If you master this unit, you should:
§ be able to compute the force of static friction between
two objects by solving Newton’s 2nd Law.
§ be able to determine the maximum force that can be
applied to an object before it starts to slide.
§ be able to compute the acceleration of one object sliding
over another in the presence of friction.
Mechanics Lecture 6, Slide 4
Friction
Mechanics Lecture 6, Slide 5
Example problem
A car of mass m drives around a circular track of radius R. The coefficient
of static friction between the car’s tires and the road is µs. The coefficient
of kinetic friction is µk.
What is the maximum speed of the car before it starts to skid?
Mechanics Lecture 6, Slide 6
1&2: Understand Problem & Describe Physics
A car of mass m drives around a circular track of radius R. The coefficient
of static friction between the car’s tires and the road is µs. The coefficient
of kinetic friction is µk.
What is the maximum speed of the car before it starts to skid?
1. What kind of friction are we dealing with?
2. Between which objects does this friction act?
3. As the car goes around the track, in which direction does the friction
force point?
Ø Which object are we talking about?
Ø Is the car maintaining constant speed?
Ø How does the situation change if the car speeds up or slows down?
4. How is the maximum friction force related to the car’s weight?
5. Should the maximum speed depend on the car’s mass?
Mechanics Lecture 6, Slide 7
Step 2: Describe the Physics
A car of mass m drives around a circular track of radius R. The coefficient
of static friction between the car’s tires and the road is µs. The coefficient
of kinetic friction is µk.
What is the maximum speed of the car before it starts to skid?
v2
a=
R
fs,max = µs N = µ
= µs mg
Mechanics Lecture 6, Slide 8
Steps 3 & 4: Plan and Execute the Solution
A car of mass m drives around a circular track of radius R. The coefficient
of static friction between the car’s tires and the road is µs. The coefficient
of kinetic friction is µk.
What is the maximum speed of the car before it starts to skid?
• The maximum possible centripetal acceleration occurs at the
maximum speed, vmax
• This is when the static friction force is maximized.
amax =
2
vmax
R
fs,max = µs N = µs Wcar
= µs mg
A) vmax = µs g
B) vmax
C)
p
R
p
= µs gR
p
vmax = µs R g
D) vmax =
p
µs gR
Mechanics Lecture 6, Slide 9
Follow up question
A car of mass m drives around a circular track of radius R. The coefficient
of static friction between the car’s tires and the road is µs. The coefficient
of kinetic friction is µk.
What is the force of static friction on the car when it travels at half the
maximum speed?
Mechanics Lecture 6, Slide 10
Stuff you asked about:
I struggled with the trigonometry on the box sliding down the ramp. i thought that this was a good lecture, very informa%ve, but could we have an example, just one li>le example, that doesn't involve a box? this lecture was very clear though. y
90
θ 90-θ
90
x
θ
mg
Mechanics Lecture 6, Slide 11
CheckPoint: Box in Accelerating Truck
A box sits on the horizontal bed of a moving truck. Sta%c fric%on between the box and the truck keeps the box from sliding around as the truck drives. µS
a
If the truck moves with constant accelera%on to the leE as shown, which of the following diagrams best describes the sta%c fric%onal force ac%ng on the box: A
B
C
Mechanics Lecture 6, Slide 12
Revote: Box in accelerating truck
A
B
C
D
µS
a
If the truck moves with constant acceleration to the left as shown,
which of the following diagrams best describes the static frictional
force acting on the box:
A
B
C
F~net
~a =
M
A) The friction is causing the box to accelerate in the same
direction as the truck.
B) The force of friction always acts in the opposite direction of
acceleration.
Mechanics Lecture 6, Slide 13
ACT!
A
B
C
D
A flower pot of mass M sits on a horizontal table. A horizontal string
having tension T applies a force on the pot, but static friction
between the pot and the table keeps the pot from moving.
What is the magnitude of the net force acting on the pot?
N
A) Mg
B) μsMg
f
T
X
Fy = M a = 0 ) N = M g
C) T
D) 0
W = Mg
X
Fx = M a x = 0 ) T = f
Since acceleration is zero, the net force MUST be zero also!
Mechanics Lecture 6, Slide 14
CheckPoint: Box on Table
A
B
C
D
A box of mass M sits on a horizontal table. A horizontal string
having tension T applies a force on the box, but static friction
between the box and the table keeps the box from moving.
What is the magnitude of the static frictional force acting on the
box?
A) Mg
T
f
M
B) µsMg
C) T
D) 0
Apply Newton’s 2nd Law:
X
=0
Fx = M a x = 0 ) T = f
Mechanics Lecture 6, Slide 15
CheckPoint: Box on Table
A box of mass M sits on a horizontal table. A horizontal string
having tension T applies a force on the box, but static friction
between the box and the table keeps the box from moving.
What is the magnitude of the static frictional force acting on the box?
Ff
X
T
M
Fx = M ax = 0 ) T = Ff
Ff depends on weight
µs does not!!!
What happens if I pull harder?
As long as v = 0: The static friction force will continue to match T…
…until Ff = µs N. After that, Ff can’t increase any more:
Ø T > Ff and the box will accelerate!
Mechanics Lecture 6, Slide 16
Example problem
A block (m2) slides on a table pulled by a string attached to a
mass (m1) hanging over the side. The coefficient of kinetic
friction between the sliding block and the table is µk. What is
the tension in the string?
m2
g
m1
Mechanics Lecture 6, Slide 17
CheckPoint: Blocks and Pulley
A block slides on a table pulled by a string a>ached to a hanging weight. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the sliding block and the table. m2
m2
Case 1
g
(No Friction)
Case 2
g
(With Friction)
m1
m1
In which case is the tension in the string larger?
A) Case 1
B) Case 2
C) Same
64% of you got this right
Mechanics Lecture 6, Slide 18
ACT: Blocks and Pulley, I
A
B
C
D
A block slides on a table pulled by a string a>ached to a hanging weight. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the sliding block and the table. +
a
a is positive when m1
m2
accelerates downward!
Case 1
(No Friction)
g
m1 g
T
m1
a
+
T = m1 a1 = m1 a
=) T = m1 (g
a)
m1 g
What is the tension in the string for Case 1?
A) T = 0
B) T = m1g
C) 0 < T < m1g
Mechanics Lecture 6, Slide 19
ACT: Blocks and Pulley, II
A
B
C
D
A block slides on a table pulled by a string a>ached to a hanging weight. In Case 1 the block slides without fric%on and in Case 2 there is kine%c fric%on between the sliding block and the table. m2
m2
Case 1
(No Friction)
g
m1
Case 2
(With Friction)
g
m1
In which case is the acceleration of the blocks larger?
A) Case 1
B) Case 2
C) Same
Mechanics Lecture 6, Slide 20
A
CheckPoint revisited: Blocks and PulleyC
m2
B
D
m2
Case 1
(No Friction)
g
m1
Case 2
(With Friction)
g
m1
In which case is the tension in the string larger? A) Case 1 B) Case 2 C) Same One of your answers: B) M1 will not be accelera%ng as fast in case two than in case one because there is the extra force of fric%on ac%ng against mass 1. Since the accelera%on is smaller in case two, there has to be more of a force ac%ng against gravity, the only other possible force is Tension. Mechanics Lecture 6, Slide 21
ACT!
A
B
C
D
A block (m2) slides on a table pulled by a string attached to a
mass (m1) hanging over the side. The coefficient of kinetic
friction between the sliding block and the table is µk. What is
the tension in the string?
T1
m2
T2
g
What is the relationship between the magnitude of
the tension of the string at block 2 and the
magnitude of the tension in the string at block 1?
A) T1 > T2 B) T1 = T2
m1
C) T1 < T2
Mechanics Lecture 6, Slide 22
ACT:
A
B
C
D
A block (m2) slides on a table pulled by a string attached to a
mass (m1) hanging over the side. The coefficient of kinetic
friction between the sliding block and the table is µk. What is
the tension in the string?
m2
g
What is the relationship between the magnitudes
of the acceleration of the two blocks?
m1
A) a1 = a2 B) a1 < a2 C) a1 > a2
Mechanics Lecture 6, Slide 23
Back to example problem
A block (m2) slides on a table pulled by a string attached to a
mass (m1) hanging over the side. The coefficient of kinetic
friction between the sliding block and the table is µk. What is
the tension in the string?
m2
g
m1
Mechanics Lecture 6, Slide 24
1) FBD
m2
N
f
m2
T
g
T
m2g
m1
m1
m1g
Mechanics Lecture 6, Slide 25
1) FBD
2) ΣF=ma
m2
N
f
T
m2
T
m2g
N = m2g
T – µ m2g = m2a
g
m1g – T = m1a
m1
m1
m1g
add
m1g – µ m2g = m1a + m2a
a=
m1g – µ m2g
m1 + m2
Mechanics Lecture 6, Slide 26
1) FBD
2) ΣF=ma
m2
N
f
T
m2
g
T
m1
m1
m2g
m1g
m1g – T = m1a
a=
m1g – µ m2g
m1 + m2
T = m1g – m1a
T is smaller when a is bigger
Mechanics Lecture 6, Slide 27