Physics 207: Lecture 2 Notes

1
Lecture 7

• Goals
• Solve 1D and 2D problems with forces in
  equilibrium and non-equilibrium (i.e.,
  acceleration) using Newton 1st and 2nd laws.
• Differentiate between Newtons 1st, 2nd and 3rd
  Laws
• Use Newtons 3rd Law in problem solving

Assignment HW4, (Chapters 6 7, due 10/1,
Wednesday) Finish Chapter 7 1st Exam Thursday,
Oct. 2nd from 715-845 PM Chapters 1-7
2
No Net Force, No accelerationa demo exercise

• In this demonstration we have a ball tied to a
  string undergoing horizontal UCM (i.e. the ball
  has only radial acceleration)
• 1 Assuming you are looking from above, draw the
  orbit with the tangential velocity and the radial
  acceleration vectors sketched out.
• 2 Suddenly the string brakes.
• 3 Now sketch the trajectory with the velocity and
  acceleration vectors drawn again.

3
Friction revisited Static friction
Static equilibrium A block with a horizontal
force F applied, As F increases so does fs
S Fx 0 -F fs ? fs F S Fy 0 - N
mg ? N mg
4
Static friction, at maximum (just before slipping)
Still equilibrium A block, mass m, with a
horizontal force F applied, Direction A
force vector ? to the normal force vector N and
the vector is opposite to the velocity.
Magnitude fS is proportional to the magnitude
of N fs ms N ms called the
coefficient of static friction
N
F
m
fs
mg
5
Kinetic or Sliding friction (fk lt fs)
Dynamic equilibrium, moving but acceleration is
still zero As F increases fk remains nearly
constant (but now there acceleration is
acceleration)
S Fx 0 -F fk ? fk F S Fy 0 - N
mg ? N mg
v
F
fk mk N
6
Case study ... big F

• Dynamics
• x-axis i max F ? ?KN
• y-axis j may 0 N mg or N mg
• so F ???Kmg m ax

fk
v
N
F
max
fk
?K mg
mg
7
Case study ... little F

• Dynamics
• x-axis i max F ? ?KN
• y-axis j may 0 N mg or N mg
• so F ???Kmg m ax

fk
v
j
N
F
i
max
fk
?K mg
mg
8
Sliding Friction Quantitatively

• Direction A force vector ? to the normal force
  vector N and the vector is opposite to the
  velocity.
• Magnitude fk is proportional to the magnitude of
  N
• fk ?k N ( ?K??mg in the previous example)
• The constant ?k is called the coefficient of
  kinetic friction
• As the normal force varies so does the frictional
  force

9
Additional comments on Friction

• The force of friction does not depend on the area
  of the surfaces in contact (a relatively good
  approximation if there is little surface
  deformation)
• Logic dictates that ?S gt ?K for any
  system

10
Coefficients 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
11
An experiment

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mS mK.

T
Static equilibrium Set m2 and add mass to m1
to reach the breaking point. Requires two FBDs
fS
T
m1
m1g
Mass 1 S Fy 0 T m1g T m1g mS m2g ?
mS m1/m2
Mass 2 S Fx 0 -T fs -T mS N S Fy 0
N m2g
12
An experiment

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mS mK.

T
Dynamic equilibrium Set m2 and adjust m1 to
find place when a 0 and v ? 0 Requires
two FBDs
fk
T
m1
m1g
Mass 1 S Fy 0 T m1g T m1g mk m2g ?
mk m1/m2
Mass 2 S Fx 0 -T ff -T mk N S Fy 0
N m2g
13
An experiment (with a ? 0)

• Two blocks are connected on the table as shown.
  The
• table has unknown static and kinetic friction
  coefficients.
• Design an experiment to find mS mK.

T
Non-equilibrium Set m2 and adjust m1 to
find regime where a ? 0 Requires two FBDs
fk
T
m1
m1g
Mass 1 S Fy m1a T m1g T m1g m1a
mk m2g m2a ? mk (m1(ga)m2a)/m2g
Mass 2 S Fx m2a -T fk -T mk N S Fy
0 N m2g
14
Sample Problem

• You have been hired to measure the coefficients
  of friction for the newly discovered substance
  jelloium. Today you will measure the coefficient
  of kinetic friction for jelloium sliding on
  steel. To do so, you pull a 200 g chunk of
  jelloium across a horizontal steel table with a
  constant string tension of 1.00 N. A motion
  detector records the motion and displays the
  graph shown. What is the value of µk for jelloium
  on steel?

15
Sample Problem

• S Fx ma F - ff F - mk N F - mk mg
• S Fy 0 N mg
• mk (F - ma) / mg x ½ a t2 ? 0.80 m
  ½ a 4 s2
• a 0.40 m/s2
• mk (1.00 - 0.20 0.40 ) / (0.20 10.) 0.46

16
Another experiment

• A block is connected to a horizontal massless
  string. The table has coefficients of kinetic
  static friction (mK mS). There is a unknown
  mass m and you apply a variable force T (by
  pulling on the rope) as shown in the plot.

T (N)
50
T
40
30
(A) On the next slide are tables and plots of
velocity vs. time (B) Can you deduce the various
coefficients of friction and the mass ?
20
10
20
30
40
10
t (sec)
17
The Experimental Data

• t lt 30 s puts constraints on ms
• (Static equilibrium)
• t gt 30 s reflects mk
• (Non-equilibrium)
• Const. accel. at 30-40 (T 40 N) and 40-50 (T50
  N) second times

speed (m/s)
time (sec) vel. 1 m/s vel. 2 m/s vel. 3 m/s
10 0.0 0.0 0.0
20 0.0 0.0 0.0
30 0.0 0.0 0.0
40 5.1 4.9 5.0
50 20.2 19.8 20.0

20
30
40
10
t (sec)
18
Another experiment

• A block is connected by a horizontal massless
  string. The table has coefficients of kinetic
  static friction (mK mS). There is a unknown
  mass m and you apply a variable force as shown in
  the plot.

N
FBD

• Static case (30 N less)
• S Fx 0 -T f -T mN
• S Fy 0 - N mg
• T m m g , 2 unknowns

f
T
mg
19
Another experiment

• A block is connected by a horizontal massless
  string. The table has coefficients of kinetic
  static friction (mK mS). There is a unknown
  mass m and you apply a variable force as shown in
  the plot.

(B) Non-equilibrium S Fx max -T f -T
mK N S Fy 0 N mg max -T mK m
g Using information at right you can identify 2
equations and 2 unknowns
T
mg
Notice that at 30 s lt t lt 40 s T 40 N ax
Dv/Dt -5/10 m/s2 and at 40 s lt t lt 50 s T 50
N ax Dv/Dt -15/10 m/s2
20
Another experiment

• A block is connected by a horizontal massless
  string. The table has coefficients of kinetic
  static friction (mK mS). There is a unknown
  mass m and you apply a variable force as shown in
  the plot. Let g 10 m/s2

N
FBD
(B) Non-equilbrium m (-0.5 m/s2) -40 N mK m
10 m/s2 m (-1.5 m/s2 ) -50 N mK m 10 m/s2 So
- m 1.0 m/s - 10 N ? m 10 kg and mK
0.35 With m we revisit the static case T
m m g ? m T/ mg At 30 N m 0.30, less than
mK At 40 N m 0.40, greater than mK So 0.35
lt mS lt 0.40
f
T
mg
At 30 s lt t lt 40 s T 40 N ax Dv/Dt
-5/10 m/s2 and at 40 s lt t lt 50 s T 50 N ax
Dv/Dt -15/10 m/s2
21
Inclined plane with Normal and Frictional
Forces

1. At first the velocity is v up along the slide
2. Can we draw a velocity time plot?
3. What the acceleration versus time?

Normal means perpendicular
Normal Force
Friction Force Sliding Down
fk Sliding Up
v
q
q
mg sin q
Weight of block is mg
Note If frictional Force Normal Force ?
(coefficient of friction) Ffriction ? Fnormal
m mg sin q then zero acceleration
22
The inclined plane coming and going (not
static)the component of mg along the surface gt
kinetic friction

• Fx max mg sin q uk N gt 0
• Fy may -mg cos q N

Putting it all together gives two different
accelerations, ax g sin q uk g cos q. A tidy
result but ultimately it is the process of
applying Newtons Laws that is key.
23
Lecture 6
Assignment HW4, (Chapters 6 7, due 10/1,
Wednesday) Finish Chapter 7 1st Exam Thursday,
Oct. 2nd from 715-845 PM Chapters 1-7