Transcript Physics 207: Lecture 2 Notes

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 Newton’s 1st, 2nd and 3rd Laws
 Use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday)
Finish Chapter 7
1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 1
No Net Force, No acceleration…a 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.
Physics 207: Lecture 7, Pg 2
Friction revisited: Static friction
Static equilibrium: A block with a horizontal force F applied,
S Fx = 0 = -F + fs
 fs = F
FBD
S Fy = 0 = - N + mg  N = mg
As F increases so does fs
N
F
m
fs
1
mg
Physics 207: Lecture 7, Pg 3
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
ms called the “coefficient of static friction”
fs = ms N
N
F
m
mg
Physics 207: Lecture 7, Pg 4
fs
Kinetic or Sliding friction (fk < fs)
Dynamic equilibrium, moving but acceleration is still zero
S Fx = 0 = -F + fk
 fk = F
S Fy = 0 = - N + mg  N = mg
As F increases fk remains nearly constant
(but now there acceleration is acceleration)
FBD
v
N
F
m
fk
1
mg
fk = mk N
Physics 207: Lecture 7, Pg 5
Case study ... big F

Dynamics:
x-axis i :
y-axis j :
so
max = F  mKN
may = 0 = N – mg or N = mg
F  mKmg = m ax
fk
v
j
N
F
max
fk
i
mK mg
mg
Physics 207: Lecture 7, Pg 6
Case study ... little F

Dynamics:
x-axis i :
y-axis j :
so
max = F  mKN
may = 0 = N – mg or N = mg
F  mKmg = m ax
fk
v
j
N
F
i
ma
fk
x
mK mg
mg
Physics 207: Lecture 7, Pg 7
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 = mk N
( = mK mg in the previous example)

The constant mk is called the “coefficient of kinetic friction”

As the normal force varies so does the frictional force
Physics 207: Lecture 7, Pg 8
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
mS > mK
for any system
Physics 207: Lecture 7, Pg 9
Coefficients of Friction
Material on Material
ms = static friction
mk = 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
Physics 207: Lecture 7, Pg 10
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.
N
T
Static equilibrium: Set
m2
m2 and add mass to
m1 to reach the
breaking point.
Requires two FBDs
Mass 1
S Fy = 0 =
T – m1g
fS
T
m1
m2g
m1g
Mass 2
S Fx = 0 = -T + fs = -T + mS N
S Fy = 0 = N – m2g
T = m1g = mS m2g  mS = m1/m2
Physics 207: Lecture 7, Pg 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
Dynamic equilibrium:
Set m2 and adjust m1
to find place when
T
a = 0 and v ≠ 0
fk
m1
Requires two FBDs
Mass 1
S Fy = 0 =
T – m1g
N
m2
m2g
m1g
Mass 2
S Fx = 0 = -T + ff = -T + mk N
S Fy = 0 = N – m2g
T = m1g = mk m2g  mk = m1/m2
Physics 207: Lecture 7, Pg 12
An experiment (with a ≠ 0)
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
N
Design an experiment to find mS & mK.
T
Non-equilibrium: Set m2
T
and adjust m1 to find
regime where a ≠ 0
m1
Requires two FBDs
m1g
Mass 1
S Fy = m1a =
T – m1g
fk
m2
m2g
Mass 2
S Fx = m2a = -T + fk
S Fy = 0 = N – m2g
= -T + mk N
T = m1g + m1a = mk m2g – m2a  mk = (m1(g+a)+m2a)/m2g
Physics 207: Lecture 7, Pg 13
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?
Physics 207: Lecture 7, Pg 14
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
Physics 207: Lecture 7, Pg 15
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
m
1
30
20
10
10
20
30
40
(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 ?
t (sec)
Physics 207: Lecture 7, Pg 16
The Experimental Data


t < 30 s puts constraints on ms
(Static equilibrium)
T (N)
t > 30 s reflects mk
50
(Non-equilibrium)
Const. accel. at 30-40 (T= 40 N)
40
and 40-50 (T=50 N) second
times
30
time
vel. #1
(sec)
m/s
vel. #2
m/s
speed (m/s)
25
20
15
vel. #3
m/s 20
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
10
5
10
10
20
30
40
t (sec)
Physics 207: Lecture 7, Pg 17
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
(A)
m
T
f
1
mg
Physics 207: Lecture 7, Pg 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
(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
m
T
f
1
mg
Notice that at 30 s < t < 40 s
T= 40 N ax= Dv/Dt = -5/10 m/s2
and at 40 s < t < 50 s
T= 50 N ax= Dv/Dt =-15/10 m/s2
Physics 207: Lecture 7, Pg 19
Inclined plane with “Normal” and Frictional Forces
1. At first the velocity is v up along the
slide
“Normal” means
perpendicular
2. Can we draw a velocity time plot?
Normal
Force
3. What the acceleration versus time?
v
Friction Force
Sliding Down
q
fk
Sliding
Up
q
mg sin q
Weight of block is mg
Note: If frictional Force = Normal Force  (coefficient of friction)
Ffriction = m Fnormal = m mg sin q then zero acceleration
Physics 207: Lecture 7, Pg 21
The inclined plane coming and going (not static):
the component of mg along the surface > kinetic friction
S Fx = max = mg sin q ± uk N > 0
S 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 Newton’s Laws that is key.
Physics 207: Lecture 7, Pg 22
Lecture 6
Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday)
Finish Chapter 7
1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 23