Transcript Physics 207: Lecture 2 Notes

Lecture 7
"Professor Goddard does not know the relation between action and
reaction and the need to have something better than a vacuum
against which to react. He seems to lack the basic knowledge ladled
out daily in high schools."
New York Times editorial, 1921,
about Robert Goddard's revolutionary
rocket work.
"Correction: It is now definitely
established that a rocket can
function in a vacuum.
The 'Times' regrets the error."
New York Times editorial, July 1969.
Physics 207: Lecture 7, Pg 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.
 Distinguish static and kinetic coefficients of friction
 Differentiate between Newton’s 1st, 2nd and 3rd Laws
Assignment: HW4, (Chapters 6 & 7, due 2/18, 9 am,
Wednesday)
Read Chapter 7
1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters 1-7
Physics 207: Lecture 7, Pg 2
Exercise, Newton’s 2nd Law
A woman is straining to lift a large crate, without success. It
is too heavy. We denote the forces on the crate as follows:
P is the upward force being exerted on the crate by the person
C is the contact or normal force on the crate by the floor, and
W is the weight (force of the earth on the crate).
Which of following relationships between these forces is
true, while the person is trying unsuccessfully to lift the
crate? (Note: force up is positive & down is negative)
A.
B.
C.
D.
P+C<W
P+C>W
P=C
P+C=W
Physics 207: Lecture 7, Pg 3
Mass
We have an idea of what mass is from everyday life.
 In physics:
 Mass (in Phys 207) is a quantity that specifies
how much inertia an object has
(i.e. a scalar that relates force to acceleration)
(Newton’s Second Law)
 Mass is an inherent property of an object.
 Mass and weight are different quantities; weight is
usually the magnitude of a gravitational (non-contact)
force.
“Pound” (lb) is a definition of weight (i.e., a force), not
a mass!

Physics 207: Lecture 7, Pg 4
Inertia and Mass

The tendency of an object to resist any attempt to
change its velocity is called Inertia
Mass is that property of an object that specifies how
much resistance an object exhibits to changes in its
velocity (acceleration)
 
If mass is constant then
a  Fnet
If force constant 






| a |
1
m
|a|

Mass is an inherent property of an object
m
Mass is independent of the object’s surroundings
Mass is independent of the method used to measure it
Mass is a scalar quantity
The SI unit of mass is kg
Physics 207: Lecture 7, Pg 5
Exercise
Newton’s 2nd Law

An object is moving to the right, and experiencing
a net force that is directed to the right. The
magnitude of the force is decreasing with time
(read this text carefully).

The speed of the object is
A.
B.
C.
D.
increasing
decreasing
constant in time
Not enough information to decide
Physics 207: Lecture 7, Pg 6
Exercise
Newton’s 2nd Law
A 10 kg mass undergoes motion along a line with a
velocities as given in the figure below. In regards to
the stated letters for each region, in which is the
magnitude of the force on the mass at its greatest?
A.
B.
C.
D.
E.
B
C
D
F
G
Physics 207: Lecture 7, Pg 7
Moving forces around
Massless strings: Translate forces and reverse their
direction but do not change their magnitude
(we really need Newton’s 3rd of action/reaction to justify)

string
T1

-T1
Massless, frictionless pulleys: Reorient force direction
but do not change their magnitude
T2
T1
-T1
| T1 | = | -T1 | = | T2 | = | T2 |
-T2
Physics 207: Lecture 7, Pg 13
Scale Problem

You are given a 1.0 kg mass and you hang it
directly on a fish scale and it reads 10 N (g is
10 m/s2).
10 N
1.0 kg


Now you use this mass in a second
experiment in which the 1.0 kg mass hangs
from a massless string passing over a
massless, frictionless pulley and is anchored
to the floor. The pulley is attached to the fish
scale.
What force does the fish scale now read?
?
1.0 kg
Physics 207: Lecture 7, Pg 14
Scale Problem


Step 1: Identify the system(s).
In this case it is probably best to treat each
object as a distinct element and draw three
force body diagrams.
 One around the scale
 One around the massless pulley (even
though massless we can treat is as an
“object”)
 One around the hanging mass
Step 2: Draw the three FBGs. (Because this
is a now a one-dimensional problem we
need only consider forces in the y-direction.)
?
1.0 kg
Physics 207: Lecture 7, Pg 15
Scale Problem
3:
T”
T’
1:
2:
T
?
W

1.0 kg
-T ’
S Fy = 0 in all cases
-T
-T
?
-mg
1: 0 = -2T + T ’
2: 0 = T – mg  T = mg
3: 0 = T” – W – T ’ (not useful here)
 Substituting 2 into 1 yields T ’ = 2mg = 20 N
(We start with 10 N but end with 20 N)
1.0 kg
Physics 207: Lecture 7, Pg 16
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 17
Static and Kinetic Friction

Friction exists between objects and its behavior has been
modeled.
At Static 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 direction of acceleration if m were
0.
Magnitude: f is proportional to the applied forces such that
fs ≤ ms N
ms called the “coefficient of static friction”
Physics 207: Lecture 7, Pg 18
Friction: 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 19
Static friction, at maximum (just before slipping)
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 direction of acceleration if m were 0.
Magnitude: fS is proportional to the magnitude of N
fs = ms N
N
F
m
mg
Physics 207: Lecture 7, Pg 20
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 21
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”

Logic dictates that
mS > mK
for any system
Physics 207: Lecture 7, Pg 22
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 23
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
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 24
A 2nd experiment
Two blocks are connected on the table as shown. The
table has unknown static and kinetic friction coefficients.
Design an experiment to find 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 25
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 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 26
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 29
Recap
 Assignment:
HW4, (Chapters 6 & 7, due 2/18, 9 am,
Wednesday)
 Read Chapter 7
 1st Exam Wednesday, Feb. 18 from 7:15-8:45 PM Chapters
1-7
Physics 207: Lecture 7, Pg 30