Announcements
1. More tutorial sessions – (feedback welcome)
schedule
2.
Physics seminar today – 4 PM – research presentations by
5 newest members of Physics Department.
3. Some comments about Tuesday’s lecture and HW 4
4. Today’s topic −
Causes of motion
Newton’s “laws” of motion
Isaac Newton 1642-1727
9/11/2003
PHY 113 -- Lecture 5
1
v
a
RE = 6.37x106 m
h = 630x103 m
∆R
∆t
v2 ˆ
a=−
R total
Rtotal
v=
9/11/2003
PHY 113 -- Lecture 5
2
The basic laws of motion
Sir Isaac Newton (1642-1727)
mass
F=ma
acceleration
net “force”
9/11/2003
PHY 113 -- Lecture 5
3
1. In absence of a net force, an object remains at constant
velocity or at rest.
2. In the presents of a net force, the motion of an object of
mass m is described by
F=ma
.
3. F12=−F21
Units of force: 1 N = 1kg m/s2 = 0.2248 lb
(quarter pound hamburger ~ 1 Newton burger)
9/11/2003
PHY 113 -- Lecture 5
4
Examples
Ftension= T j
Fgravity=-mg j
F=Fnet=Ftension+Fgravity=(T-mg) j =0
9/11/2003
PHY 113 -- Lecture 5
5
Examples
Ftension= T j
Fgravity=-mg j
Fgravity=-mg j
F=Fnet=Ftension+Fgravity=(T-mg) j =0
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F=Fnet=Fgravity=(-mg) j = ma
PHY 113 -- Lecture 5
a=-gj
6
On line quiz
An object having a mass of 2 kg is moving at constant velocity of 3
m/s in the upward direction near the surface of the Earth. What is the
net force acting on the object?
a. 0 N.
b. 6 N in the upward direction.
c. 25.6 N in the upward direction.
d. 19.6 N in the downward direction.
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PHY 113 -- Lecture 5
7
Example:
(assume surface is frictionless)
Suppose T measured
a observed & measured
T=ma
T = 5 N; a = 0.1 m/s 2
5N
T
= 50kg
m= =
2
a 0.1m/s
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PHY 113 -- Lecture 5
8
Example – support forces
Fg = − mgyˆ
Fsupport = −Fapplied Îacts in direction
⊥ to surface
(in direction of surface “normal”)
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PHY 113 -- Lecture 5
9
Winston-Salem
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PHY 113 -- Lecture 5
10
Peer instruction question
Realizing that the Earth rotates about its axis once every 24 hrs.
and that in Winston-Salem, this means that you are moving
in a circular at a speed of v~400 m/s at a radius of
r~5000 km from the axis of rotation. What can you say
about your centripetal acceleration?
(a) It is too small to measure.
(b) Is is measureable.
(c) There is no effect, since you are moving together with
everything else.
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PHY 113 -- Lecture 5
11
Newton’s second law
F=ma
Types of forces:
Fundamental
Gravitational
Approximate
F=−mg j
Empirical
Friction
Electrical
Support
Magnetic
Elastic
Elementary
particles
9/11/2003
PHY 113 -- Lecture 5
12
dv
d 2r
F = ma = m = m 2
dt
dt
Examples (one dimension):
F = F0 (constant)
=>
F = F0 sin ωt
=>
F = − kx
F = F0 − mbv
9/11/2003
1 F0 2
t
2m
F0
x(t ) = x0 + v0t −
sin ωt
2
mω
x(t ) = x0 + v0t +
=>
=>
x(t ) = x0 cos
k
t
m
g g
v(t ) = − + e −bt
b b
PHY 113 -- Lecture 5
13
Newton’s third law
F12 = - F21
T
1
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FP
T
2
T=m1a
a=FP/(m1+m2)
FP-T=m2a
T=FP (m1/(m1+m2))
PHY 113 -- Lecture 5
14
a=
T-m1g=m1a
T-m2g=-m2a
m2 − m1
a=
g
m2 + m1
9/11/2003
=> after some algebra:
T=
PHY 113 -- Lecture 5
2m1m2
g
m2 + m1
15
Fgravity=-mg j
Fscale
Fscale + Fgravity = 0; Fscale = mg
Question: What if you step on the scale in the elevator?
Fscale + Fgravity = m a; Fscale = mg +m a
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PHY 113 -- Lecture 5
16
Exercise
For this exercise, you should ride the elevator from the first floor
of Olin Physical Laboratory up to the second or third floor. While
in the elevator, stand on the scale and record the scale readings to
answer the following questions.
1. What is the scale reading when the elevator is stationary? x
2. What is the scale reading when the elevator just starts moving
upward? >x
3. What is the scale reading when the elevator is coming to a stop
just before your destination floor? <x
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PHY 113 -- Lecture 5
17
No
rm
al
Normal forces:
N- mg cos θ = 0
Ta
ng
ent
ial
N
Tangential forces:
mg sin θ = ma
a=g sinθ
mg sin θ
mg
mg cos θ
θ
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PHY 113 -- Lecture 5
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N
mg
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PHY 113 -- Lecture 5
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