PHY 113 A General Physics I
11 AM-12:15 PM MWF Olin 101
Plan for Lecture 5:
Chapter 5 – NEWTON’S GENIUS
1. Concept of force
2. Relationship between force and acceleration
3. Net force
Note: today we are going to neglect friction.
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Webassign question from Assignment #4
North
1
East
4
3
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Webassign question from Assignment #4
vi
x=
Given: h,H,x,q
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Determine: vi
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Problem solving steps
1. Visualize problem – labeling
variables
2. Determine which basic physical
principle(s) apply
3. Write down the appropriate
equations using the variables
defined in step 1.
4. Check whether you have the correct
amount of information to solve the
problem (same number of knowns
and unknowns).
5. Solve the equations.
6. Check whether your answer makes
sense (units, order of magnitude,
etc.).
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Two-dimensional motion
with constant acceleration
a   gˆj
r t    xi  vi cos q i t ˆi


 yi  vi sin q i t  12 gt 2 ˆj
g  x  xi 




 y x  yi  tan q i x  xi 
vi cos q i 2
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1
2
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Isaac Newton, English physicist and mathematician
(1642—1727)
1. In the absence of a
net force, an object
remains at constant
velocity or at rest.
2. In the presence of a
net force F, the
motion of an object of
mass m is described
by the form F=ma.
3. F12 =– F21.
http://www.newton.ac.uk/newton.html
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Detail: “Inertial” frame of reference
Strictly speaking, Newton’s laws work only in a reference
frame (coordinate system) that is stationary or moving at
constant velocity. In a non-inertial (accelerating) reference
frame, there are some extra contributions.
iclicker question:
Are we (here in Winston-Salem)
A. In an inertial frame of reference (exactly)
B. In a non-inertial frame of reference
Newton’s laws are only approximately true in WinstonSalem, but the corrections are very small.
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Example force – weight
1 lb = 4.45 N
1 N = 0.225 lb
Newton
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Gravitational mass versus inertial mass –
The concept of mass (“inertial” mass m) is
distinct from the concept of weight mg. In fact
the gravitation “constant” g varies slightly
throughout the globe. The weight of mass m
on the surface of the earth is different from its
weight measured on the surface of the moon or
on the surface of Mars. (Stay tuned to Chapter
13 – “Universal Gravitation”.)
iclicker question
Will we always need to take gravity into account?
A. Yes
B. No
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Isaac Newton, English physicist and mathematician
(1642—1727)
1. In the absence of a
net force, an object
remains at constant
velocity or at rest.
2. In the presence of a
net force F, the
motion of an object of
mass m is described
by the form F=ma.
3. F12 =– F21.
http://www.newton.ac.uk/newton.html
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Examples
Ftension= T j
Fgravity=-mg j
F=Fnet=Ftension+Fgravity=(T-mg) j =0
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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
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a=-gj
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Two-dimensional motion with constant acceleration
a   gˆj
r t    xi  vi cos q i t ˆi


 yi  vi sin q i t  12 gt 2 ˆj
g  x  xi 
 y  x   yi  tan q i  x  xi  
vi cos q i 2
1
2
2
vi
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x= PHY 113 C Fall 2013 -- Lecture 5
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iclicker question:
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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Example:
(assume surface is frictionless)
Suppose T measured
a observed & measured
T=ma
T  5 N; a  0.1 m/s 2
T
5N
m 
 50kg
2
a 0.1m/s
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Example of two dimensional motion on a frictionless
horizontal surface
F1  F2  ma
F1  F2
a
m
m
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Example of two dimensional motion on a frictionless
horizontal surface -- continued
F1  F2  52  82  2  5  8  cos100o N
m
 10.14 N
a
F1  F2
m
10.14 N

 33.8 m/s 2
0.3 kg
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Example – support forces
Fg  mgyˆ
Fsupport  Fapplied
Fsupport acts in direction  to surface
(in direction of surface “normal”)
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Another example of support force:
n  Fg  F  0
n  F
g
 F ˆj  0
n  Fg  F
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iclicker question:
Why do we care about forces in equilibrium?
A. Normal people don’t care.
B. Physicist care because they are weird.
C. It might be important if the materials
responsible for the support forces are
near their breaking point.
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Example of forces in equilibrium
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Example of forces in equilibrium -- continued
T3  Fg  122 N
3
T  0
i 1
i
ˆi component :  T cos q  T cos q  0
1
1
2
2
ˆj component : T sin q  T sin q  T  0
1
1
2
2
3
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Example of forces in equilibrium -- continued
Solving equations using algebra :
 T1 cos q1  T2 cos q 2  0
T1 sin q1  T2 sin q 2  T3  0
cos q 2
T1  T2
cos q1
cos q 2
T2
sin q1  T2 sin q 2  T3  122 N
cos q1
T2 
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T3
cos q 2
sin q1  sin q 2
cos q1
 97.4 N
cos q 2
T1  T2
 73
.4 N
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cos q1
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Newton’s second law
F=ma
Types of forces:
Fundamental
Approximate
Gravitational
F=-mg j
Empirical
Friction
Electrical
Support
Magnetic
Elastic
Elementary
particles
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dv
d 2r
F  ma  m  m 2
dt
dt
Examples (one dimension):
F  F0 (constant) 
F  F0 sin t
F  kx
F  F0  mbv
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1 F0 2
t
2m
F0
x(t )  x0  v0t 
sin t
2
m
x(t )  x0  v0t 



k
x(t )  x0 cos
t
m
g g bt
v(t )    e
b b
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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))
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a
T-m1g=m1a
T-m2g=-m2a
m2  m1
a
g
m2  m1
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=> after some algebra:
T
2m1m2
g
m2  m1
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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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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?
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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x
Ffloor = mg
mg
T
T
Mg=12 lb
mg=14.5 lb
T=?
Ffloor = mg-T
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Mg
mg
iclicker exercise
A. mg
B. Mg
C. Not enough info
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Example: 2-dimensional forces
A car of mass m is on an icy (frictionless)
driveway, inclined at an angle t as shown.
Determine its acceleration.
Conveniently tilted
coordinate system:
Along y : n  mg cos q  0
Along x : mg sin q  ma x
a x  g sin q
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Another example:
Note: we are using a tilted coordinate frame
i
vf=0
vi
Along incline : x(t )  xi  vi t  12 g sin q t 2
v(t )  vi  g sin q t
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