9/10/2013phy 113 c fall 2013 -- lecture 51 phy 113 a general physics i 11 am-12:15 pm mwf olin 101...
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PHY 113 C Fall 2013 -- Lecture 5 19/10/2013
PHY 113 A General Physics I11 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.
PHY 113 C Fall 2013 -- Lecture 5 29/10/2013
PHY 113 C Fall 2013 -- Lecture 5 39/10/2013
Webassign question from Assignment #4
East
North
3
1
2
4
PHY 113 C Fall 2013 -- Lecture 5 49/10/2013
Webassign question from Assignment #4
Given: h,H,x,q Determine: vi
vi
x=
PHY 113 C Fall 2013 -- Lecture 5 59/10/2013
Problem solving steps1. Visualize problem – labeling
variables2. Determine which basic physical
principle(s) apply3. 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.).
Two-dimensional motion with constant acceleration
2
221
221
cos tan
ˆ sin
ˆ cos
ˆ
ii
iiii
iii
iii
v
xxgxxyxy
gttvy
tvxt
g
j
ir
ja
PHY 113 C Fall 2013 -- Lecture 5 69/10/2013
Isaac Newton, English physicist and mathematician (1642—1727)
http://www.newton.ac.uk/newton.html
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.
PHY 113 C Fall 2013 -- Lecture 5 79/10/2013
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 Winston-Salem, but the corrections are very small.
PHY 113 C Fall 2013 -- Lecture 5 89/10/2013
PHY 113 C Fall 2013 -- Lecture 5 99/10/2013
Example force – weight
1 lb = 4.45 N 1 N = 0.225 lb
Newton
PHY 113 C Fall 2013 -- Lecture 5 109/10/2013
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 questionWill we always need to take gravity into account?
A. YesB. No
PHY 113 C Fall 2013 -- Lecture 5 119/10/2013
Isaac Newton, English physicist and mathematician (1642—1727)
http://www.newton.ac.uk/newton.html
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.
PHY 113 C Fall 2013 -- Lecture 5 129/10/2013
Examples
Fgravity=-mg j
Ftension= T j
F=Fnet=Ftension+Fgravity=(T-mg) j =0
PHY 113 C Fall 2013 -- Lecture 5 139/10/2013
Examples
Fgravity=-mg j
Ftension= T j
F=Fnet=Ftension+Fgravity=(T-mg) j =0
Fgravity=-mg j
F=Fnet=Fgravity=(-mg) j = ma
a=-gj
PHY 113 C Fall 2013 -- Lecture 5 14
vi
x=9/10/2013
Two-dimensional motion with constant acceleration
2
221
221
cos tan
ˆ sin
ˆ cos
ˆ
ii
iiii
iii
iii
v
xxgxxyxy
gttvy
tvxt
g
j
ir
ja
PHY 113 C Fall 2013 -- Lecture 5 159/10/2013
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.
PHY 113 C Fall 2013 -- Lecture 5 169/10/2013
Example: (assume surface is frictionless)
Suppose T measured
a observed & measured
T = m a
kg50m/s1.0N5
m/s 0.1 N;5
2
2
aT
m
aT
PHY 113 C Fall 2013 -- Lecture 5 179/10/2013
Example of two dimensional motion on a frictionless horizontal surface
m
m
m
21
21
FFa
aFF
PHY 113 C Fall 2013 -- Lecture 5 189/10/2013
Example of two dimensional motion on a frictionless horizontal surface -- continued
mN 10.14
N100cos85285 2221
oFF
2m/s 8.33kg 3.0
N 10.14
m21 FF
a
PHY 113 C Fall 2013 -- Lecture 5 199/10/2013
Example – support forces
appliedsupport
ˆ
FF
yF
mgg
Fsupport acts in direction to surface
(in direction of surface “normal”)
PHY 113 C Fall 2013 -- Lecture 5 209/10/2013
Another example of support force:
FFn
FFn
g
g
g
0ˆ
0
j
FFn
PHY 113 C Fall 2013 -- Lecture 5 219/10/2013
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.
PHY 113 C Fall 2013 -- Lecture 5 229/10/2013
Example of forces in equilibrium
PHY 113 C Fall 2013 -- Lecture 5 239/10/2013
Example of forces in equilibrium -- continued
0sinsin :component ˆ
0coscos :component ˆ
0
122
32211
2211
3
1
3
TTT
TT
NFT
ii
g
j
i
T
PHY 113 C Fall 2013 -- Lecture 5 249/10/2013
Example of forces in equilibrium -- continued
0sinsin
0coscos
:algebra using equations Solving
32211
2211
TTT
TT
NTT
NT
T
NTTT
TT
4.73cos
cos
4.97sinsin
coscos
122sinsincos
cos
cos
cos
1
221
211
2
32
32211
22
1
221
PHY 113 C Fall 2013 -- Lecture 5 259/10/2013
Newton’s second law
F = m a
Types of forces:
Fundamental Approximate Empirical
Gravitational F=-mg j Friction
Electrical Support
Magnetic Elastic
Elementary
particles
PHY 113 C Fall 2013 -- Lecture 5 269/10/2013
2
2
dt
dm
dtd
mmrv
aF
Examples (one dimension):
0FF (constant)
tm
Ftvxtx
sin)(
20
00 tFF sin0
2000 2
1)( t
m
Ftvxtx
kxF tmk
xtx cos)( 0
mbvFF 0
btebg
bg
tv )(
PHY 113 C Fall 2013 -- Lecture 5 279/10/2013
Newton’s third law
F12 = - F21
1 2
FPT T
T=m1a a=FP/(m1+m2)
FP-T=m2a T=FP (m1/(m1+m2))
PHY 113 C Fall 2013 -- Lecture 5 289/10/2013
T-m1g=m1a T-m2g=-m2a => after some algebra:
a
gmmmm
a12
12
g
mmmm
T12
212
PHY 113 C Fall 2013 -- Lecture 5 299/10/2013
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
PHY 113 C Fall 2013 -- Lecture 5 309/10/2013
PHY 113 C Fall 2013 -- Lecture 5 319/10/2013
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?
3. What is the scale reading when the elevator is coming to a stop just before your destination floor?
x
<x
>x
PHY 113 C Fall 2013 -- Lecture 5 329/10/2013
mg
Ffloor = mg
mgFfloor = mg-T
T T
Mg
Mg=12 lbmg=14.5 lb
T=?
iclicker exerciseA. mgB. MgC. Not enough info
PHY 113 C Fall 2013 -- Lecture 5 339/10/2013
Example: 2-dimensional forcesA car of mass m is on an icy (frictionless) driveway, inclined at an angle t as shown. Determine its acceleration.
Conveniently tilted coordinate system:
sin
sin : Along
0cos : Along
ga
mamgx
mgny
x
x
PHY 113 C Fall 2013 -- Lecture 5 349/10/2013
vi
vf=0
tgvtv
tgtvxtx
i
ii
sin)(
sin)( :incline Along 221
Another example: Note: we are using a tilted coordinate frame
i
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