PHY 113 A Fall 2012 -- Lecture 20 110/17/2012
PHY 113 A General Physics I9-9:50 AM MWF Olin 101
Plan for Lecture 20:
Chapter 12 – Static equilibrium
1. Balancing forces and torques; stability
2. Center of gravity
3. Note: May not have time to cover elastic properties of materials
PHY 113 A Fall 2012 -- Lecture 20 210/17/2012
PHY 113 A Fall 2012 -- Lecture 20 310/17/2012
Comment on exam question #4
PHY 113 A Fall 2012 -- Lecture 20 410/17/2012
PHY 113 A Fall 2012 -- Lecture 20 510/17/2012
From Webassign #17 (11.34)
f
iif
fStudentf
iStudenti
ffii
fi
I
I
mdII
mdII
II
LL
2
2
2
2
:momentumangular
ofon Conservati
22
2
1
2
1
:energy kinetic Rotational
fffiii IKIK
PHY 113 A Fall 2012 -- Lecture 20 610/17/2012
Conditions for stable equilibrium
0 : torqueof Balance
0 :force of Balance
ii
ii
τ
F
PHY 113 A Fall 2012 -- Lecture 20 710/17/2012
Stability of “rigid bodies”
N
mig
PHY 113 A Fall 2012 -- Lecture 20 810/17/2012
Center-of-mass
ii
iii
CM m
m rr
Torque on an extended object due to gravity (near surface of the earth) is the same as the torque on a point mass M located at the center of mass.
miri
jrjrτ Mggm CMi
ii
rCM
PHY 113 A Fall 2012 -- Lecture 20 910/17/2012
Notion of stability:
mg(-j)
r
Tq
F=ma T- mg cos q 0
mg sin q maq
t=I a r mg sin q = mr2 a mraq
Notion of equilibrium: i
i 0F i
i 0τ
Example of stable equilibrium.
PHY 113 A Fall 2012 -- Lecture 20 1010/17/2012
Unstable equilibrium:
mg(-j)
r
Tq
Support above CM:
Support below CM:
PHY 113 A Fall 2012 -- Lecture 20 1110/17/2012
Analysis of stability: i
i 0F i
i 0τ
PHY 113 A Fall 2012 -- Lecture 20 1210/17/2012
0)1( :Torques
0 :Forces
gxmmgM
gmgmgMn
cD
PcD
**X
PHY 113 A Fall 2012 -- Lecture 20 1310/17/2012
PHY 113 A Fall 2012 -- Lecture 20 1410/17/2012
**X
Fg1
mg
RCM
0)()2( :Torques 1 CMg RmgmF
PHY 113 A Fall 2012 -- Lecture 20 1510/17/2012
iclicker question:
F1F2
Consider the above drawing of the two supports for a uniform plank which has a total weight Mg and has a weight mg at its end. What can you say about F1 and F2?
(a) F1 and F2 are both up as shown.
(b) F1 is up but F2 is down.
(c) F1 is down but F2 is up.
L/3
LMg mg
PHY 113 A Fall 2012 -- Lecture 20 1610/17/2012
F1F2
L/3
LMg mg
023
:Torques
0 :Forces
2
21
mgLL
MgL
F
mgMgFF
**X
mgMgFmgMgF 22
1 3
2
312
PHY 113 A Fall 2012 -- Lecture 20 1710/17/2012
iclicker question:The fact that we found F1<0 means:
A. We set up the problem incorrectlyB. The analysis is correct, but the
direction of F1 is opposite to the arrowC. Physics makes no sense
iclicker question:What would happen if we analyzed this problem by placing the pivot point at F1 ?:
A. The answer would be the same.B. The answer would be different.C. Physics makes no sense
PHY 113 A Fall 2012 -- Lecture 20 1810/17/2012
PHY 113 A Fall 2012 -- Lecture 20 1910/17/2012
mg
Mg
Fwall
N
T
Mg = 120 N
mg = 98 N
T < 110 N
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