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