PHY 113 C Fall 2013 -- Lecture 14 110/14/2013

PHY 113 C General Physics I11 AM-12:15 PM MWF Olin 101

Plan for Lecture 14:

Chapter 12 – Static equilibrium

1. Balancing forces and torques; stability

2. Center of gravity

3. Will discuss elasticity in Lecture 15 (Chapter 15)

PHY 113 C Fall 2013 -- Lecture 14 210/14/2013

PHY 113 C Fall 2013 -- Lecture 14 310/14/2013

Newton’s law of gravitation:

Earth’s gravity:

Stable circular orbits of gravitational attracted objects:

RES

F

a

vMsat

Summary of gravity:

RE m

2226

2411

2

2

m/s8.9m/s)1037.6(

1098.51067.6

E

E

E

E

RGMg

mgR

mGMF

212

122112

ˆrmGm rF

ESsat

ES

satE

RvM

RMGMF

2

2

PHY 113 C Fall 2013 -- Lecture 14 410/14/2013

From Webassign Assignment #12:

When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration due to the Earth's gravitation?m/s2 towards earth

r=4.4RE 22

2

4.4 E

EE

E

RGM

rGMa

mar

mGM

m

PHY 113 C Fall 2013 -- Lecture 14 510/14/2013

From Webassign Assignment #12:

An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 6.79 m/s2. Determine the orbital period of the satellite.

r

s

T

aGM

arT

rTr

Trrv

rvmma

rmGM

E

E

6675 79.6

1098.51067.62

22

2/2

4/1

3

2411

4/1

3

222

2

2

PHY 113 C Fall 2013 -- Lecture 14 610/14/2013

From Webassign Assignment #12:

How much work is done by the Moon's gravitational field as a 1090 kg meteor comes in from outer space and impacts on the Moon's surface?

M

M

RM

RM

RmGM

rmGMdr

rmGMW

MM

2

RM

iclicker questionA. W>0B. W<0

PHY 113 C Fall 2013 -- Lecture 14 710/14/2013

From Webassign Assignment #12:

A space probe is fired as a projectile from the Earth's surface with an initial speed of 1.74 104 m/s. What will its speed be when it is very far from the Earth? Ignore atmospheric friction and the rotation of the Earth.

vi22

21

21

fE

Ei

ffii

mvR

mGMmv

UKUK

PHY 113 C Fall 2013 -- Lecture 14 810/14/2013

From Webassign Assignment #12:Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)

GTvM

vTRTRv

MTvR

GMRvM

24

:algebra someAfter 2

2 know o want t;, Know

2

3

2

22

PHY 113 C Fall 2013 -- Lecture 14 910/14/2013

From Webassign Assignment #12:Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v = 190 km/s and the orbital period of each is 10.7 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)

iclicker exercise:Who might pose a question like this?

A. A mean professor.B. A puzzle master.C. An observational astronomer.

PHY 113 C Fall 2013 -- Lecture 14 1010/14/2013

Meanwhile – back on the surface of the Earth:

Conditions for stable equilibrium

0 : torqueof Balance

0 :force of Balance

ii

ii

τ

F

PHY 113 C Fall 2013 -- Lecture 14 1110/14/2013

Stability of “rigid bodies”

N

mig

PHY 113 C Fall 2013 -- Lecture 14 1210/14/2013

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 C Fall 2013 -- Lecture 14 1310/14/2013

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 for q 0.

PHY 113 C Fall 2013 -- Lecture 14 1410/14/2013

Unstable equilibrium:

mg(-j)

r

Tq

Support above CM:

Support below CM:

PHY 113 C Fall 2013 -- Lecture 14 1510/14/2013

Nik Wallenda walking on high wire across Grand Canyon

PHY 113 C Fall 2013 -- Lecture 14 1610/14/2013

Analysis of stability: i

i 0F i

i 0τ

PHY 113 C Fall 2013 -- Lecture 14 1710/14/2013

0)1( :Torques0 :Forces

gxmmgMgmgmgMn

cD

PcD

**X

PHY 113 C Fall 2013 -- Lecture 14 1810/14/2013

PHY 113 C Fall 2013 -- Lecture 14 1910/14/2013

**X

Fg1

mg

RCM

0)()2( :Torques 1 CMg RmgmF

PHY 113 C Fall 2013 -- Lecture 14 2010/14/2013

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 C Fall 2013 -- Lecture 14 2110/14/2013

F1F2

L/3

LMg mg

023

:Torques

0 :Forces

2

21

mgLLMgLF

mgMgFF

**X

mgMgFmgMgF 221 3

23

12

PHY 113 C Fall 2013 -- Lecture 14 2210/14/2013

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 C Fall 2013 -- Lecture 14 2310/14/2013

T

Mgmg

**X

2/

x

t

sin2//

0sin2

0

MgmgxT

TMgmgx

NTNMgNmgmxmo

313200 600

2 8 53For

PHY 113 C Fall 2013 -- Lecture 14 2410/14/2013

d

**X

hrhrr

mghr

mgdF

hrFmgd

22

02 0

22

t

PHY 113 C Fall 2013 -- Lecture 14 2510/14/2013

mg

Mg

Fwall

N

T

Mg = 120 N

mg = 98 N

T < 110 N

**X

q

qq

qq

sincos2/

0sin

cos2

cos

00

MgmgxT

LF

MgLmgxL

MgmgNFT

wall

wall

PHY 113 C Fall 2013 -- Lecture 14 2610/14/2013

x

A ladder of weight Mg and of length L is supported by the ground with static friction force f and by a frictionless wall as shown. The firefighter has weight mg and is half-way up the ladder. Find the force that the ladder exerts on the wall.

q

wallF

f**X

hamg

LhaxMgfF

amgxLaMghF

fF

wall

wall

wall

2

021

0