Chapter 11ww2.odu.edu/~agodunov/teaching/notes231/Chapter_11.pdf · Chapter 11 Equilibrium 2 Two Types of equilibrium 9Static equilibrium ... The center of gravity The gravitational

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Chapter 11

Equilibrium

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Two Types of equilibrium

Static equilibriumA body is at rest (no translation or rotation)example: a book resting on a desk

DynamicA body is moving without accelerationexample: airplane flying with constant speed

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3

Stable static equilibriumIf a body returns to a state of static equilibrium after having been displaced from it by a forceexample: a marble at the bottom of a hemispherical bowl

Unstable static equilibriumIf a small force can displace the body and end the equilibriumexample: a marble at the top of a hemispherical surface

Static equilibrium: stable or unstable

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Two Conditions for Equilibrium

1. The vector sum of all the external forces that act on the body must be zero

2. The vector sum of all the external torques that act on the body, measured about any possible point, must be zero

0=netFr

0=netτr

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5

the conditions in the component form

balance of forces balance of torques

000000

==

==

==

net,znet,z

net,ynet,y

net,xnet,x

τFτFτF

in physics 231 we consider only situations in which the forces that act on the body lie in the (x,y) plane, then

⎪⎩

⎪⎨

⎧

=

=

=

000

net,z

net,y

net,x

τFF

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Choice of a reference point for torque

find torque about point O if torque about point P is given

It does not matter what point to choose, but choose wisely

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The center of gravityThe gravitational force on an extended body is the vector sum of the gravitational forces acting on the individual elements (atoms) of the body. Instead of counting all those individual elements, we can say

The gravitational force acting on a body effectively acts at a single point, called the center of gravity of the body

“effectively” means that if the forces on individual elements were turned off and force at the center of gravity were turned on the net force and the net torque (about any point) acting on the body would not change

gFr

gFr

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Finding the center of gravityIf is the same for all elements of a body, then the body’s center of gravity is coincident with the body’s center of mass.

Proof:

gr

∑∑

∑∑∑

∑∑

==

×=

=×=×=×=

×=

===

iiicm

ii

cmnet

iii

iii

iiinet

iii

ii

iig

rmM

rmM

Mr

gMgrmM

grmgmr

gmr

gMmggmF

rr

rr

rrrrrrrr

rrr

rrrr

1

1

and where

torque net the

element single a on torque

:force naltranslatio net the

τ

τ

τ

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9

A uniform beam on two scalesA uniform beam, of length L and mass m is at rest with it’s ends on two scales. A uniform block, of mass M, is at rest on the beam, with it’s center a distance x from the beam’s left end. What do the scales read?

example

left scale right scale

Mm

xL

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A uniform beam on two scalesexample

x L/2

gMrgmr

lFr

rFr

02/00

=+−−

=−−+

rl

rl

LFLmgxMgFmgMgFF

:point left the about : yalong

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A uniform beam on two scalesexample

⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

=

⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛=

=

⎟⎠⎞

⎜⎝⎛ −

+=

⎟⎠⎞

⎜⎝⎛ +=

MmgFMmgF

Lx

MmgFmgF

x

MLxLmgF

MLxmgF

lr

lr

l

r

21

21

21

21

2/21

21

02121

and

:2 case

and

:1 case

forces unknown two for system the solving

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Lifting a carYou are about to lift a car of mass M using a lever. If you mass is m and the lever is L meters long and negligible mass comparing to m and M . Where must the support point be placed?How much load must the support hold?

example

gMmNmM

mLx

xMgmgxLMgmgN

)(

0)(0

+=+

=

=−−

=−−

and then

:O :y

lift a planet?

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13

in a museum …

Two identical pictures are hanging from a ceiling. Where is the greatest tension in the string?

question

1T 2T 3T 4T

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A picture on a wall

A picture can be hung on a wall in three different ways, as shown. Where is the greatest tension in the string?

example

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Not too contemporary art

Toy penguins are hanging from a ceiling. Each crossbar is horizontal, has negligible mass, and extends three times as far to the right of the wire supporting it as to the left. Penguin 1 has mass 48 kg. What are the masses of the other penguins?

example

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A problem with a safeA safe of mass M hanging by a rope from a boom with dimensions a and b. The boom consist of a hinged beam and a horizontal cable that connects the beam to a wall. The uniform beam has a mass m. The masses of the cable and the rope are negligible.

a) what is the tension in the cable

b) find the magnitude of the net force on the beam from the hinge.

example

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Equilibrium: comments on problem solvingChoose ONE object in a time for consideration

Draw a free-body diagram (show ALL forces acting ON that object)

Choose (wisely) a coordinate system and resolve forces in their components

“Generate” equilibrium equations using the conditions for equilibrium

⎪⎩

⎪⎨

⎧

=

=

=

000

net,z

net,y

net,x

τFF

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A problem with a safeexample

⎟⎠⎞

⎜⎝⎛ +=

=−−

+==

mMabgT

bmgbMgaT

MgmgFTF

c

c

v

ch

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021

equation third the from

:O

:y :x

beam :object

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An advertising signA metal sign of mass M is suspended from the end of a rod of mass m and length L. The rod is supported by a cable with negligible mass.

a) what is the tension in the cable

b) find the magnitude of the net force on the rod from the hinge.

example

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An advertising signexample

gMrgmr

Tr

xFryF

r

αh

L

02

sin

0sin0cos

=−−

=−−+=−

LMgmgLLT

MgmgTFTF

y

x

α

ααrod :object

⎟⎠⎞

⎜⎝⎛ += MmgT

21

sinα

solution

22sin

Lhh+

=α where

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what if …example

gMrgmr

Tr

xFryF

r

αh

L

( ) 02

sin

0sin0cos

=−−−

=−−+=−

LMgmgLTxL

MgmgTFTF

y

x

α

αα

⎟⎠⎞

⎜⎝⎛ +

−= Mm

xLLgT

21

sinα

solution

( )22sin

xLh

h

−+=α where

x

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An advertising sign (more complicated)A metal sign of mass M is suspended by two wires from a rod of mass m and length L. The rod is supported by a cable with negligible mass.

What is the tension in the cable

example

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An advertising sign (more complicated)example

gM r

2

gmr

TryF

r

xFr

gM r

2

( ) 02

)(2

sin

022

sin

0cos

=+−−−

=−−−+

=−

gMLaLmgLLT

gMgMmgTF

TF

y

x

α

α

αrod :object

( )

for

solution

⎟⎠⎞

⎜⎝⎛ +==

−+=

MmgTa

aLMmLLgT

2sin0

)2(sin2

α

α

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A pickup truckA pickup truck has a wheelbase of L meters. Ordinary M1 kg rests on the front wheels, and M2 on the rear wheels, when the truck is parked on a level road. A box of m kg is now placed on the tailgate, x meters behind the rear axel.

How much total weight now rests on the front wheels?

On the rear wheels?

How much weight would need to be placed on the tailgate to make the front wheels come off the ground?

example

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example

gM r1

gM r2

gmr

1Nr

2Nr

L x

0)(

00

11

11

2121

=−=

=−−

=−−−+

mLxMgN

xmgLNgLMmggMgMNN

for solution

axle) second the (aroung

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Climbing ladderexample

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Climbing ladder IIexample

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Climbing very light ladderexample

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Empty ladderexample