lecture 23 ch12 f19 static equilibrium · phys.1410 lecture 23 a.danylov the 2nd condition for...

DepartmentofPhysicsandAppliedPhysicsPHYS.1410Lecture23A.Danylov

Lecture 23

Chapter 12

Static Equilibrium

Physics I

My favorite subject

… Static Equilibrium

Course website:https://sites.uml.edu/andriy-danylov/teaching/physics-i/


Today we are going to discuss:

Chapter 12:

Static Equilibrium: Section 12.8

IN THIS CHAPTER, you will discuss static equilibrium of an object


Angular Momentum Conservation helps to solve many problems

I11 I22

IL For a rigid body21 LL

Launch: leg and hands are out to make I large

Flight: leg and hands are in to make large

Landing: leg and hands are out to “dump” large

1212 )( II

large1 I mallI s2

Example Figure Skater’s Jump

Angular Momentum stays constant throughout the whole jump

1 2

small-1 large-2

ConcepTest Figure Skater

A) thesame

B) largerbecauseshe’srotatingfaster

C)smallerbecauseherrotationalinertiaissmaller

Afigureskaterspinswithherarmsextended.Whenshepullsinherarms,shereducesherrotationalinertiaandspinsfastersothatherangularmomentumisconserved.Comparedtoherinitialrotationalkineticenergy,herrotationalkineticenergyaftershepullsinherarmsmustbe:

KErot =I2 =(I) =L (usedL=I ).BecauseL isconserved,largermeanslargerKErot.

12

12

12

12

The“extra”energycomesfromtheworkshedoesonherarms.


Bicycle wheel/turntable as a demo of Angular Momentum Conservation


Bicycle wheel/Ang. Mom. Conservation

xL

wzL

pzL

Initial situation Final situation

finz

inz LL

Angular Momentum Conservation (z comp):

pz

wz

finz LLL 0 w

zpz LL wwpp II

wp

wp I

I )( counterclockwiseclockwise

0inzL w - wheel

p - person

ConcepTest Spinning Bicycle Wheel

You are holding a spinning (CCW) bicyclewheel while standing on a stationaryturntable. If you suddenly flip the wheelover so that it is spinning in the oppositedirection, the turntable will:

ThetotalangularmomentumofthesystemisL upward,anditisconserved.Soifthewheelhas−L downward,youandthetablemusthave+2L upward.

A)remainstationary

B)starttospininthesamedirectionasbeforeflipping(CCW)

C)starttospininthesamedirectionasafterflipping(CW)

finz

inz LL

tablemeLLL zLL tableme 2

L

L2L?


Static Equilibrium


The 1st Condition for Equilibriumprevents translational motion

0F

0 xF 0 yF 0 zF

amF

N. 2nd law describes translational motion

He doesn’t want to have any sliding of a ladder, i.e. 0a

Force equilibrium


The 2nd Condition for Equilibriumprevents rotational motion

0

I

Rotational N. 2nd law describes rotational motion:v

He doesn’t want to have any rotation of a ladder, i.e. 0

There must be no net torque around any axis (the choice of axis is arbitrary).

0 x 0 y 0 z

Torque equilibrium

Six conditions!


Reduce#ofEquilibriumEquations

For simplicity, we will restrict the applications to situations in which all the forces lie in the xy plane.

1st condition:

2nd condition:

0 xF 0 yF 0 zF

0 x 0 y 0 z

0)1 xF 0)2 yF 0)3 z

There are three resulting equations, which we will use

Fr

Fr ,

Simplification!


Axisofrotationforthe3rd equation

Does it matter which axis you choose for calculating torques?

0 z

0F

Anyaxisofrotationworks.Itisuptoyouwhichonetochoose

If an object is in a force equilibrium and the net torqueis zero about one axis, then the net torque must be zero about any other axis

We should be smart to choose a rotation axis to simplify problems

So.Thechoiceofanaxisisarbitrary

EH


Anyaxisofrotationworksforthe3rd equation(proof)(Read only if you want)

The choice of an axis is arbitrary 0 z

End of Class


Concurrent/Nonconcurrent forces

1F 2F

3F

Concurrentforces:whenthelinesofactionoftheforcesintersectatacommonpoint,therewillbenorotation.So

1F 2F

3F

0 xF 0 yF0 z 0 xF 0 yF0 z

Nonconcurrent forces:whenthelinesofactionoftheforcesdonotintersectatacommonpoint,therewillberotation.So

Automaticallysatisfied

What’s a difference?


Recallhowtocalculatetorque(shortcut)

Fr

F

r

• Draw a line of force • Find the perpend. distance (r┴) from the axis of rotation

to that line.• Magnitude of torque is r┴F• Torque is positive if it tries to rotate an object CCW

sinrF

Line of force (action)

r Fr

Lecture 20

ConcepTest Static equilibriumConsideralightrodsubjecttothetwoforcesofequalmagnitudeasshowninfigure.Choosethecorrectstatementwithregardtothissituation:

(A) The object is in force equilibrium but not torque equilibrium.(B) The object is in torque equilibrium but not force equilibrium(C) The object is in both force equilibrium and torque equilibrium(D) The object is in neither force equilibrium nor torque equilibrium

F

extforce equilibrium

torque equilibrium

Here,the1stconditionissatisfiedbutthe2nd isn’t,sotherewillberotation.So,tohavestaticsituation,bothconditionsmustbesatisfied.

origin

Fr

1

2

0

FF

0 X21


Example

0 z

0 xF

A 3.0‐m‐long ladder leans against africtionless wall. The coefficient ofstatic friction between the ladder andthe floor is 0.40.

What is the minimum angle the laddercan make with the floor withoutslipping?

Ladder stability (12.58)

0 yF

The forces are nonconcurrent, so we need all equilibrium

conditions


Find the tension in the two wires supporting the traffic light

Example Traffic light


1. Choose one object at a time, and make a free-body diagram by showing all the forces on it and where they act.

2. Choose a coordinate system and resolve forces into components.

3. Write equilibrium equations for the forces.

4. Choose any axis perpendicular to the plane of the forces and write the torqueequilibrium equation. A clever choice here can simplify the problem enormously.

5. Solve.

Solving Statics Problems


That’s all, Folks!!

lecture 23 ch12 f19 static equilibrium · phys.1410 lecture 23 a.danylov the 2nd condition for...

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