17/29/2004

Midterm 2 – Tomorrow (7/30/04) Material from Chapters 7-12 Room where recitation meets

Practice Exam available on-line or in Davey library

Some good practice problems 7.13, 8.7, 8.21, 9.10, 9.23, 9.35, 10.4, 10.23, 10.29,

11.33, 11.56, 12.19, 12.23, 12.40

Announcements

Chapter 13

Equilibrium and Elasticity

37/29/2004

Equilibrium

An object is in equilibrium if:

constP

constL

and

Examples: fan blades, ball rolling across the floor…

An object is in static equilibrium if it is at rest in our reference frame

0P

0L

and

Examples: bridges, skyscrapers…

47/29/2004

Stable vs. Unstable Static Equilibrium

Stable Equilibrium:

A small perturbation of the objects position results in the object being pushed back to its original position by restoring forces

Unstable Equilibrium:

A small perturbation of the objects position results in the object being pushed away from its original position.

Neutral Equilibrium:

When displaced, the object stays in new position

57/29/2004

Types of Equilibrium

U

x

stable

unstable

neutral

67/29/2004

Stable vs Unstable Equilibrium

Unstable Equilibrium: easy to knock things overExamples: Dominos or a Rube Goldberg machine

77/29/2004

Demo: Equilibrium

If I stack two blocks on top of each other, what is the condition for equilibrium?

L

½ L

Center of mass of top block must be supported by the bottom block

87/29/2004

Demo: Equilibrium

What about 3 blocks?

CM of top two blocks must be supported by bottom block

L

½ L

??

LM

LMLMx cm 4

3

2

)()2/(2,

L2

1

L4

3

L4

1

97/29/2004

Demo: Equilibrium

4 blocks?

LM

LMLMLMx cm 12

11

3

)()()( 45

21

3,

L2

1

L4

5

L

???

L4

1

L2

1

L4

3

L6

1L2

1

A pattern is developing…

i

Lxx ii

1

21

1

1

1

2

n

i i

Lx

107/29/2004

Demo: Equilibrium

1

1

1

2

n

i i

Lx

This is known as the geometric series, and it never converges!

With enough blocks, we can make the overhang as big as we like!

With 5 blocks:

LL

i

Lx

i

04.14

1

3

1

2

11

2

1

2

4

1

Complete overhang!

117/29/2004

Requirements for Equilibrium

If an object is in equilibrium, then:

0netF

0net

So we can solve statics problems using only the physics we already know!

127/29/2004

Problem Solving Strategy:

1)Draw and label a diagram

2)Pick an appropriate origin

a)torques sum to zero about any choice of origin

3)Sum forces and torques

4)Solve for unknown quantities

137/29/2004

Example: Teeter-Totter

mgMg

d=2mD

A 80 kg parent and a 20 kg child are balancing on a see-saw. If the child sits 2 m from the pivot, where does the parent need to sit, and what is the force on the pivot?

147/29/2004

Example:

Pick origin at pivot (makes the torque from the force at the pivot = 0) mgMg

d=2mD

Fp

Summing the torques:

0 childfather

0 mgdMgD

mkg

mkg

M

mdD 5.0

80

)2)(20(

+

157/29/2004

Example:

mgMg

d=2mD

Fp

Summing the forces:

0 childfatherp FFF

0 mgMgFp

Nkgkgsm

mMgFp

980)2080)(/8.9(

)(2

167/29/2004

Example: Another Way…

mgMg

d=2mD=0.5m

Fp

Pick dad as origin, then sum torques:

0 childpivot

Nm

msmkg

D

dDmgFp

9805.0

)5.2)(/8.9)(20(

)(

2

0)( dDmgDFp

Same result!

+

177/29/2004

Example:

10 in

2 in 3 in

3 N

2 N0.8 N

5 in

T1=? T2=?

Weights are suspended from a rod which is suspended by two ropes at its ends. Find the tension in these ropes, T1 and T2.

187/29/2004

Example: (continued)10 in

8 in

3 in

3 N 2 N0.8 N

5 in

T1=? T2=?

Pick origin at right edge

Sum torques:

0)10()3)(8()8.0)(5()2)(3( 1 TinNinNinNin

+

NT 4.31

197/29/2004

Example: (continued)10 in

8 in

3 in

3 N 2 N0.8 N

5 in

T1=? T2=?

Sum forces:

028.034.3 2 TNNNN

NT 4.22

207/29/2004

Does it work?

NT 4.22

NT 4.31

T1 T2

217/29/2004

Standing on a Ladder

θL

d

FN1

FN2

Mg

mg

Ffr

A 5 m ladder rests against a frictionless wall at an angle 30º from vertical. The weight of the ladder is 200 N, uniformly distributed along its length. A 1000 N person climbs the ladder.

What coefficient of friction must the ladder have with the floor so the ladder does not slip when the person has climbed 4 m along the ladder?

227/29/2004

Standing on a Ladder

L

d

FN1

FN2

Mg

mg

Ffr

Pick the origin at the bottom of the ladder:

Summing forces in the x-direction:

01 frN FF

frN FF 1

Summing forces in the y-direction:

02 mgMgFN

NNNgmMFN 12002001000)(2

237/29/2004

Standing on a Ladder

L

d

FN1

FN2

Mg

mg

Ffr

Summing the torques:

0)50sin(30sin30sin2 1

LFMgdL

mg N

)50sin(

30sin)2/(1

L

MgdLmgFN

N

m

mNmNFN 587

50sin)5(

30sin)4)(1000()2/5)(200(1

NFF Nfr 5871

489.01200

587

2

N

N

F

F

N

fr

247/29/2004

Towing a Trailer Safely

d1=3m

d2=5m

x d=4m

NhN1 N2

½Mg ½Mg

A trailer hitch has a rated ‘tongue weight’ of 2000 N. A lightweight two-axle trailer, with axles 3 m and 5 m behind the hitch, is pulled behind the vehicle. A 4 m car weighing 10000 N is parked on the trailer. How far back must the car be parked, and what are the forces on the axles?

257/29/2004

Towing a Trailer Safely

d1=3m

d2=5m

x d=4m

NhN1 N2

½Mg ½MgSum forces:

Pick the hitch as the origin.

Sum torques:

021 MgNNNh

0)( 221121

21 NdNdMgdxMgx

Don’t know x, N1, N2

2 equations and 3 unknowns!Not enough info!

267/29/2004

Indeterminate Structures

Sometimes the force and torque equations lead to more unknown forces than equations.

Example: Four-legged table

Detailed material properties and history determine the forces.

277/29/2004

Elasticity

When a real object is subjected to a force (or a stress), it will deform (strain).

Over a range of deformation, objects usually obey a form of Hooke’s law

strainmodulusstress

The “modulus” is like the spring constant. It varies for different materials and different stresses

287/29/2004

Types of Stress

Tensile (and compressive):

L

F

A

L+LL

LE

A

F

E = Young’s modulus

Not all materials are as good under tension as under compression (example: concrete)

297/29/2004

Types of Stress

Shear stress:

L

A

L

xG

A

F

G = shear modulus

x F

V

V

Hydraulic stress:

V

VBp

B = Bulk modulus

p = pressure

307/29/2004

Elasticity

Objects return to original shape if deformation is small enough.

The point at which a deformation becomes permanent is called the yield strength

The point at which on object breaks is called its ultimate strength

317/29/2004

Tensile Stress: