• Main points of today’s lecture:

Physics 231 Lecture 23

• Gravitation potential energy

1 2grav

GM MPEr

= −

• Tensile stress and strainLYF Δ=Δ

12r

• Bulk stress and strain:0LA

VBPF ΔΔΔ

• Pressure in fluids:F

VBP

A=Δ=

• Buoyancy

ghPPAFP topbot ρ+== ;

gVWF displacedfdisplacedB ρ==

Gravitational Potential Energy• ΔPE = mgh is valid only near the earth’s surface

F bj t hi h b th th’ f• For objects high above the earth’s surface, an alternate expression is needed (You would need calculus to derive it.) mMGPE E=

– Zero potential energy is defined to be the value infinitely far from the earth

rGPE −=2

y• If the total mechanical energy of an object exceeds

zero, the object can escape the gravitation field of the earth. The escape velocity is when the Emech=0.At the Earth's surface:

2 Earthmech

Earth

GM m1E mv2 R

= −

mech escapeIf E 0 corresponding to v v , the object

can escape the Earth's gravitational field

≥ ≥

2 Earth Earthescape escape

Earth Earth

GM m 2GM10 mv v2 R R

= − = you will have loncapaproblems on this. 9

Conceptual question

A k i iti ll t t ith t t E th d l t d• A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth Neglecting friction the rock’srock as it plummets to Earth. Neglecting friction, the rock s speed when it hits the ground is– a) twice

b) th ti

f f 0 0 0KE PE KE PE PE+ = + =EGM mPE PE 0= − =– b) three times

– c) four times– d) six times

0PE PE 0r

= =

GM m0also: KE 0= f fKE PE 0 + =

– e) eight times

its speed at the top of the tower.

Ef f

f

GM mKE PEr

= − =E

d d

GM mKE r r E4r 4r

1its speed at the top of the tower. ground ground

Etower

tower

KE rGM mKE

r

= tower

ground

rr

= E

E

4r

= =

tower

ground

r

r1

22

21

4 groundgroundground vmvKE2

2

2124

tower

ground

towertower

ground

vmvKE===

Stress and strain in solids• All materials deform when under stress ieAll materials deform when under stress, ie.

when a force is applied to them. The atoms in a solid are held in place by forces that act like springs. p g

• If you pull on a bar of cross-sectional area A with a force F, the bar will stretch a distance ΔL. In this case, the force is perpendicular to

ΔΔ

. t s case, t e o ce s pe pe d cu a tothe surface, and the force F and distance ΔL are related by Young’s modulus Y:

StrainLLStress

AF

LLY

AF =Δ=Δ=

00

; ;

• The amount of force you need to apply to achieveThe amount of force you need to apply to achieve the same strain increases with the cross sectional area of the bar. The force will be linearly proportional to the strain up to the elastic limit. Beyond this point, the bar will be permanently deformed. Stress units Pa=N/m2. It is the force divided by the area, and has the units of pressure.

Example

B h Y ’ d l f b t 18 109 P U d• Bone has a Young’s modulus of about 18 x 109 Pa. Under compression, it can withstand a stress of about 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long and calculate the length by which this bone can belong and calculate the length by which this bone can be compressed before breaking.– a) 4.4 mm

b) 6 4

Y 18 x 109 PaF/A 160 x 106 Pa

– b) 6.4 mm– c) 1.6 cm– d) 2.0 cm

6 2F LΔ

L0 0.50 m

ΔL ?6 2F 160x10 N / m

A=

0

LYLΔ=

0LFLΔ 6 2 0.5m160 10 N / 4 40LA Y

Δ = 6 29 2160x10 N / m 4.4mm

18x10 N / m= =

If this force is uniformly applied, then F/A=P, where P is theIf this force is uniformly applied, then F/A P, where P is thepressure or force per unit area applied to the end of the femur. Units for pressure are N/m2 = Pa. (Pa = Pascals)

Bulk modulus

If d t l bl k i th it ill i k t b tt h th• If you drop a metal block in the ocean, it will sink to bottom where the pressure is much greater than that of the atmosphere. This will compress the block; the amount of compression is governed by the bulk modulus Bbulk modulus B.

strain Volume ;change pressure ; =Δ=ΔΔ=ΔVVP

VVBP

Reading Quiz1. The SI unit of pressure is1. The SI unit of pressure is

A. NB. kg/m2

C. PaD kg/m3D. kg/m

Slide 13-6

Density

Slide 13-12

Nuclear matter 2.66x1017

Pressure in fluids, Pascal’s principle• Consider a vessel filled with liquid and a smaller rectangular

l V f th li id t i d ithi th l Ifvolume V of the same liquid contained within the vessel. If the liquid is at rest, there can be no shear force on V, because it would cause the liquid to move, but there is a pressure force F=P A perpendicular to all 6 surfaces

y

force F=P⋅A perpendicular to all 6 surfaces.

zxIn equilibrium: F 0=

left rightx : F F 0; − = left left x right right xF P A ;F P A= = left rightP P =

front backz: F F 0; − =

bot topy: F F W 0; − − =

back back zF P A= front backP P =front front z F P A ;=

bot bot y F P A ;= top top yF P A= bot topWP PA

= +

• More detailed considerations show that pressure always exerts a force perpendicular to a surface and that pressure is the same everywhere at the same height i e P =P and P =P and P = P etc

bot topy ; bot bot y p p y pyA

height, i.e. Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc. • Defining density ρ of the liquid by ρ=M/V , and using V=Ayh we can simplify the

dependence of P on height as follows:yW Mg Vg A hgρ ρ= = =

ybot top

y

A hgP P

Aρ

= + bot topP P ghρ = +

Pressure and Depth equation

ghPP ρ+=

• If the top is open, Pbot is normal

ghPP topbot ρ+=

atmospheric pressure– 1.013 x 105 Pa = 14.7 lb/in2

• The pressure does not depend p pupon the shape of the container

• Gauge pressure is the difference between the pressure in the fluid and atmospheric pressure. If you connected a gauge which measures the pressure relative to atmospheric pressure to the fluid at depth h, then

• for this specific case:

guage bottom,h _ below _ surface atmosphereP P P ghρ= − =

B F PA k h d li

Pascal’s principal and hydraulic machines

• Because F=PA, one can make hydraulic machines that can generate large forces.

• Units: One atmosphere (1 atm) =– 76.0 cm of mercury– 1.013 x 105 Pa = 1.013 x 105 N/m2

– 14.7 lb/in

• Quiz: A simple hydraulic lift is employed to lift a truck that weighs 30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift the truck? (Hint: Are the pressures equal under each piston?)– a) 3N P P ; this is pascal's principal=– b) 300N– c) 30N– d) 0 33N

1 2P P ; this is pascal s principal=

21 1

FF AA

= 22

30000N1cm 30N1000cm

= =1 2F FA A

=d) .0.33N2A 1000cm1 2A A