Phys211C11 p1

Equilibrium and Elasticity

Equilibrium:

pointanyabout

forcesexternal

F

F

F

z

y

x

0

0

0

0

0

=

=

=

=

⇒=

∑

∑∑∑

∑

τ

F

r

r

Phys211C11 p2

Center of Gravity:

gr

gr

grgrgr

grgrgr

g

rrrr

rr

rr

rrL

rrrr

Lrrrrrr

r

r

L

Lrr

r

M

mmm

mmm

mmm

gravityconstantofeffect

m

m

mm

mm

cm

iiii

ii

ii

i

iicm

×=τ

Σ×ΣΣ=

×Σ=+×+×=

+×+×=×Σ=τ

Σ

Σ=

++

++=

)()(

)()()(

)()()(

2211

2211

21

2211

The Center of Gravity of an object

The point of an object from which it could be suspended without tending to rotate.

The point where all the mass of an object can be considered to be located.

CG does not need to be located within the physical object!

Horseshoe, for example

usually easily identified from symmetry.

Phys211C11 p3

phys phys

phys

cgcg

cg

w⃗ w⃗ w⃗

area of support area of support area of support

aieeee!!!!

Phys211C11 p4

Example: A child walks on a massive (90 kg) 6.0 m plank which rests on two saw

horses separated by D = 1.5m equally spaced from the center of the plank. If the plank

is not to tip over when the child walks to the end of the plank, what is the most mass

the child can have?

D

L

Phys211C11 p5

How to approach an equilibrium problem: For each object

• Draw the forces that act on the object (i.e. draw a free-body diagram)

• Choose a convenient set of coordinate axis and resolve all forces into components.

Watch carefully for appropriate use of +/- signs.

• Pick a convenient axis to calculate your torques about.

• Set the sum of the force components along each axis equal to 0.

• Set the sum of the torques equal to 0

• Solve the resulting equations for the unknown quantity or quantities.

objectfor0

0

0

each

F

F

y

x

=

=

=

∑

∑

∑

τ

Phys211C11 p6

Example: A ladder 5.00 m long is leaning against a frictionless wall with its lower end

3.00 m away from the wall. The ladder weighs 180 N, and an 800 N man is a third of

the way up the ladder. What forces do the wall and the ground exert on the ladder?

What is the minimum coefficient of friction necessary for the ladder to not slip?

θ

Phys211C11 p7

Example : A 60 kg woman stands at the end of a uniform 4m, 50 kg diving board

supported as shown. Determine the forces exerted by the two supports.

.800 m

4.00 m

Phys211C11 p8

Stress, Strain and Elastic Moduli

material property of “stretchiness/springiness”

−> how materials respond to stress

compression tension shear

Elastic modulus = stress/strain (approximately constant)

strain: deformation = fractional change

stress: force per area

property of type of material

Phys211C11 p9

Young’s Modulus: how things stretch (elastically)

tensile stress: force per area = F⊥/A

strain: fractional change in length

= change in length per original length = ∆l/l0

Elastic modulus = stress/strain

Young’s modulus (for stretching in one direction)

compression tension

l A

l0A l0

A

l A

0ll

AFY

∆=

Phys211C11 p10

Bulk Modulus: compression of solids, liquids and gases

bulk stress: force per area = F⊥/A = pressure p ( 1 N/m2 = 1 Pa)

1 atm = 1.013 E5 Pa = 14.7 lb/in2

bulk strain: ∆V/V0

Bulk Modulus B relates small change in pressure to bulk strain

0VV

pB

∆

∆−=

Phys211C11 p12

φ≅

φ==

∆

∆−=

∆=

AFAF

hx

AFS

VV

pB

ll

AFY

||||||

0

0

tan

Elastic Moduli

7.516.020.0Steel

0.64.11.6Lead

7.716.021.0Iron

4.414.011.0Copper

3.56.09.0Brass

2.57.57.0Aluminum

S

1010 Pa

B

1010 Pa

Y

1010 Pa

Material

45.8Water

3.70Mercury

21.0Glycerin

110.0Ethyl Alcohol

93.0

Compressibility k = 1/B

1E-11 Pa

Carbon disulfide

Liquid

Phys211C11 p13

Example: A steel cable 2.0 m long has a cross section area of 0.30 cm2. A 550 kg mass is

suspended from the cable. Calculate the stress, the resulting strain, and the elongation of

the cable.

Example: A hydraulic press contains 0.25 m3 of oil is subjected to a pressure increase of ∆p

= 1.6E7 Pa B = 5.0E9 Pa Find the decrease in volume.

Phys211C11 p14

Elastic Limit: the maximum stress (force) which can be applied to an object without resulting in permanent deformation.

Plastic Deformation or Plastic Flow: the permanent deformation which results when a material’s elastic limit has been exceeded.

Ultimate strength: greatest tension (or compression or shear) the material can withstand with breaking *snap*, tearing, fracturing etc. a.k.a. Breaking Stress or Tensile Strength.

A malleable or ductile material has a large range of plastic deformation.

Fatigue: small defects reduce materials strength well below original strength.

Elastic Hysteresis: strain depends on material “history”

stre

ss

strain

Phys211C11 p15

Example: A copper wire1.0 mm in diameter and 2.0 m long is used to support a mass of

5.0 kg. By how much does this wire stretch under this load? What is the maximum

mass which can be supported without exceeding copper’s elastic limit?

Y = 1.1x1011 Pa elastic limit = 1.5x108 Pa

Phys211C11 p16

From Lab

Force: suspended weight

Area: cross section of wire

elongation causes cylinder to rotate

2θ

θ ∆L

d