Chapter 11
Equilibrium and Elasticity
Equilibrium
Two Conditions for Equilibrium

Fext  0

 ext  0

(about any point! )
• To motivate these, recall:




dL
dpcm
 ext 
Fext 
dt
dt
Defining Equilibrium
• Equilibrium
= no net external force or torque
= no change in translation or rotation)
• your text says L=0; others allow nonzero L:

dpcm
0
dt

pcm  constant

dL
0
dt

L  constant
Defining Static Equilibrium
• ‘Static’ Equilibrium
= the special case of
no translation or rotation at all

dpcm
0
dt

pcm  constant  0

dL
0
dt

L  constant  0
Two Conditions for Equilibrium

Fext  0

 ext  0

(about any point! )
• When applying these, we must consider all
external forces
• But the gravitational force is rather subtle
Center of Gravity (cg)
• Gravity acts at every
point of a body
• Let  = the torque on a
body due to gravity
• Can find  by treating
the body as a single
particle (the ‘cg’)
Center of Mass (cm)

rcm 

 mi ri
i
 mi
i
• it can be shown:
if g = constant
everywhere, then:
• center of gravity =
center of mass
Using the Center of Gravity
Pressent some more explanatory notes
Solving
Equilibrium Problems
Two Conditions for Equilibrium

Fext  0
•
•
•
•
 ext  0
From now on, in this chapter/lecture:
center of mass = center of gravity
‘equilibrium’ means ‘static equilibrium’
write: F and  for Fext and ext
First Condition
for Equilibrium

Fext  0
Fx  0
Fy  0
Fz  0
Second Condition
for Equilibrium

 ext  0
 x  0
 y  0
 z  0
Exercise 11-11
Work through Exercise 11-11
Exercise 11-14
Work through Exercise 11-14
A different version of
Example 11-3
The ‘Leaning Ladder’ Problem
Work through the variation the the text’s leaning ladder problem
Problem 11-62
‘Wheel on the Curb’ Problem
Work through Problem 11-62
Elasticity
Elasticity
• Real bodies are not perfectly rigid
• They deform when forces are applied
• Elastic deformation:
body returns to its original shape
after the applied forces are removed
Stress and Strain
• stress: describes the applied forces
• strain: describes the resulting deformation
• Hooke’s Law: stress = modulus × strain
• modulus: property of material under stress
• (large modulus means small deformation)
Hooke’s Law and Beyond
• O to a :
• small stress, strain
• Hooke’s Law:
stress=modulus×strain
• a<b:
• stress and strain are
no longer proportional
Units
• stress = modulus × strain
• stress (‘applied force’): pascal= Pa=N/m2
• strain (‘deformation’): dimensionless
• modulus:
same unit as stress
Types of Stress and Strain
• Applied forces are perpendicular to surface:
•
tensile stress
•
bulk (volume) stress
• Applied forces are parallel to surface:
•
shear stress
Tensile Stress and Strain
• tensile stress = F/A
• tensile strain = Dl/l0
• Young’s modulus = Y
Tensile Stress and Strain
stress  Y  strain
F
Dl
Y
A
l0
Work through Exercise 11-22
Compression vs. Tension
• tension (shown):
pull on object
• compression:
push on object
(reverse direction
of F shown at left)
• Ycompressive = Ytensile
Work through Exercise 11-26
Tension and Compression at once
Bulk Stress and Strain
• pressure: p=F/A
• bulk stress = Dp
• bulk strain = DV/V0
• bulk modulus = B
Bulk Stress and Strain
stress  B  strain
DV
Dp   B
V0
• B>0
• negative sign above:
Dp and DV have
opposite signs
Work through Exercise 11-30
Shear Stress and Strain
Shear Stress and Strain
• shear stress
= F7/A
• shear strain = x/h = tanf
• shear modulus = S
Shear Stress and Strain
stress  S  strain
F| |
Do Exercise 11-32
x
S
A
h
or
F| |
A
 S tan f  Sf
Regimes of Deformation
•
•
•
•
O to a :
(small stress, strain)
stress=modulus×strain
elastic, reversible
• a<b:
• elastic, reversible
• but stress and strain
not proportional
Regimes of Deformation
• From point O to b :
• elastic, reversible
• from point b to d:
• plastic, irreversible
• ductile materials have
long c–d curves
• brittle materials have
short c–d curves
Demonstation
Tensile Strength and Fracture