Chapter 9
Static Equilibrium; Elasticity and Fracture

Torque and Two Conditions For Equilibrium
An object in mechanical equilibrium must satisfy the following conditions:
1.
2.
The net external force must be zero:
ΣF = 0
The net external torque must be zero:
Σ τ = 0

Two Conditions of Equilibrium
 First Condition of Equilibrium
○ The net external force must be zero
0 or
 F x
 0 and  F y
 0
 Necessary, but not sufficient
 Translational equilibrium

Two Conditions of Equilibrium
 Second Condition of Equilibrium
○ The net external torque must be zero
τ
τ x
τ y
 Rotational equilibrium
 Both conditions satisfy mechanical equilibrium

Two Conditions of Equilibrium
 Objects in mechanical equilibrium
 Rock on ridge
 See-saw

Examples of Objects in Equilibrium
1.
2.

Draw a diagram of the system
Include coordinates and choose a rotation axis

Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object
For systems containing more than one object, draw a separate free body diagram for each object

Examples of Objects in Equilibrium
3.
4.
4.

Apply the Second Condition of Equilibrium
Σ τ = 0
This will yield a single equation, often with one unknown which can be solved immediately
Apply the First Condition of Equilibrium
ΣF = 0


This will give you two more equations
Solve the resulting simultaneous equations for all of the unknowns
Solving by substitution is generally easiest

Examples of Objects in Equilibrium
 Examples of Free Body Diagrams
(forearm)
• Isolate the object to be analyzed
• Draw the free body diagram for that object
• Include all the external forces acting on the object

Examples of Objects in Equilibrium
 FBD - Beam
• The free body diagram includes the directions of the forces
• The weights act through the centers of gravity of their objects
Fig 8.12, p.228
Slide 17

Examples of Objects in Equilibrium
 FBD - Ladder
• The free body diagram shows the normal force and the force of static friction acting on the ladder at the ground
• The last diagram shows the lever arms for the forces

Examples of Objects in Equilibrium
 Example:

 So far, studying rigid bodies
 the rigid body does not ever stretch, squeeze or twist
 However, we know that in reality this does occur, and we need to find a way to describe it.
 This is done by the concepts of stress , strain and elastic modulus .

 All objects are deformable
 All objects are spring-like!
 It is possible to change the shape or size
(or both) of an object through the application of external forces
 When the forces are removed, the object tends to its original shape
 This is a deformation that exhibits elastic behavior (spring-like)

 Stress is the force per unit area causing the deformation
 Strain is a measure of the amount of deformation

 The elastic modulus is the constant of proportionality between stress and strain
 For sufficiently small stresses, the stress is directly proportional to the strain
 The constant of proportionality depends on the material being deformed and the nature of the deformation
 Can be thought of as the stiffness of the material
 A material with a large elastic modulus is very stiff and difficult to deform
○ Analogous to the spring constant

 Tensile stress is the ratio of the external force to the crosssectional area
 Tensile is because the bar is under tension
 The elastic modulus is called
Young’s modulus

 SI units of stress are Pascals, Pa
 1 Pa = 1 N/m 2
 The tensile strain is the ratio of the change in length to the original length
 Strain is dimensionless
F
A
 Y
 L
L o s t r e s s = E la s t ic m o d u lu s × s t r a in




A bar of material, with a force F applied, will change its size by:
Δ L/L =

=

/Y = F/AY
Strain is a very useful number, being dimensionless





Example: Standing on an aluminum rod:
 Y = 70

10 9 N·m 
2 (Pa)
 say area is 1 cm 2 = 0.0001 m 2 say length is 1 m weight is 700 N

= 7

10 6 N/m 2

= 10

4
 ΔL = 100
 m compression is width of human hair
F
A

= F/A

= Δ L/L

= Y
· 
L
F

L


Young’s modulus applies to a stress of either tension or compression
 It is possible to exceed the elastic limit of the material
 No longer directly proportional
 Ordinarily does not return to its original length

 If stress continues, it surpasses its ultimate strength
 The ultimate strength is the greatest stress the object can withstand without breaking
 The breaking point
 For a brittle material, the breaking point is just beyond its ultimate strength
 For a ductile material, after passing the ultimate strength the material thins and stretches at a lower stress level before breaking





Forces may be parallel to one of the object’s faces
The stress is called a shear stress
The shear strain is the ratio of the horizontal displacement and the height of the object
The shear modulus is
S

shear stress shear strain 

F
A
 x h
F
A
 S
 x h


S is the shear modulus
A material having a large shear modulus is difficult to bend

 Bulk modulus characterizes the response of an object to uniform squeezing
 Suppose the forces are perpendicular to, and act on, all the surfaces
○ Example: when an object is immersed in a fluid
 The object undergoes a change in volume without a change in shape


Volume stress, ΔP, is the ratio of the force to the surface area
 This is also the
Pressure
 The volume strain is equal to the ratio of the change in volume to the original volume

P B
 V
V
 A material with a large bulk modulus is difficult to compress
 The negative sign is included since an increase in pressure will produce a decrease in volume
 B is always positive
 The compressibility is the reciprocal of the bulk modulus


Solids have Young’s, Bulk, and Shear moduli
 Liquids have only bulk moduli, they will not undergo a shearing or tensile stress
 The liquid would flow instead

 The ultimate strength of a material is the maximum force per unit area the material can withstand before it breaks or factures
 Some materials are stronger in compression than in tension

 Example: