Transcript Equilibrium

Chapter 12:
Equilibrium and Elasticity
Conditions Under Which a Rigid
Object is in Equilibrium
Problem-Solving Strategy
Elasticity
Equilibrium:
An object at equilibrium is either ...
• at rest and staying at rest (i.e., static equilibrium) , or
• in motion and continuing in motion with the constant
velocity and constant angular momentum.
For the object in equilibrium,
• the linear momentum (
constant.


P  mv ) of its center of mass is
  
 
L  r  P  m(r  v ) ) about its
• the angular momentum (
center of mass, or any other point, is constant.
Conditions of Equilibrium:



dP
Net force: Fnet   ma
dt

Net torque:  net


dL 

 r  ma
dt
Conditions of equilibrium:


Fnet  0 (balance of forces) and  net  0 (balance of torques)
Fnet , x  0
Fnet , y  0
Fnet , z  0
 net , x  0
 net , y  0
 net , z  0
Another requirements for static equilibrium:

P0
The center or gravity:
The gravitational force on a body effectively acts at a
single point, called the center of gravity (cog) of the
body.
•the center of mass of an object depends on its shape and its density
•the center of gravity of an object depends on its shape, density, and the
external gravitational field.
Does the center of gravity of the body always
coincide with the center of mass (com)?
Yes, if the body is in a uniform gravitational field.
How is the center of gravity of an object determined?
The center of gravity (cog) of a regularly shaped body of
uniform composition lies at its geometric center.
The (cog) of the body can be located by suspending it from
several different points. The cog is always on the line-ofaction of the force supporting the object.
cog
Problem-Solving Strategy:
• Define the system to be analyzed
• Identify the forces acting on the system
• Draw a free-body diagram of the system and show all
the forces acting on the system, labeling them and
making sure that their points of application and lines of
action are correctly shown.
• Write down two equilibrium requirements in
components and solve these for the unknowns
Sample Problem 12-1:
• Define the system to be analyzed: beam & block
• Identify the forces acting on the system:
the gravitational forces: mg & Mg,
the forces from the left and the right scales: Fl & Fr
• Draw a force diagram
• Write down the equilibrium requirements in
components and solve these for the unknowns
O
balance of forces : Fl  Fr  Mg  mg  0
balance of torques : LFr  ( 12 L)mg  ( 14 L) Mg  0 Fl  0
Elasticity
Some concepts:
• Rigid Body:
• Deformable Body:
elastic body: rubber, steel, rock…
plastic body: lead, moist clay, putty…
• Stress: Deforming force per unit area (N/m2)
• Strain: unit deformation
Stress
Elastic modulus 
Strain
Young’s Modulus:
Elasticity in Length
The Young’s modulus, E, can be calculated by dividing the stress
by the strain, i.e.
stress F / A F L
E


strain L / L A L
where (in SI units)
E is measured in newtons per square metre (N/m²).
F is the force, measured in newtons (N)
A is the cross-sectional area through which the force is applied,
measured in square metres (m2)
L is the extension, measured in metres (m)
L is the natural length, measured in metres (m)
Table 12-1:
Some elastic properties of selected material of engineering interest
Density 
(kg/m3)
Young’s Modulus E
(109N/m2)
Ultimate Strength Su
(106N/m2)
Yield Strength Sy
(106N/m2)
Steel
7860
200
400
250
Aluminum
2710
70
110
90
Glass
2190
65
50

Concrete
2320
30
40

Wood
525
13
50

Bone
1900
9
170

Polystyrene
1050
3
48

Material
Shear Modulus:
Elasticity in Shape
The shear modulus, G, can be calculated by dividing the shear
stress by the strain, i.e.
shear stress F / A F L
G


shear strain x / L A x
where (in SI units)
G is measured in newtons per square metre (N/m²)
F is the force, measured in newtons (N)
A is the cross-sectional area through which the force is applied,
measured in square metres (m2)
x is the horizontal distance the sheared face moves, measured in
metres (m)
L is the height of the object, measured in metres (m)
Bulk Modulus:
Elasticity in Volume
The bulk modulus, B, can be calculated by dividing the hydraulic
stress by the strain, i.e.
B
hydraulic pressure
p
V

p
hydraulic strain
V / V
V
where (in SI units)
B is measured in newtons per square metre (N/m²)
P is measured in in newtons per square metre (N/m²)
V is the change in volume, measured in metres (m3)
V is the original volume, measured in metres (m3)
Young’s modulus
Shear modulus
Bulk modulus
Under tension and
Under shearing
Under hydraulic
compression
L / L
Strain is
Stress F L

Strain A L
G
Strain is
E
stress
x / L
Stress F L

Strain A x
Strain is
B
V / V
Stress
V
p
Strain
V
Summary:


Fnet  0 and  net  0
• Requirements for Equilibrium:
• The cog of an object coincides with the com if the object is in a uniform
gravitational field.
• Solutions of Problems:
•Define the system to be analyzed
• Identify the forces acting on the system
• Draw a force diagram
• Write down the equilibrium requirements in components and solve these
for the unknowns
• Elastic Moduli:
tension and compression
shearing
V
pB
hydraulic stress
V
stress  modulus  strain
F
L
E
L
F
x A
G
A
L
Sample Problem 12-2:
• Define the object to be analyzed: firefighter & ladder
• Identify the forces acting on the system:
the gravitational forces: mg & Mg,
the force from the wall: Fw
the force from the pavement: Fpx & Fpy
• Draw a force diagram
• Write down the equilibrium requirements in
components and solve these for the unknowns
balance of forces : Fw  Fpx  0
Fpy  Mg  mg  0
balance of torques : ( 12 a) Mg  ( 13 a)mg  hFw  0
where, a  L2  h 2
Sample Problem 12-3:
• Define the object to be analyzed: Beam
• Identify the forces acting on the system:
the gravitational force (mg),
the force from the rope (Tr)
the force from the cable (Tc), and
the force from the hinge (Fv and Fh)
• Draw a force diagram
• Write down the equilibrium requirements in components and solve these
for the unknowns
Balance of torques:  net , z  aTc  bTr  ( 12 b)( mg )  0
gb( M  12 m)
 Tc 
 6093 N
a
Balance of forces:
Fnet , x  Fh  Tc  0
 Fh  Tc  6093N
Fnet , y  Fv  mg  Tr  0
 Fv  mg  Mg  5047 N
F  Fh2  Fv2  7900 N
Sample Problem 12-6:
• Define the system to be analyzed:
table plus steel cylinder.
• Identify the forces acting on the object:
the gravitational force (Mg),
the forces on legs from the floor (F1= F2=
F3 and F4).
• Draw a force diagram
F1  F2  F3  F4
 3F3  F4
Fg  Mg
• Write down the equilibrium requirements in components and solve these
for the unknowns
Balance of forces:
Fnet  3F3  F4  Mg  0
L3
 F3
 A E L
If table remains level: 
L4
 F4
E

L
 A
L4  L3  d

F4 L F3 L


d
AE AE
F3  550 N
F4  1200 N
Chapter 12: Recitation
Problem-Solving Tips:
• Try to guess the correct direction for each force
• The choice of the origin for the torque equation is
arbitrary. Choose an origin that will simplify your
calculation as much as possible.
• You must have as many independent equations as you
have unknowns in order to obtain a complete solution.
Homework: 12-22
• Define the system to be analyzed: the rod plus the uniform square sign
• Identify the forces acting on the system:
the gravitational force (mg), the force from the cable (Tv and Th), and the force
from the hinge (Fv and Fh)
• Draw a force diagram
• Write down the equilibrium requirements
balance of forces :
Fv  Tv  mg  0
4
Fh  Th  0
blanance of torques :
3Tv  2mg  0
other equations :
Tv 4

Th 3
Tv
Fv
Fh
O
Th
2
3
mg
balance of forces :
Fv  Tv  mg  0
Fh  Th  0
blanance of torques :
3Tv  2mg  0
other equations :
Tv 4

Th 3
Tv  23 mg  326 N
Th  34 Tv  245 N
T  Tv2  Th2  408 N
Fh  Th  245 N (rightward )
Fv  mg  Tv  164 N (upward)
Homework: 12-34
• Define the system to be analyzed: the beam plus the package
• Identify the forces acting on the
system:
the gravitational force (mbg & mpg),
Fv
Tsin
the force from the cable (T),
and the force from the hinge (Fv and Fh)
Fh
Tcos
o
• Draw a force diagram
• Write down the equilibrium requirements
mbg
mpg
balance of torques :
mp g x
mb g
LT sin   Lmb g  xmp g  0  T 

( )
2 sin  sin  L
x
x
From the -T plot :
T  500  200( )
L
L
1
2
other equations :
mb  m p  61.22
sin  
(mb  m p ) g
1200
sin   30o
mb  51.0kg
m p  10.2kg
 0.5
mb g
 1000
sin 
mp g
 200
sin 
mb  m p  61.22
Homework: 12-35
• Define the system to be analyzed: two sides of the ladder are considered separately
• Identify the forces acting on the system:
Left side: the gravitational force of the man (mg), the tension force of the tie rod
(T), the force of the floor on the left feet (FA), and the force exerted by the right
side of the ladder (Fv and Fh)
Right side: the tension force of the tie rod (T), the force of the floor on the right
feet (FE), and the force exerted by the right side of the ladder (Fv and Fh)
Fv
Fh
O
• Draw a force diagram
O
Fh
Fv
mg
T
T
FE
FA


• Write down the equilibrium requirements in components and solve these
for the unknowns
Right side :
FE  Fv  0
left side :
Fv  FA  mg  0
Fh  T  0
T  Fh  0
FA L cos   mg ( L / 4) cos   T ( L / 2) sin   0 FE L cos   T ( L / 2) sin   0
Other equations :
2.44 2  0.762 2
tg 
 3.04
0.762
2.44

0.762
(a) First we solve for T by eliminating the other unknowns. The first
equations of two sides give FA=mg-Fv and FV=FE . Substituting them into the
remaining three equations to obtain:
T  Fh  0
mg (3L / 4) cos   FE L cos   T ( L / 2) sin   0
F E L cos   T ( L / 2) sin   0
T sin 
F E
 (T / 2)tg
The last of these gives
2 cos 
We substitute this expression into the second equation and solve for T.
3mg
The result is:
T
4tg
 211N
T sin  3mg
(b) We now solve for FA. Fv  FE 

2 cos 
8
FA  mg  Fv  85 mg  534 N
(c) We now solve for FE. We have already obtained an expression for FE.
FE 
T sin  3
 8 mg  320 N
2 cos 
Homework: 12-40
(a) , (b). Since the brick rests horizontally on cylinders
now and the cylinders had identical length l before the
brick was placed on them, then both cylinders have
been compressed an equal amount l. Thus,
EA 
FA l
AA l
 FA  E A AA
l
l
EB 
FB l
AB l
 FB  E B AB
l
l
FA E A AA

4
FB E B AB
FA  FB  W
FA  0.8W
FB  0.2W
(c) Computing torques about the center of mass, we
found,
d A FB 1
FA d A  FB d B

dB

FA

4
Homework: 12-43
x
Choose x-axis is parallel to the
incline (positive uphill)
y
N
Balance of forces in x-direction:


mg
T
T cos   mg sin 
sin 
T  mg
 76 N
cos 
Basic Pulley Physics:
Fixed Pulley
2P
Movable Pulley
Homework: 12-47
Cable 3
T=8F
Cable 2
4F
2F
Cable 1
4F 2F F
4 F  2 F  F  mg  F  17 mg  8.96 N
mg
T  8F  71.68 N
Homework: 12-54
2Ff
(a) With F=ma=-kmg, the
magnitude of the deceleration is:
a  k g  3.92 m/s 2
(b), (c) :
2Fr
2ff
blanance of torques :
hf k  ( L  d )2 Fr  d 2 F f  0
mg
balance of forces in y - direction :
2 Fr  2 F f  mg  0
Fr  1964 N
total friction force :
F f  3536 N
f k   k mg
(d), (e) : f r  k Fr  786 N
f f   k F f  1414 N
2fr