Transcript Ch12

Chapter 12: Static Equilibrium
Conditions for equilibrium
 First
condition for equilibrium
• When a particle is in equilibrium (no acceleration), in an inertial
frame of reference, the vector sum of all the forces acting on the
particle is zero.
A frame of reference where Newton’s first law is valid

F  0
F
x
 0,  Fy  0,  Fz  0
The sum includes external forces only.
Conditions for equilibrium (cont’d)
 Second condition for equilibrium
• For an extended body to be in equilibrium the body must have no
tendency to rotate.


L  0 , dL / dt  0
about any point in the body

  0 about any point in the body
• The sum of the torques due to all external forces acting on the
body with respect to any specified point must be zero.
• Note that the above conditions apply to a rigid body in uniform
translational motion or to constant rotation (if L is constant) as well.
But in the following we will mostly consider situations in which a
rigid body is at rest, i.e., in a static equilibrium.
Center of gravity
 The center of gravity (= the center of mass)
• The gravitational torque about any point can be found by assuming
that all the weight of the body acts at its center of gravity
 Definition of the center of mass
• For a collection of particles with masses mi and coordinates
xi , yi ,, zi the coordinates of the center of mass are given
by:
Although the gravity changes with
xcm  imi xi / imi the altitude, if we can ignore this
change, this statement is valid.
ycm
zcm
 
 m y / m
 m z / m
i
i
i
i
i
i
i i
i
i


rcm  imi ri / imi
The center of gravity coincides
with the center of mass when the
gravitational field is uniform.
Center of gravity (cont’d)
 Newton’s third law
• For a collection of particles with masses mi and the center of
mass coordinates xcm , ycm , zcm, Newton’s third law is given
by:
Fx ,net  i f x ,i  imi ax ,i  ax imi  ax M
Fy ,net  i f y ,i  imi a y ,i  a y imi  a y M
Fz ,net  i f z ,i  imi az ,i  az imi  az M


Fnet  Ma
 
ai  a
For a rigid
body
M
Center of gravity (cont’d)
 Total torque

• Assume that the acceleration due to gravity g
has the same
magnitude and direction at every point in the body
• For a collection of particles with masses mi and coordinates
the total torque is given by:

ri
z
y 
rcm
mi

ri
O

cg=cm ?


 
  i i  iri  mi g  (imi ri )  g


rcm  imi  rcm  M


w  Mg


wi  mi g

x
M


 

  rcm  Mg  rcm  w
• If g has the same value at all points on a body, its center of gravity
is identical to its center of mass
How to find and use center of gravity
 How to find cg
• The center of gravity of a homogeneous sphere, cube, circular
sheet, or rectangular plate is at its geometric center.
• The center of gravity of a right circular cylinder or cone is on its
axis of symmetry.
• For a body of more complex shape, the center of gravity can be
located approximately by dividing the body into symmetrical pieces.
• When a body acted on by gravity is supported or suspended at a
single point, the center of gravity is always at, or directly above or
below the point of suspension because there should not be any
torque.
• A body supported at several points must have its center of gravity
somewhere within the area bounded by the supports.
How to find and use center of gravity (cont’d)
 How to use cg
Ex.1
Ex.2
+

Ex.3
stable :
cg is within the area of support
area of support
unstable:
cg is outside the area of support
area of support
Stability
 Classification of stability
U(x)
dU ( x)
0
dx
d 2U ( x)
Stable equilibrium condition:
0
2
dx
dU/dx
Equilibrium condition
:
All positions above are in equilibrium but:
1
Conditionally stable or metastable
(weakly stable)
2
Unstable
3
Conditionally stable or metastable
(strongly stable)
1
3
2
x
Example from Biomechanics
W=Fxd
D = distance covered by
motion
Example for biomechanics (Study Example 12.3)
A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps
muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find
the upward force exerted by the biceps on the forearm and the downward force exerted
by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.
FB
Since the system is in equilibrium, from
the translational equilibrium condition
F  0
O
l
mg
 F  F  F  mg  0
F
From the rotational equilibrium condition   F  0  F  d  mg  l  0
d
x
U
y
B
U
Thus, the force exerted by
the biceps muscle is
Force exerted by the upper arm is
U
B
FB  d  m g l
mg  l 50.0  35 .0

 583 N
FB 
3.00
d
FU  FB  m g  583  50.0  533 N
Examples of objects in equilibrium

Example : Walking a horizontal beam
Examples of objects in equilibrium
Example : Walking a horizontal beam (cont’d)

2nd condition:

i
 0  wB (L / 2)  wM (1.50 m)  TL sin(53)  0
T  413 N
1st condition:
F
F
x
 Rx  T cos35.0  0
y
 R y  wB  wM  T sin 53.0  0
Rx  249 N, Ry  5.70102 N
wB
wM
L T
Exercises
30.0o
(a)
Problem 1
Find the tension T in each cable and
the direction of the force exerted on the strut
by the pivot in each of the arrangements in
Figs. In each case let w be the weight of the
suspended crate full of priceless art objects.
The strut is uniform and also has weight w.
Solutions
(a)
TL tan30.0  wL  w( L / 2)
w
w
T
(b)
L
30.0o
(torque=0)
w
w
45.0o
T  2.60 w
The pivot exerts an upward vertical force of 2w and a horizontal
2.60w so that the magnitude is 3.2w, dir. 37.6o from the horizon.
(b) TL sin 15.0  wL sin 45.0  ( w / 2) L sin 45.0
(torque=0)
T  4.10 w
The horizontal force by the pivot on the strut is Tcos30.0o =3.55w &
the vertical 2w+Tsin30o=4.05w so that the mag. 5.38w and dir. 48.8o.
Problem 2
A uniform ladder of length l and weight mg=50 N rests against a smooth,
vertical wall. If the coefficient of static friction between the ladder and the
ground is ms=0.40, find the minimum angle qmin at which the ladder does
not slip.
First the translational
P
equilibrium, using components
l
FBD
q
n
O
F
F
mg
f
Thus, the normal force is
x
 f P  0
y
  mg  n  0
n  mg  50 N
The maximum static friction force f smax  m s n  0.4  50 N  20 N  P
just before slipping is, therefore,
From the rotational equilibrium

O
l
 mg cos q min  Pl sin q min  0
2
 m g
1  50N 

  tan 
  51
 40N 
 2P 
q min  tan1 
Problem 3
Three vertical forces act on an airplane when it is flying at a constant
altitude and with a constant velocity. These are the weight of the airplane,
aerodynamic force on the wing of the airplane, an aerodynamic force on
the airplane’s horizontal tail. For a particular light airplane with weight of
6,700 N, the center of gravity is 0.30 m in front of the point where the wing’s
vertical aerodynamic force acts and 3.66 m in front of the point where the
tail’s vertical aerodynamic force acts. Determine the magnitude and direction
of each of the two vertical aerodynamic forces.
Fwing
Solution
.3 m
Ftail
F  0 :
  0 :
 Ftail  W  Fwing  0
3.66 m
 (3.66 m)Ftail  (0.3 m)Fwing  0
Fwing  7,300 N (up), Ftail  600 N (down)
w
Problem 4
You are trying to raise a bicycle wheel of mass m
and radius R up over a curb of height h. To do this
you apply a horizontal force F. What is the least
magnitude of the force F that will succeed in raising
the wheel onto the curb when the force is applied
(a) at the center of the wheel?
(b) at the top of the wheel?
R
(c) In which case is less force required?
(a) Perp. distance from mg to the upper corner :
Perp. distance from F to the upper corner :
R 2  ( R  h )2  2 Rh  h 2
Rh
mg 2 Rh  h2  F ( R  h)
F  mg
(b) torque by mg = torque by F
h
mg
Solutions
torque by mg = torque by F

F
2 Rh  h 2
Rh
mg 2Rh  h2  F (2R  h)
2 Rh  h 2
F  mg
2R  h
(c) Less force is required when the force is applied to the top of the wheel.
Problem 5
Max. overhang at L/2
(a) max. overhang?
(b) max. overhang?
max.overhang
Solutions
(a) The center of gravity of the top block can be as far out as the edge
of the lower block. The center of gravity then is at the mid point
between the center of the top and bottom block. The combined center
L/4 3L/4
of gravity is at the edge of the table.
xmg-(L/2-x)mg=0 for bottom block
x=L/4
mg
mg
x
Problem 5
Solutions
Answer to
part (a)
(b) Repeating the argument above the max. overhang for three
stacked block is (3L/4)+(L/6)=11L/12.
xmg-(L/2-x)2mg=0 for bottom block
x=L/3L/2-L/3=L/6
L/4 3L/4
mg
2mg
x