Transcript No Slide Title

Chapter 11
Statics
In this chapter we will study the conditions under
which a rigid body does not move. This part of
Mechanics is called “Statics” . Even though it is
rather restricted in scope it is nevertheless a very
important topic Engineers. After all, nobody
appreciates buildings, bridges, or other structures that
move.
(11-1)
Conditions under which a rigid body does not move
1. The acceleration Acm of the center of mass is zero
Acm = 0 
dVcm/dt = 0  Velocity of the
center of mass Vcm is constant. We consider the
special case: Vcm = 0
2. The angular acceleration cm of the center of mass is zero
 dcm/dt = 0  the angular velocity cm of the
center of mass is constant. We consider the special
case:
cm= 0
(11-2)
F2
F1
MAcm = Fnet .
FN
Equilibrium Condition I Consider a
rigid body on which N external forces
F1, F2 , …, FN are applied. The net
external force Fnet = F1 + F2 + …+ FN
Newton’s second law connects Fnet
with the acceleration of the center of
mass Acm :
In Statics Acm = 0 
Fnet = 0
The net external force must be zero
This vector equation can be broken down into three separate equations,
one for each coordinate axis:
Fxnet = 0 or
F1x + F2x + … + FNx = 0
Fynet = 0 or
F1y + F2y + … + FNy = 0
Fznet = 0 or
F1z + F2z + … + FNz = 0
(11-3)
F2
FN
F1
Newton’s second law connects net
with the acceleration of the center of
mass cm :
r1
Iocm = net .
Equilibrium Condition II
The net external torque on the rigid
body of the figure is:
net = 1 + 2 + …+ N
1 = R1F1, 2 = R2F2, ...
In Statics cm = 0 
net = 0
The net torque by all external forces must vanish
net = 0
(11-4)
In statics the value of net does
not depend on the choice of the
origin
So how do we go about choosing the origin O for a statics
problem? The answer is that since the choice of O does not
affect the result, as a practical matter we choose the origin so
as to make the equations simpler. This will become clearer
after we work on a few examples.
(11-5)
Example (11-1)
page 301
Consider figure (b) which shows the book resting on the table
so that the book’s center of mass lies above the table. In this
case FN -mg = 0 because Fynet must be zero. We take the
origin to be the edge of the table because when the book starts
to fall it will rotate about this point. The net torque net about
O is also zero and the book is at rest. If we perturb slightly the
book by lifting it up, FN = 0 and the torque of mg brings it
back
(11-6)
Consider figure (c) which shows the book resting on the table
so that the book’s center of mass lies beyond the table. In
this case FN -mg = 0 because Fynet must be zero. The net
torque net in this case about O is provided my mg only and it
is not zero. Thus the second equilibrium condition is not
satisfied and the book will rotate about point O and fall off the
table
(11-7)
.
. .
A
CM
.
A
B
mg
mg
.
.
B
CM
B = 0
A = 0
In statics the extension of the gravitational force must pass
through the point of suspension. This is indicated in the
two figures. The flat object is suspended first from point A
and then from point B. In each case the extension of the
gravitational force passes though the point of suspension.
In each case the torque of mg is zero and the second
equilibrium condition is satisfied.
(11-8)
.
A
A
 r
mg
.
. .
.
CM
Fig.a
CM
B
mg
Fig.b
What happens if we
perturb the object from
its equilibrium
condition as in fig.a?
A = 0
In this case the torque of mg is not zero any more.
Instead  = -mgrsin  0 Under the action of mg the object
will rotate about A in the clockwise direction till it reaches the
position of fig.b. In this position, the torque is zero and the
object can be at equilibrium
(11-9)
A
B
CM
C
CM
Practical method for the determination of the center of mass of a flat
object with a complicated shape (such as the map of the US). Suspend
the object from a point A (other than the CM). Mark the position of the
plumb line. Repeat the process for a second point B. The intersection of
the two plumb lines gives the CM. If the process is repeated for a third
point C the plumb line passes also from the Center of mass.
(11-10)
Statics Problem Recipe
1.
Draw a force diagram
2.
Choose a convenient origin O. A good choice is to
have one of the unknown forces acting at O
3.
Sign of the torque  for each force:
+ If the force induces clockwise (CW) rotation
- If the force induces counter-clockwise (CCW)
rotation
4.
Equilibrium conditions:
Fxnet = 0 ,
Fynet = 0 ,
5.
Make sure that:
numbers of unknowns = number of equations
O = 0
(11-11)
Example (11-2) page 303
Find the forces FA and FB on the
man’s hands if:
m = 5 kg, L = 3 m,  = 0.75 m
y
Fynet = FA + FB - mg = 0
eqs.1
net = FB - Lmg = 0
eqs.2
We solve eqs.2 for FB : 
FB = Lmg/ = 35 9.8/0.75 = 196 N
CCW
CW
We solve eqs.1 for FA: FA = mg - FB= 59.8 - 196 = -147 N
Note: The direction of FA is opposite to that shown in the
figure
(11-12)
Example (11-3) page 304
(11-13)
A crane whose cabin and
engine are fixed to earth is
used to lift a mass m =
5300 kg. The arm of the
crane is supported at its
base (point B), by a
friction free pivot, and at
its top (point A) by a cable.
The arm of the cable makes angles arm = 45 and
T = 32, respectively, with the horizontal and the arm has
a length L = 10 m. The mass m is lifted from a point on
the arm at a distance  = 0.52 m from point A. Ignore the
mass of the arm. Determine the tension T of the cable.
net = -mgd2 + Td1 = 0 
CCW
E

T = mg(d2/d1)
eqs.1
From triangle ABC:
d1 = Lsin eqs.2
CW

Angle  = arm - T
 = 45 -32 = 13
C
d1
D
d2
From triangle BED: 
d2 = (L - )cosarm eqs.3
We substitute d1 from eqs.2 and d2 from eqs.3 into eqs.1 
T = mg (L - )cosarm / Lsin = 1.55105 N
(11-14)
Fxnet = FNx - TcosT = 0 
FNx = TcosT
FNx = 1.55105 cos32
FNx = 1.31105 N
Fynet = FNy -mg - TsinT = 0  FNy = mg + TsinT
FNy = 53009.8 + 1.55105sin32 = 1.34105 N
FN = [FNx2 + FNy2 ]1/2 = 1.87105 N
(11-15)
Example (11-4) page 306
A ladder of length L = 3 m is
leaning against the wall making an
angle  = 58  with the horizontal.
The mass of the ladder is
insignificant compared to the mass
m = 85 kg of a window washer
who is climbing the ladder.
The coefficient of static friction between the ladder feet and
the floor s = 0.51. There is no friction between the top of
the ladder and the wall. Determine the maximum distance d
the man can climb up the ladder so that the ladder starts
slipping
(11-16)
(11-17)
CW
d2
d1
CCW
System = ladder + window washer
Fxnet = Fwall – f = 0

Fwall = f
eqs.1
Fynet = Ffloor – mg = 0 
Ffloor = mg eqs.2
f = sFfloor = smg
f = smg eqs.3
net = -Fwalld2 + mgd1 = 0
d1 = dcos and d2 = Lsin 
-Fwall Lsin +mg dcos = 0  d = Fwall Lsin/mgcos eqs.4
(11-18)
d = Fwall Lsin/mgcos
From eqs. 1 
Fwall = f
eqs.4
From eqs.3  f = smg

Fwall = smg We substitute Fwall into eqs.4 
d = smgLsin/mgcos = sLtan = 0.513tan(58) = 2.4 m
(11-19)
y
CCW
O
x
Example (11-5) page 307
The biceps muscle bends
the arm. Lengths a = 30
cm and x = 4 cm. If a
mass M is held in your
hand with the forearm
horizontal and the upper
arm vertical, determine
the upward force FB the
biceps exerts on the
forearm bones
CW
net = xFB – aMg = 0  FB = (a/x)Mg = (30/4)Mg = 7.5Mg
The force FB exerted by the muscle is 7.5 times greater than
the weight itself!