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Chapter 1: Static Forces
“Mechanics” is a branch of physics which concerns with the effect of forces.
Force
m : mass

a : accelerati on


F  ma
Dimensions:
m

[m]  kg; [a ] 
sec 2
Fundamental quantities in physics:
1. length
2. mass
3. time
Aristotle (384 ~ 322 BC)
“The animal that moves makes its change of position by pressing against that
which is beneath it…Runners run faster if they swing their arms for in
extension of the arms there is a kind leaning upon the hands and the wrist.”
“The animal that moves makes its change of position by pressing against that
which is beneath it…Runners run faster if they swing their arms for in
extension of the arms there is a kind leaning upon the hands and the wrist.”
1. The law of physics are expressed in terms basic quantities.
2. In mechanics, the three basic quantities are length (L), mass (M), and
time (T). All other quantities in mechanics can be expressed in terms of these
three.
3. An international committee has agreed on a system of definitions and
standards to describe fundamental physical quantities.
4. It is called the SI system (Système International) of units.
Length
Length
Length: meter (m)
Mass: kilogram (kg)
Time: second (s)
1. In A.D. 1120. King Henry I of England decreed that the standard of length in
his country would be the yard and that the yard would be precisely equal to
the distance from the tip of his nose to the end of his outstretched arm.
2. Similarly, the original standard for the foot adopted by the French was the
length of the royal foot of King Louis XIV. This standard prevailed until 1799.
3. In 1799 the legal standard of length in France became the meter, defined as
one ten-millionth of the distance from the equator to the North Pole.
4. In 1960, the length of the meter was defined as the distance between two
lines on a specific bar of platinum-iridium alloy stored under controlled
condition.
5. With the strong demand of scientific accuracy, the definition of the meter
was modified to be equal to 1 650 763.73 wavelengths of orange-red light
emitted from a kryton-86 lamp.
6. In October 1983, the meter was redefined to be the distance traveled by light
in a vacuum during a time interval of 1/299 792 458 second.
1. The SI unit of mass, the kilogram, is
defined as the mass of a specific
platinum-iridium alloy cylinder kept at
the International Bureau of Weights and
Measures at Sèvres, France.
2. The cylinder is 39 cm in height and in
diameter.
3. The mass of a carbon-12 atom is taken
to be 12 atomic mass units (12 u).
1u = 1.660 540 2 (± 0.000 001 0) × 10-27 kg
Table 1.2, p. 5
1. Before 1960, the standard of time was defined in terms of the average length of
a solar day in the year 1900.
2. The basic unit of time, the second, was defined to be 1/86 400 of the average
solar day.
3. In 1967, the second was redefined to take advantage of the great precision with
an atomic clock. The clock will neither gain nor lose a second in 20 million
years.
4. The second is now defined as 9 192 631 770 times the period of oscillation of
radiation from the cesium atom.
Cesium fountain atomic clock
Table 1.3, p. 6
Torque
Equilibrium and Stability
A body is in equilibrium if the vector sum of the
forces and the torques acting on the body is zero.

 Fi  0
i

 i  0
i
The position of the center of mass with respect to
the base of support determines whether the body
is stable or not.
The wider the base on which the body rests, the
more stable it is; that is the more difficult it is to
topple it.
In Fig. 1.2a the torque produced by its weight
tends to restore it to its original position.
In Fig. 1.2b the same amount of angular
displacement of a narrow-based body results in a
torque that will topple it.
Equilibrium Considerations for the Human Body
The center of gravity of an erect person with arms at
the side is at approximately 56% of the person’s
height measured from the soles of the feet. The
center of gravity shifts as the person moves and
bends. The act of balance requires maintenance of
the center of gravity above the feet.
56%
When carrying an uneven
load, the body tends to
compensate by bending and
extending the limbs so as to
shift the center of gravity
back over the feet.
Stability of the Human Body under the Action of an External Force
The counterclockwise torque Ta about the point A
produced by the applied force Fa is
Ta  Fa  1.5 m
The opposite restoring torque Tw due to the person’s
weight is
Tw  W  0.1 m
Assuming that the mass of the person is 70 kg,
his weight W is
W  mg  70  9.8  686 newton (N)
The restoring torque produced by the weight is therefore
68.6 newton-meter (N-m). The person is on the verge of
toppling when the magnitudes of these two torques are
just equal; that is Ta = Tw or
Fa  1.5 m  68.6 N - m
Therefore, the torque required to topple an erect person is
Fa 
68.6
 45.7 N
1.5
By bending the torso the center of
gravity will be shifted away from the
point A and as the result will the
restoring torque be increased.
One can also increase the stability against
a topple force by spreading the legs.
Skeletal Muscles
The skeletal muscles producing skeletal movements
consist of many thousands of parallel fibers wrapped
in a flexible sheath that narrows at both ends into
tendons. The tendons, which are made of strong tissue,
grow into the bone and attach the muscle to the bone.
1. Most muscles taper to a single tendon. But some
muscles end in two or three tendons; these
muscles are called, respectively, biceps and
triceps.
2. Each end of the muscle is attached to a different
bone.
3. In general, the two bones attached by muscles
are free to move with respect to each other at the
joints where they contact each other.
There is a great variability in the pulling force that a given
muscle can apply. The force of contraction at any time is
determined by the number of individual fibers that are
contracting within the muscle.
When an individual fiber receives an electric stimulus, it
tends to contract to its full ability.
If a stronger pulling force is required, a larger number of
fibers are stimulated to contract.
Experiments have shown that the maximum force a
muscle is capable of exerting is proportional to its
cross section.
From measurements, it has been estimated that a
muscle can exert a force of about 7 × 106 dyn/cm2
of its area.
7 × 106 dyn/cm2 = 7 × 105 Pa
1 atm = 1.013 × 105 Pa
Levers
To compute the forces exerted by muscles, the
various joints in the body can be conveniently
analyzed in terms of levers.
Such a representation implies some simplified
assumptions.
We will assume that the tendons are connected to
the bones at well-defined points and that the
joints are frictionless.
A lever is a rigid bar free to rotate
about a fixed point called the fulcrum.
GIVE ME A PLACE TO STAND AND I WILL MOVE THE EARTH
A remark of Archimedes quoted by Pappus of Alexandria
Collection or Synagoge, Book VIII, c. AD 340
There are three classes of levers.
In a Class 1 lever, the fulcrum is located between the applied force and the load.
In a Class 2 lever, the fulcrum is at one end of the bar; the force is applied to the other end;
and the load is situated in between.
A Class 3 lever has the fulcrum at one end and the load at the other. The force is applied
between the two ends.
Many of the limb movements of animals are performed by Class 3 levers.
For all three types of levers, the force F required to balance a load of weight W is given by
F
Wd1
d2
The mechanical advantage M of the lever is defined as
M
W d2

F d1
M (Class 1): > or < 1
M (Class 2): > 1
M (Class 3): < 1
A force slightly greater than what is required to
balance the load will lift it. As the point at which the
force is applied moves through a distance L2, the
load moves a distance L1.
L1 d1

L2 d 2
1

M
v1 d1

v2 d 2
These relationships apply to all three classes of levers.
It is evident that the excursion and velocity of the load are inversely
proportional to the mechanical advantage.
The Elbow
This is a Class 3 lever.
Model of the Elbow
The direction of the reaction force Fr shown
is a guess. The exact answer will be
provided by the calculations.
x component of the forces
Fm cos   Fr cos 
y component of the forces
Fm sin   W  Fr sin 
The torque about the fulcrum must be zero
Fm sin   10 W
4 cm  Fm sin   40 cm  W
With θ = 72.6◦, the muscle force Fm is
With a 14 kg weight in hand,
Fm 
10 W
10 W

 10.5 W

sin 72.6
0.954
Fm  10.5 14  9.8  1440 N
Model of the Elbow
The direction of the reaction force Fr shown
is a guess. The exact answer will be
provided by the calculations.
Now we are in a position to evaluate Fr and ϕ.
1440  cos 72.6  Fr cos 
1440  sin 72.6  14  9.8  Fr sin 
Fr cos   430 N
Fr  1320 N
Fr sin   1240 N
  70.9
or
The Hip
This figure shows the hip joint and its
simplified lever representation,
giving dimensions that are typical for
a male body.
The hip is stabilized in its socket by a
group of muscles, which is
represented in figure b as a single
resultant force Fm.
When a person stands erect, the angle
of this force is about 71◦with respect
to the horizon.
WL represents the combined weight
of the leg, foot, and thigh. Typically,
this weight is a fraction (0.185) of the
total body weight W (WL = 0.185W).
The Hip
θ
We will now calculate the magnitude of the muscle force
Fm and the force FR at the hip joint when the person is
standing erect on one foot as in a slow walk.
The force W acting on the bottom of the lever is the
reaction force of the ground on the foot of the person. This
is the force that supports the weight of the body.
Fm cos 71  FR cos   0
(x components 0f the
force = 0)
Fm sin 71  W  WL  FR sin   0
(y components 0f the
force = 0)
( FR sin )  7 cm  WL  10 cm
 W  18 cm  0
FR sin   2.31W
Since WL = 0.185 W,
Fm  1.59 W
(torque about point A = 0)
  77.3
FR  2.37 W
For a person with mass 70 kg the weight is 686 N and the force
on the hip join is 1625 N.
Limping
Persons who have an injured hip limp by
leaning toward the injured side as they step on
that foot.
Injured Case
As a result, the center of gravity of the body
shifts into a position more directly above the
hip joint, decreasing the force on the injured
area.
Calculations for the case show that the muscle
force Fm = 0.47W ( as opposed to the normal
case of Fm = 1.59W) and that the force on the
hip joint is FR = 1.28W (as opposed to the
normal case of FR = 2.37W)
Normal Case
“24And Jacob was left alone, and a man
wrestled with him until the break of the
dawn. 25And when the man saw that He
did not prevail against him, He touched
the socket of his hip; and the socket of
Jacob’s hip was dislocated as he wrestled
with Him. 26And the man said, Let Me go,
for the dawn is breaking. But Jacob said, I
will not let You go unless You bless me.”
(Genesis 32: 24 ~ 26)
Cervical vertebrae
Thoracic vertebrae
Lumbar vertebrae
Pivot
Sacrum
Vertebral column
The Back
When the trunk is bent forward, the spine
pivots mainly on the fifth lumbar vertebra.
the fifth lumbar
vertebra.
A
We will analyze the forces involved when
the trunk is bent at 60˚ from the vertical with
the arms hanging freely.
The pivot point A is the fifth lumbar vertebra.
The lever arm AB represents the back.
The weight of the trunk W1 is uniformly
distributed along the back; its effect can be
represented by a weight suspended in the
middle.
The weight of the head and arms is
represented by W2 suspended at the end of
the lever arm.
The rector spinalis muscle, shown as the
connection D-C attached at a point two-third
up the spine, maintains the position of the
back.
The Back
the fifth lumbar
vertebra.
For a 70-kg man, W1 and W2 are
typically 320 N and 160 N,
respectively.
A
W1  320 N
W2  160 N
To hold up the body weight, the
muscle must exert a force of
2000 N and the compressional
force of the fifth lumbar vertebra
is 2230 N.
This example indicates that large
forces are exerted on the fifth
lumbar vertebra.
It is not surprising that backaches originate most
frequently at this point, the fifth lumbar vertebra.
When standing on tiptoe, the total weight of
the body is supported by the reaction force at
point A.
This is a Class 1 lever with the fulcrum at the
contact of the tibia.
The balancing force is provided by the muscle
connected to the heel by the Achilles tendon.
Calculations show that while standing tiptoe
on one foot the compressional force on the
tibia is 3.5 W and the tension force on the
Achilles tendon is 2.5 W.
Standing on tiptoe is a fairly strenuous position.
Ballet Dancing