The effect forces have on an object depends on the magnitude,
point of application, and direction of each force. When force is
applied in line with a freely moving object’s center of
gravity, linear motion occurs. When the direction of force is not
in line, a combination of rotary and translatory motion is likely
to occur. This relationship between force application and direction
and the resulting motion are apparent when a book is pushed along
a table. Linear motion occurs when sufficient force is applied in
line with the book’s center of gravity, and a combination
of linear and rotary motion results from a force directed left or
right of center. Similarly, an object with a fixed axis, like a
door or one of the body’s limbs, rotates when the force
is applied off center but does not rotate when the force is in line
with the axis of rotation. A force whose direction is not in line with
the center of gravity of a freely moving object or the center of
rotation of an object with a fixed axis of rotation is called an eccentric force. There must be an
eccentric force for rotation to occur. Some examples of the application
of eccentric force are shown in Figure 13.1.
Examples of the application of eccentric force.
The turning effect of an eccentric force is called torque, or moment of force. The torque
about any point equals the product of the force magnitude and the perpendicular
distance from the line of force to the axis of rotation. The perpendicular
distance between the force vector and the axis is called the moment arm, or torque arm. In this
text, the turning effect will be referred to as torque and the perpendicular distance
as moment arm. An example of torque is illustrated in Figure 13.2.
Neglecting the weight of the arm, the torque or turning effect caused
by a 9-N weight held in the hand with the forearm flexed at the
elbow to a horizontal position is the product of the weight times
its perpendicular distance, or moment arm length, from the elbow.
If the moment arm d is 0.3 meters,
the amount of torque or downward turning force is 3 newton-meters
(N · m). To keep the arm motionless in this position,
the elbow flexors must exert an equal and opposite torque or upward-directed
turning force of 3 N · m.
A weight applied at some distance (d) from the axis (elbow)
Because torque is the product of force and the length of the
moment arm, it may be modified by changing either the force or the
moment arm. Torque may be increased by increasing the magnitude
of the force or by increasing the length of the moment arm. Decreasing
either of these factors will produce a decrease in torque. In the
preceding example, for instance, if the magnitude of the weight
held in the hand were decreased to 6 N, the torque would also decrease,
in this case to 2 N · m. Conversely, increasing the
weight would increase the torque. If, however, the weight is increased
but is moved closer to the axis (the elbow), the torque produced
by the heavier weight would be reduced through reducing the length
of the moment arm.
It is important to emphasize that the moment arm is the perpendicular distance from the direction
of force to the axis of rotation. In the example just given, the
moment arm length was the same as the length of the forearm because
the horizontal forearm was perpendicular to the vertical direction
of the force, and therefore the forearm length was equal to the
perpendicular distance from the force direction to the axis. If
the arm were in some position other than the horizontal, its length
would no longer be equal to the moment arm distance between the force
direction and axis. Suppose the forearm’s position is shifted
from the horizontal to one of 45 degrees of flexion with the horizontal.
Now the force line of the weight is not at right angles to the forearm
(Figure 13.3). This means that the length of the moment arm is no
longer the length of the forearm because, by definition, the moment
arm is the perpendicular distance from the direction or line of
force to the axis of motion. The moment arm now is considerably
shorter. Using trigonometric relations (see Chapter 10), we find
that the moment arm length is 0.2 meters. Consequently, the torque
decreases from 3 N · m to 1.9 N · m,
and the amount of muscular effort needed to counteract this downward
force decreases proportionately.
The torque of the 9-N weight about the elbow joint is
the product of the rotary component of the weight and the distance
from the rotary component to the axis of rotation.
In the human body, the mass or weight of a segment cannot be
altered instantaneously. Therefore, the torque of a segment that
is due to gravitational force can be changed only by changing the length
of the moment arm in relation to the axis. This is done by moving
a body segment so that the force line of the weight is closer to
or farther from the axis. The effect of doing this is quite apparent
in the stages of the sit-up. Compare the gravitational torque on
the trunk when the trunk is just leaving the floor with the torque
when the trunk is at a 30-degree angle from the floor (Figure 13.4).
Gravitational torque d decreases
in going from (a) to (b) as the force line moves closer to the axis.
The torque is 0 at 90 degrees.
Muscle forces also exert torque on rotating segments. The amount
of torque depends on both the magnitude of the muscle force and
the moment arm length. Unlike gravitational torque, each of these
factors can be altered. The moment arm length depends on the point
of insertion of the muscle and the position of the body segment
at any point in a given motion. Figure 13.5 shows how the moment
arm of the biceps changes for every change of arm position during
elbow flexion. The magnitude of the force of the muscle contributing
to the torque changes also as its length, tension, and angle of
The biceps at various points of elbow flexion, showing
variations in the moment arm. From B. LeVeau, Williams
and Lissner: Biomechanics of Human Motion, 2nd ed. Copyright © 1977
W. B. Saunders, Orlando, FL. Reprinted by permission.
With changes in muscle angle of pull, the magnitude of the force
effectively producing torque also changes. When muscle force vectors
are resolved into their two components, it can be seen that only
the rotary component—that which is perpendicular to the
mechanical axis of bone—is actually a factor in torque
production. The stabilizing component of the muscle force acts along the
mechanical axis of the bone through the axis of rotation; thus this
component is not an eccentric, or off-center, force and cannot contribute
to torque. It has a moment arm length equal to zero. Once a muscle
force is resolved into a rotary and a stabilizing component, the
moment arm is the distance between the axis of rotation and the
point of application of the rotary component of the muscle force.
Because the rotary component of any known force can be calculated
trigonometrically, given the angle between the line of force and
the mechanical axis, torque can be calculated as the product of
the rotary component of the force (Fy)
times the moment arm along the lever (d)
(see Figure 13.9b). In the example used earlier, the angle between
the line of force for the weight and the mechanical axis of the
bone is 45 degrees. Using the angle and the 9-N magnitude of the force,
the rotary component of the force is calculated to be 6.4 N. When
multiplied by the 0.3-meter length of the moment arm, the torque
is 1.9 N · m, the same value found when calculating
the length of the true moment arm in Figure 13.3. Either method
of finding torque is satisfactory.
Summation of torques using trigonometric functions. The
amount of muscle force needed to counteract the force of gravity
in this example is 933.3 N.
The important thing to remember from this discussion is that
the turning effect, or torque, produced on objects free to rotate
depends on two factors: the magnitude of the applied force and the perpendicular
distance from the force vector to the axis. Change in turning effect
occurs when either of these factors is altered. Shortening the moment
arm distance or decreasing the force magnitude will decrease the
torque. Lengthening the moment arm or increasing the force magnitude
will increase the turning effect and consequently the necessary
effort to resist it.
The sum of two or more torques may result in no motion, linear
motion, or rotary motion. When equal parallel eccentric forces are
applied in the same direction on opposite sides of the center of rotation
of an object, either no motion or linear motion will occur. An example
of no motion is two children balanced on a seesaw. When the equal
parallel forces are adequate to overcome the resistance of the object,
linear motion will occur. This could be the situation with paddlers
in a canoe, with one paddling on the port side and the other on
the starboard side. When equal and opposite parallel forces are
exerted on opposite sides of the axis of rotation, rotary motion
will occur. The effect of equal parallel forces acting in opposite
directions is called a couple, or force couple. An example of this type
of movement is steering a car when both hands are used on opposite sides
of the wheel (Figure 13.6). Another example is that of turning in
a rowboat by simultaneously pulling on the handle of one oar while
pushing on the handle of the other.
When one steers with two hands, the hands act as a force
There are many examples in the human body in which two muscles
rotate a bone by acting cooperatively as a force couple; for instance,
trapezius II and the lower portion of serratus anterior are a force
couple, rotating the scapula upward. Trapezius II and IV, acting
on the two extremities of the spine of the scapula, similarly serve
as a force couple to help rotate the scapula upward, as do the oblique
abdominals in acting to rotate the trunk (Figure 13.7).
(a) Two force couples acting on the scapula to rotate
it upward. Trapezius II and lower serratus anterior are an excellent
example. Trapezius II and the lower fibers of IV also tend to act
as a force couple although their pulls are not in opposite directions.
(b) A cross section of the trunk in which the oblique abdominal
muscles act as a force couple to rotate the trunk.
It is common for more than one torque to act on a body at any
given time. The resultant effect of these torques is expressed in
the principle of the summation of torques, which states that the resultant torques of a force system must
be equal to the sum of the torques of the individual forces of the
system about the same point. Because torques are vector quantities,
the summation must consider both magnitude and direction. The direction
of rotation is expressed as either clockwise or counterclockwise
direction. Clockwise torques are usually labeled as negative, and
counterclockwise torques as positive. Their signs must be accounted
for when they are summed.
When the sum of counterclockwise torques equals the sum of clockwise
torques, no turning will occur. This idea may also be expressed
by stating that rotation will be absent when the sum of the torques
of all forces about any point or axis equals zero. When the sum
of clockwise torques does not equal the sum of the counterclockwise
torques, the resultant turning effect (torque) will be the difference
between the two opposing forces and in the direction of the larger.
In Figure 13.8 all the forces are applied perpendicular to the
lever. Force A is 1.5 m from the axis
and B is 3 m away. Both A and B are
clockwise forces. Force C is a counterclockwise
force that also is 3 m away from the axis. The sum of these torques
produces a resultanttorque of 22.5 N · m in a clockwise
direction about the center of rotation.
Summation of moments. The resultant moment about a point
is equal to the sum of the moments of the individual forces about
the same point. Clockwise moments are negative, and counterclockwise
moments are positive. The sum of the moments about the fulcrum on
this diagram is 22.5 Newton-meters in a clockwise direction.
When an eccentric force is applied at an angle to the lever other
than 90 degrees, the moment arm is not along the lever line. Before
the torque can be found, the length of the moment arm must be determined.
This can be done using trigonometric functions. An example of this
procedure is shown in the following problem:
What muscular force F pulling
at an angle of 25 degrees would be required to keep the abducted
arm in a position of 20 degrees with the horizontal? The muscle
inserts 10 cm from the shoulder joint. The arm weighs 50 N and its
center of gravity is located 30 cm from the shoulder. A 45-N weight
is held in the hand 60 cm from the shoulder joint. Note: For the arm to be held stationary,
the sum of the counterclockwise torques must equal the clockwise
torques (∑CCW = ∑CW).
(See Figure 13.9 for solution.)