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(c) Determine a · b, b · a, a — b and b — a.
(d) Determine |a|, |b|, |a — b| and |b — a|.
(e) Determine the smallest angle between a and b.
(f) Determine a unit normal vector on the plane de¬ned by a and b.
(g) Determine a — b · c and a — c · b.
(h) Determine ab · c, ( ab)T ·c and ba · c.
(i) Do the vectors a, b and c form a suitable vector basis? If the answer
is yes, do they form an orthogonal basis? If the answer is yes, do they
form an orthonormal basis?
Consider the basis {a, b, c} with a, b and c de¬ned as in the previous
exercise. The following vectors are given: d = a + 2b and e = 2a ’ 3c.
(a) Determine d + e.
(b) Determine d · e.
The basis {ex , ey , ez } is right-handed and orthonormal. The vectors ax , ay
and az are given by: ax = 4ex + 3ey ; ay = 3ex ’ 4ey and az = ax — ay .
(a) Determine az expressed in ex , ey and ez .
(b) Determine |ai | for i = x, y, z.
(c) Determine the volume of the parallelepiped de¬ned by ax , ay and az .
(d) Determine the angle between the lines of action of ax and ay .
(e) Determine the vector ± x from ai = |ai |± i for i = x, y, z. Is {± x , ± y , ± z }
a right-handed, orthonormal vector basis?
9 Exercises

Consider the vector b = 2ex + 3ey + ez . Determine the column rep-
resentation of b according to the bases {ex , ey , ez }, {ax , ay , az } and
{± x , ± y , ± z }.
(g) Show that: ax — ay · b = ax · ay — b = ay · b — ax .
Assume {ex , ey , ez } is an orthonormal vector basis. The following vectors
are de¬ned:
a = 4ex + 3ey ’ ez
b = 6ey ’ ez
c = 8ex ’ ez .
Are a, b and c linearly independent? If not, what is the relationship between
the vectors?
The vector bases {ex , ey , ez } and { x , y , z } are orthonormal and do not
(a) What is the effect of ex x + ey y + ez z acting on a vector a?
(b) What is the effect of x ex + y ey + z ez acting on a vector a?
The vector basis {ex , ey , ez } is orthonormal. What is the effect of the
following dyadic products if they are applied to a vector a?
(a) ex ex .
(b) ex ex + ey ey .
(c) ex ex + ey ey + ez ez .
(d) ex ey ’ ey ex + ez ez .
(e) ex ex ’ ey ey + ez ez .
2 The concepts of force and moment

2.1 Introduction

We experience the effects of force in everyday life and have an intuitive notion
of force. For example, we exert a force on our body when we lift or push an
object while we continuously (fortunately) feel the effect of gravitational forces,
for instance while sitting, walking, etc. All parts of the human body in one way
or the other are loaded by forces. Our bones provide rigidity to the body and can
sustain high loads. The skin is resistant to force, simply pull on the skin to witness
this. The cardiovascular system is continuously loaded dynamically due to the
pulsating blood pressure. The bladder is loaded and stretched when it ¬lls up. The
intervertebral discs serve as ¬‚exible force transmitting media that give the spine its
¬‚exibility. Beside force we are using levers all the time in our daily life to increase
the ˜force™ that we want to apply to some object, for example by opening doors
with the latch, opening a bottle with a bottle-opener. We feel the effect of a lever
arm when holding a weight close to our body instead of using a stretched arm.
These experiences are the result of the moment that can be exerted by a force.
Understanding the impact of force and moment on the human body requires us to
formalize the intuitive notion of force and moment. That is the objective of this

2.2 Definition of a force vector

Imagine pulling on a thin wire that is attached to a wall. The pulling force exerted
on the point of application is a vector with a physical meaning, it has
• a length: the magnitude of the pulling force
• an orientation in space: the direction of the wire
• a line-of-action, which is the line through the force vector.
The graphical representation of a force vector, denoted by F, is given in Fig. 2.1.
The ˜shaft™ of the arrow indicates the orientation in space of the force vector. The
point of application of the force vector is denoted by the point P.
11 2.2 Definition of a force vector

f actio
line o

Figure 2.1
The force vector F and unit vector e.

F = ’|F |e2
F = |F |e1


Figure 2.2
Force vector F written with respect to e1 and written with respect to e2 .

The magnitude of the force vector is denoted by |F|. If e denotes a unit vector
the force vector may be written as
F = Fe, (2.1)
where F may be any rational number (i.e. negative, zero or positive). The absolute
value |F| of the number F is equal to the magnitude of force vector:
|F| = |F|. (2.2)
In Fig. 2.2 the force vector F is written with respect to either the unit vector e1 or
with respect to the unit vector e2 that has the same working line in space as e1 but
the opposite direction. Since the unit vector e1 has the same direction as the force
vector F:
F = |F|e1 . (2.3)
In contrast, the unit vector e2 has a direction that is opposed to F, therefore
F = ’|F|e2 . (2.4)

Example 2.1 Let the force vector F be given by
F = 2e1 .
If the unit vectors e1 and e2 have opposite direction:
e2 = ’e1 ,
then the force vector may also be written as
F = ’2e2 .
The concepts of force and moment

2.3 Newton™s Laws

The concepts in this biomechanics textbook are based on the work of Sir Isaac
Newton (1643“1727). In his most famous work ˜Philosophiae Naturalis Principia
Mathematica™ he described the law of gravity and what are currently known as the
three laws of Newton, forming the basis for classical mechanics. These laws are:
• Every object in a state of uniform motion tends to remain in that state of motion unless
an external force is applied to it. This is often termed simply the ˜Law of Inertia™.
• In a one-dimensional context the second law states that the force F on an object equals
the mass m, with SI unit [kg], of the object multiplied by the acceleration a, with
dimension [m s’2 ], of the object:

F = ma. (2.5)

Consequently, the force F has the dimension [N] (Newton), with

1 [ N] = 1 [kg m s’2 ].

This may be generalized to the three-dimensional space in a straightforward manner.
Let the position of a material particle in space be given by the vector x. If the particle
moves in space, this vector will be a function of the time t, i.e.

x = x( t) . (2.6)

The velocity v of the particle is given by
v( t) = , (2.7)
and the acceleration a follows from
d2 x
a( t) = = 2. (2.8)
dt dt
Newton™s second law may now be formulated as

F = ma. (2.9)

• The third law states that for every action there is an equal and opposite reaction.
This law is exempli¬ed by what happens when we step off a boat onto the bank of
a lake: if we move in the direction of the shore, the boat tends to move in the opposite

Example 2.2 Let the position of a particle with mass m for t ≥ 0 be given by
x( t) = 1 + x0 ,

13 2.4 Vector operations on the force vector

where x0 denotes the position of the particle at t = 0 and „ is a constant,
characteristic time. The velocity of this particle is obtained from
d( 1+( t/„ )2 )
dx d
v= = ( 1+( t/„ ) ) x0 = x0 = ( 2t/„ 2 ) x0 ,
dt dt dt
while the acceleration follows from
a= = ( 2/„ 2 ) x0 .
The force on this particle equals
F = ( 2m/„ 2 ) x0 .

2.4 Vector operations on the force vector

Suppose that a force vector is represented by
F 1 = F1 e, (2.10)
then another force vector, say F 2 may be obtained by multiplying the force by a
factor ±, see Fig. 2.3(a):
F 2 = ±F1 e = F2 e . (2.11)
The force vector F 2 has the same orientation in space as F 1 , but if ± = 1 it will
have a different length, and it may have a direction sense (if ± < 0).
The net result of two force vectors, say F 1 and F 2 , acting on the same point P
is obtained by the vector sum, graphically represented in Fig. 2.3(b):
F3 = F1 + F2 . (2.12)
The vector F 3 is placed along the diagonal of the parallelogram formed by the
vectors F 1 and F 2 . This implicitly de¬nes the orientation, sense and magnitude of
the resulting force vector F 3 .

F3 = F 1 + F 2
F2 = ±F 1


(a) F2 = ±F 1 (b) F3 = F 1 + F 2

Figure 2.3
Graphical representation of the scalar multiplication of a force vector with ± < 0 (a) and the sum
of two force vectors (b).
The concepts of force and moment

Clearly, if two force vectors F 1 and F 2 are parallel, then the resulting force
vector F 3 = F 1 + F 2 will be parallel to the vectors F 1 and F 2 as well. If F 1 = ’ F 2 ,
then the addition of these two force vectors yields the so-called zero vector 0,
having zero length.

2.5 Force decomposition

Suppose that a bone is loaded with a force F as sketched in Fig. 2.4. The principal
axis of the bone has a direction indicated by the unit vector e. The smallest angle
between the force vector F and the unit vector e is denoted by ±. It is useful to
know, which part of the force F acts in the direction of the unit vector e, indicated
by F t and which part of the force acts perpendicular to the bone, indicated by the
force vector F n . The force vector F may, in that case, be written as

F = Ft + Fn . (2.13)

To determine the vectors F t and F n , vector calculus will be used. The inner product
of two vectors, say a and b, is de¬ned as

a · b = |a| |b| cos( ±) , (2.14)

where ± is the smallest angle between the two vectors a and b, see Fig. 2.5 and
Chapter 1 for further details on the properties of the vector inner product. Com-
putation of the inner product requires knowledge of the length of both vectors


± e


Figure 2.4


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