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Bone loaded by the force vector F. The orientation of the bone is indicated by the unit vector e.



b
±
a

Figure 2.5
De¬nition of the angle ±.
15 2.5 Force decomposition

(i.e. |a| and |b|) and the smallest angle between the two vectors (i.e. ±), all physi-
cal quantities that can easily be obtained. If the vectors a and b are perpendicular
to each other, hence if ± = π/2, then√ inner product equals zero, i.e. a · b = 0.
the
The length of a vector satis¬es |a| = a · a.
Now, consider the inner product of an arbitrary vector b with a unit vector e (i.e.
|e| = 1), then

b · e = |b| cos( ±) . (2.15)

Let the vector b be written as the sum of a vector parallel to e, say bt , and a vector
normal to e, say bn , such that:

b = b t + bn , (2.16)

as depicted in Fig. 2.6(a).
If the angle ± between the unit vector e and the vector b is acute, hence if
± ¤ π/2, it is easy to show that this inner product is equal to the length of the
vector bt , the component of b parallel to the unit vector e, see Fig. 2.6(a). By
de¬nition

|b t |
cos( ±) = . (2.17)
|b|
However, from Eq. (2.15) we know that

b·e
cos( ±) = , (2.18)
|b|
hence

|bt | = b · e. (2.19)



bn
b
b
bn
±
± e
e
bt bt
|bt| |bt|


(a) Acute angle ±, the length of b t (b) Obtuse angle ±, the length of b t

Figure 2.6
Vector decomposition in case of (a) an acute and (b) an obtuse angle between the vectors.
The concepts of force and moment
16

Since the angle ± is acute, the vector bt has the same sense as the unit vector e
such that:

bt = |bt |e = ( b · e) e . (2.20)

If the angle ± is obtuse, see Fig. 2.6(b), hence if ± > π/2, we have

|b t |
cos( π ’ ±) = ’ cos( ±) = . (2.21)
|b|

With, according to Eq. (2.15), cos( ±) = b · e this leads to
|b|

|bt | = ’b · e. (2.22)

In this case the sense of the vector bt is opposite to the unit vector e, such that

bt = ’|bt |e . (2.23)

So, clearly, whether the angle ± is acute or obtuse, the vector bt parallel to the unit
vector e is given by

bt = ( b · e) e . (2.24)

Recall, that this is only true if e has unit length! In conclusion, the inner product
of an arbitrary vector b with a unit vector e de¬nes the magnitude and sense of a
vector bt that is parallel to the unit vector e such that the original vector b may be
written as the sum of this parallel vector and a vector normal to the unit vector e.
The vector bn normal to e follows automatically from

b n = b ’ bt . (2.25)

This implicitly de¬nes the unique decomposition of the vector b into a component
normal and a component parallel to the unit vector e.
Based on the considerations above, the force vector F in Fig. 2.4 can be
decomposed into a component parallel to the bone principal axis F t given by

F t = ( F · e) e, (2.26)

where e denotes a vector of unit length, and a component normal to the principal
axis of the bone:

Fn = F ’ Ft . (2.27)
17 2.6 A vector with respect to a vector basis

2.6 Representation of a vector with respect to a vector basis

Two vectors are called mutually independent if both vectors are non-zero and
non-parallel. In a two-dimensional space any vector can be expressed as a linear
combination of two mutually independent vectors, say a and b, by using the scalar
product and vector sum, for example:
F = ±a + β b, (2.28)
see, Fig. 2.7. Clearly, in a three-dimensional space three mutually independent
vectors are needed:
F = ±a + β b + γ c . (2.29)
The above mutually independent vectors a and b are called basis vectors that
form a so-called vector basis {a, b} in a two-dimensional space, while in a three-
dimensional space three mutually independent vectors, say a, b and c, are needed
to form a basis. It is convenient to have such a vector basis because it facilitates
vector manipulation. For example, let
F 1 = 2a + 5b, (2.30)
and
F 2 = ’a + 3b, (2.31)
then
F 1 + F 2 = a + 8b, F 1 + 3F 2 = ’a + 14b. (2.32)
The inner product of F 1 and F 2 yields
F 1 · F 2 = ( 2a + 5b) · ( ’a + 3b)
= ’2a · a + 6a · b ’ 5b · a + 15b · b
= ’2a · a + a · b + 15b · b, (2.33)


a



±a b
βb


F

Figure 2.7
An arbitrary vector F as a linear combination of two vectors: F = ± a + β b.
The concepts of force and moment
18

since a · b = b · a. This demonstrates that if vectors are expressed with respect to
a vector basis, typical vector operations (such as vector addition and vector inner
product) are relatively straightforward to perform. If the vector sum of the basis
vectors and the inner products of the basis vectors with respect to each other and
themselves are known, vector operations on other vectors are straightforward.
However, expressing an arbitrary vector with respect to the basis vectors may
be cumbersome. For a given vector F this requires to determine the coef¬cients ±
and β in

F = ±a + β b. (2.34)

One possibility to realize this, is to take the inner product of F with respect to both
the basis vectors:

F·a = ±a·a+βb·a
F · b = ± a · b + β b · b. (2.35)

Recall, that each of the above inner products can be computed and simply yield
a number. Therefore, the set of Eqs. (2.35) provides two linear equations from
which the two unknown coef¬cients ± and β can be obtained. A similar operation
(involving three basis vectors) is needed in a three-dimensional space.
Solving for the coef¬cients ± and β would be easy if the basis vectors a and
b have unit length (such that e.g. a · a = 1) and if the basis vectors are mutu-
ally perpendicular, hence if a · b = 0. In that case, the set of Eqs. (2.35) would
reduce to:

F·a=±
F · b = β. (2.36)

If the vectors of a vector basis are mutually perpendicular but do not have unit
length, the vector basis is called orthogonal. If the vectors of an orthogonal vector
basis have unit length, then it is called an orthonormal vector basis. If the basis
vectors of an orthonormal basis have a so-called right-handed orientation with
respect to each other and are independent of the location in three-dimensional
space, it is called a Cartesian vector basis.
The Cartesian vector basis {ex , ey , ez } is used to uniquely specify an arbitrary
vector, see Fig. 2.8. An arbitrary force vector in two-dimensional space, say F,
can be expressed with respect to the Cartesian vector basis as

F = F x e x + Fy e y . (2.37)

The use of a Cartesian vector basis substantially simpli¬es vector manipulation as
is illustrated next.
19 2.6 A vector with respect to a vector basis

ez
ey
F
Fz


Fy
F ey
Fy Fx

ex
Fx ex
(a) 2D Cartesian vector basis: (b) 3D Cartesian vector basis:
F =Fx ex + Fy ey F = Fx ex + Fy ey + Fz ez

Figure 2.8
Decomposition of F in a two- or three-dimensional Cartesian basis.



Example 2.3 Clearly, vector addition is straightforward, for example if
F 1 = 2ex + 5ey , F 2 = ’ex + 3ey ,
then
F 1 + F 2 = ex + 8ey , F 1 + 3F 2 = ’ex + 14ey .
This is similar to using an arbitrary, non-orthonormal, vector basis, see Eq. (2.32).
Taking the inner product of the two vectors is substantially simpli¬ed:
F 1 · F 2 = ( 2ex + 5ey ) · ( ’ex + 3ey )
= ’2 ex · ex + 6ex · ey ’ 5ex · ey +15 ey · ey
1 0 1
= 13.



Example 2.4 In the foot, the tendons of the tibialis anterior and the tibialis posterior may be
identi¬ed, see Fig. 2.9. Let the magnitude of the force vectors be given by:
Fa = |F a | = 50 [ N] , Fp = |F p | = 60[ N] ,
while the angles ± and β are speci¬ed by:
π

±= β= .
,
11 6
What is the net force acting on the attachment point Q of the two muscles on the
foot?
First, the force vectors F a and F p are written with respect to the Cartesian
coordinate system. Clearly:
The concepts of force and moment
20

Fp

Fa
A

β
P

±
ey Q




ex

Figure 2.9
Forces of the tendons of the tibialis anterior F a and posterior F p , respectively.




F a = Fa cos( ± + β) ex + sin( ± + β) ey
≈ ’18.6ex + 46.4ey [ N] ,

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