and

F p = Fp cos( ±) ex + sin( ±) ey

≈ 8.5ex + 59.4ey [ N] .

Therefore, the net force due to F a and F p acting on point Q equals

F = F a + F p = ’10.1ex + 105.8ey [ N] .

Example 2.5 The decomposition of a force vector F into a component parallel to a unit vector

e and a component normal to this vector is also straightforward. For example, let

F = 2ex + 6ey ,

and

1

e = √ ( 2ex ’ 3ey ) .

13

Notice that |e| = 1. Then, the component of F parallel to e is obtained from

21 2.7 Column notation

F t = ( F · e) e

14 1

= ’√ √ ( 2ex ’ 3ey )

13 13

e

F·e

14

=’ ( 2ex ’ 3ey ) .

13

2.7 Column notation

An arbitrary two-dimensional vector is written as

F = Fx ex + Fy ey , (2.38)

and in a three-dimensional space:

F = F x e x + Fy e y + Fz e z . (2.39)

The numbers Fx , Fy and, in three dimensions Fx , Fy and Fz may be collected in a

column F :

∼

⎡ ¤

Fx

⎢ ⎥

Fx

F= F = ⎣ Fy ¦.

, (2.40)

∼ ∼

Fy

Fz

These numbers only have a meaning when associated with a vector basis, in this

case the Cartesian vector basis. There is a distinct difference between the vector F

and the column F . The vector F is independent of the choice of the vector basis,

∼

while the numbers that are stored in the column F depend on the vector basis that

∼

has been chosen. An example is given in Fig. 2.10. Both the vector basis {ex , ey }

——

and the vector basis {ex , ey } are Cartesian, but they have a different orientation

in space. Hence, the column associated with each of these bases is different. With

respect to the {ex , ey } basis it holds that

Fx

F= , (2.41)

∼

Fy

while with respect to {ex— , ey— } this column is given by

—

Fx

—

F= . (2.42)

—

∼

Fy

According to Fig. 2.10 the {ex— , ey— } basis is rotated by an angle ± with respect to

the {ex , ey } basis. In that case:

The concepts of force and moment

22

ey* ey*

ey ey

F

Fy F

ex* ex*

Fy*

± Fx*

ex ex

Fx

(a) (b) (c)

Figure 2.10

(a) Basis rotation by angle ± (b) F with respect to basis {ex , ey } (c) F with respect to basis {ex— , ey— }.

ex = cos( ±) ex— ’ sin( ±) ey—

ey = sin( ±) ex— + cos( ±) ey— , (2.43)

or, alternatively:

ex— = + cos( ±) ex + sin( ±) ey

ey— = ’ sin( ±) ex + cos( ±) ey . (2.44)

Therefore, if F is known with respect to the {ex , ey } basis, i.e.

F = Fx ex + Fy ey , (2.45)

this vector can also be expressed with respect to the {ex— , ey— } basis:

F = Fx cos( ±) ex— ’ sin( ±) ey— + Fy sin( ±) ex— + cos( ±) ey—

= Fx cos( ±) + Fy sin( ±) ex— + ’Fx sin( ±) + Fy cos( ±) ey— .

(2.46)

Therefore, in terms of Fx and Fy :

Fx cos( ±) + Fy sin( ±)

F— = . (2.47)

’Fx sin( ±) + Fy cos( ±)

∼

The magnitude |F| of the force vector is obtained from

|F| = F·F. (2.48)

In a three-dimensional context, it follows immediately that with respect to a

Cartesian vector basis:

|F| = Fx + F y + F z ,

2 2 2 (2.49)

or, equivalently

|F| = FT F . (2.50)

∼∼

23 2.7 Column notation

P F

Q

Figure 2.11

The force vector F resulting from the two force vectors P and Q.

Suppose that two thin wires are connected to a point. The ¬rst wire is loaded with

force P and on the second wire a force Q is applied. The total force vector F

exerted on the point is calculated from

F =P+Q. (2.51)

See Fig. 2.11 for a visualization.

The force vectors may be written as

P = Px ex + Py ey + Pz ez (2.52)

Q = Qx ex + Qy ey + Qz ez , (2.53)

such that

F = ( Px + Qx ) ex + ( Py + Qy ) ey + ( Pz + Qz ) ez . (2.54)

The magnitude of the resulting force vector is given by

|F| = ( Px + Qx )2 +( Py + Qy )2 +( Pz + Qz )2 . (2.55)

Using the column representation the components de¬ning the force vectors P and

Q with respect to the Cartesian vector basis may be collected into, respectively:

⎡ ¤ ⎡ ¤

Px Qx

⎢ ⎥ ⎢ ⎥

P = ⎣ Py ¦, Q = ⎣ Qy ¦, (2.56)

∼ ∼

Pz Qz

The concepts of force and moment

24

such that the force vector F is represented by

⎡ ¤⎡ ¤⎡ ¤⎡ ¤

Px + Qx

Fx Px Qx

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥

F = ⎣ Fy ¦ = ⎣ Py ¦ + ⎣ Qy ¦ = ⎣ Py + Qy ¦ . (2.57)

∼

Pz + Qz

Fz Pz Qz

2.8 Drawing convention

Consider two force vectors, F 1 and F 2 , both parallel to the unit vector e as

sketched in Fig. 2.12. In this case the two vectors are identi¬ed by numbers F1

and F2 , rather than by the vector symbols F 1 and F 2 . These numbers denote the

magnitude of the force vector, while the orientation of the arrow denotes the direc-

tion of the vector. Consequently this way of drawing and identifying the vectors

implicitly assumes

F 1 = F1 e , (2.58)

while

F 2 = ’F2 e . (2.59)

This drawing convention is generally used in combination with a certain vector

basis. In this course the Cartesian vector basis is used only. In that case, forces

acting in the horizontal plane, hence in the ex direction, are frequently identi¬ed

by Hi (from Horizontal), while forces acting in vertical direction, hence in the ey

F1

e

“F2

F2

Figure 2.12

Force vectors identi¬ed by their magnitude (F1 and F2 ).

V3

ey

H1

H2 ex

Figure 2.13

Force vectors.

25 2.9 The concept of moment

direction are identi¬ed by Vi (from Vertical). For example the vectors drawn in

Fig. 2.13 indicate that

H 1 = H1 e x , H 2 = ’H2 ex , V 3 = V3 e y . (2.60)

2.9 The concept of moment

A simple example of the effect of a moment is experienced when holding a tray

with a mass on it that exerts a (gravity) force on the tray, see the schematic drawing

in Fig. 2.14(a). This force, which acts at a certain distance d, causes a moment at

the position of our hand as is shown in Fig. 2.14(b), where the tray has been

removed from the drawing and the resulting load on the hand is indicated by the

arrow F, representing the force, and additionally the curved arrow M, representing

the moment.

Increasing the distance of the mass with respect to our hand or increasing the

mass, will increase the moment that we experience. In fact, the moment (or torque

if you like) that is felt on our hand equals the distance d multiplied by the force

due to the mass F:

M = dF. (2.61)

The moment has a certain orientation in space. Changing the direction of the force

F, as visualized in Fig. 2.15(a), will change the orientation of the moment. If the

force acts at a certain angle on the tray, as indicated in Fig. 2.15(b), only the force

normal to the tray will generate a moment with respect to the hand:

M = dFn . (2.62)