d

F

F

M

(a) (b)

Figure 2.14

(a) Weight of an object on a tray (b) Loading on the hand.

The concepts of force and moment

26

F d

d

Ft

F Ft F

Fn

M = dFn

M = dF

Fn

(a) Moment due to reversed force F (b) Moment due to oriented force F

Figure 2.15

Moment due to various forces F.

2.10 Definition of the moment vector

A point in space may be identi¬ed by its position vector x, see for instance the

three-dimensional example in Fig. 2.16, where O denotes the location of the origin

of the Cartesian vector basis {ex , ey , ez }.

Assume that a force F is applied to a point Q with location xQ . The moment

vector is de¬ned with respect to a point in space, say P having location xP . The

moment exerted by the force F with respect to point P is de¬ned as

M = ( xQ ’ xP ) — F = d — F. (2.63)

For an interpretation of Eq. (2.63) it is useful to ¬rst focus on a two-dimensional

con¬guration. Consider the situation as depicted in Fig. 2.17, where we focus our

attention on the plane that is formed by the vector d and the force vector F. De¬ne

a Cartesian vector basis {ex , ey , ez } with the basis vectors ex and ey in the plane

and ez perpendicular to the plane. In this case, vector ez is pointing towards the

reader. With respect to the basis the column representations of the vectors d and

F can be given by

⎡ ¤ ⎡ ¤

dx Fx

⎢ ⎥ ⎢ ⎥

d = ⎣ dy ¦ , F = ⎣ Fy ¦ . (2.64)

∼

∼

0 0

By using Eqs. (2.63) and (1.35) we immediately derive that

M = ( dx Fy ’ dy Fx ) ez . (2.65)

From this analysis several items become clear:

• The moment vector points in a direction perpendicular to the plane that is formed by

the vectors d and F.

• The total moment vector can be written as an addition of the moments M 1 = dx Fy ez and

M 2 = ’dy Fx ez . For both composing moments the force is perpendicular to the working

27 2.10 Definition of the moment vector

F

ez

Q

d

P

xP xQ

ey

ex

Figure 2.16

A point in space identi¬ed by its position vector x.

y

Fy F

Q Fx

d

dy

P

dx

ey

x

ez ex

Figure 2.17

The moment of a force acting at point Q with respect to point P.

distance of the forces, i.e. dx ex is perpendicular to Fy ey and dy ey is perpendicular to

Fx ex .

• The directions of the composing moments M 1 = dx Fy ez and M 2 = ’dy Fx ez follow

from the corkscrew rule. To apply this corkscrew rule correctly, place the tails of the

two vectors (e.g. dx ex and Fy ey ) at the same location in space, see Fig. 2.18. In the

case of the combination dx ex and Fy ey , the rotation of the arm to the force is a coun-

terclockwise movement, leading to a vector that points out of the plane, i.e. in positive

ez -direction. In the case of the combination dy ey and Fx ex , rotating the arm to the force

is a clockwise movement resulting in a moment vector that points into the plane, i.e. in

negative ez -direction.

In the de¬nition of the moment vector the location of the force vector along the

line-of-action is not relevant since only the magnitude of the force, the direction

and the distance of the point P to the line-of-action are of interest. This is illus-

trated in Fig. 2.19. We can decompose the vector d pointing from the point P to

Q in a vector dn , perpendicular to the line of action of the force F, and a vector

The concepts of force and moment

28

y y

Fy ey

dy ey

dx ex Fx ex

z z

x x

Figure 2.18

Application of the corkscrew rule.

F

d

P

dn dt

ey

ex

ez

Figure 2.19

The moment of a force acting at point Q with respect to point P.

dt , parallel to the line of action. Then, we can write for the moment M of vector F

with respect to point P:

M = d — F = ( dn + dt ) — F = dn — F. (2.66)

The de¬nition in Eq. (2.63) also assures that the resulting moment is the zero

vector if the point P is located on the line-of-action of the force vector (in that

case dn = 0).

In the general three-dimensional case, see Fig. 2.16, the procedure to determine

the moment of the force F with respect to the point P is comparable. The column

representations of the vectors d and F in this case are given by

⎡ ¤ ⎡ ¤

dx Fx

⎢ ⎥ ⎢ ⎥

d = ⎣ dy ¦ , F = ⎣ Fy ¦ , (2.67)

∼

∼

dz Fz

and the resulting column representation of the moment follows from Eq. (1.35):

⎡ ¤

dy Fz ’ dz Fy

⎢ ⎥

M = ⎣ dz Fx ’ dx Fz ¦ . (2.68)

∼

dx Fy ’ dy Fx

29 2.11 The two-dimensional case

Example 2.6 Let the origin of the Cartesian coordinate system be the point with respect to which

the moment vector is computed, i.e.

xP = 0.

The point of application of the force vector F, is denoted by:

xQ = 2ex + ey ,

which means that this point is located in the xy-plane. The force vector is also

located in this plane:

F = 5ey .

The moment of the force F with respect to the point P follows from

M = ( x Q ’ xP ) — F

= ( 2ex + ey ) — 5ey

= 10 ex — ey + 5 ey — ey

ez 0

= 10ez .

Example 2.7 Let, as before:

xQ = 2ex + ey ,

and

F = 5ey ,

but

xP = 3ez .

Then

M = ( xQ ’ xP ) — F

= ( 2ex + ey ’ 3ez ) — 5ey

= 10ez + 15ex .

2.11 The two-dimensional case

If all forces act in the same plane, the resulting moment vector with respect to

any point in that plane is, by de¬nition, perpendicular to this plane. However,

The concepts of force and moment

30

it is common practice in this case to indicate a moment as a curved arrow that

is showing a clockwise or counterclockwise direction, see Fig. 2.20. Using the

notation:

M = Mez , (2.69)

and de¬ning the orientation vector ez = ex — ey to be pointing out of the plane into

the direction of the viewer, a counterclockwise moment corresponds to M > 0,

while a clockwise moment corresponds to M < 0. Fig. 2.20(a) shows a two-

dimensional body with a force F acting on it at point Q. We can de¬ne an arbitrary

point P in the body. The moment M with respect to P as a result of the force F will

be a vector perpendicular to the plane of drawing. Using the drawing convention as

proposed above Fig. 2.20(a) can be replaced by Fig. 2.20(b). In this case the line-

of-action of force F is drawn through point P and the resulting moment is given

by a curved arrow in counterclockwise direction. The loading of the body accord-

ing to Figures 2.20(a) and 2.20(b) is statically, completely equivalent. The same

is true for Figures 2.20(c) and 2.20(d) for a clockwise direction of the moment.

F

M

P

d

F

Q

Q

(a) (b)

F

M

P

d

F

Q

Q

(c) (d)