Drawing convention of the moment vector for different force vector orientations. Figures (a) and (b)

indicate a statically equivalent load for a counterclockwise orientation of the moment vector. Figures

(c) and (d) are equivalent for a clockwise orientation of the moment vector.

ey

F1

3

4

P

ex

F2 2

5

2

F3

2

Figure 2.21

Resulting moment in two dimensions.

31 2.11 The two-dimensional case

Example 2.8 Resulting moment using scalar notation. Following the drawing convention of

Section 2.2, the force vectors in Fig. 2.21 are given by

F 1 = F1 e x

F 2 = ’F2 ex

F 3 = F3 e x .

Similarly,

d1 = 3ex + 4ey

d2 = ’2ex ’ 2ey

d3 = 2ex ’ 5ey .

Each of the force vectors F i generates a moment vector with respect to the point P:

M i = di — Fi .

Clearly, given the fact that all force vectors are in the plane spanned by the ex and

ey vectors, the moment vectors are all in the ez direction:

M i = Mi e z .

Either using the formal de¬nition of the moment vector or the drawing convention

for two-dimensional problems as given above, it follows that

M1 = ’4F1

M2 = ’2F2

M3 = 5F3 .

The force vectors F 1 and F 2 produce a clockwise, hence negative, moment, while

F 3 produces a counterclockwise, hence positive, moment. The resulting moment

with respect to point P equals

M = M1 + M2 + M3 = ’4F1 ’ 2F2 + 5F3 .

Example 2.9 Resulting moment using vector notation. Consider the forces as depicted in

Fig. 2.22. The forces are given by

F 1 = 3ex + ey

F 2 = 4ex ’ ey

F 3 = ’2ex ’ 3ey ,

The concepts of force and moment

32

F1

x3

x1

ey

ex

x2

F3 xP

P

F2

Figure 2.22

Forces and moment.

while the points of application are, respectively:

x1 = 2ex + 2ey

x2 = 3ex ’ 2ey

x3 = ’4ex + 2ey .

The point P has location:

xP = ’2ex ’ 2ey .

The resulting moment of the forces with respect to the point P follows from

M = ( x1 ’ x P ) — F 1 + ( x2 ’ x P ) — F 2 + ( x3 ’ x P ) — F 3 ,

hence

M = ( 4ex + 4ey ) —( 3ex + ey ) +5ex —( 4ex ’ ey )

+ ( ’2ex + 4ey ) —( ’2ex ’ 3ey )

= ’8ez ’ 5ez + 14ez

= ez .

2.12 Drawing convention of moments in three dimensions

An arrow drawn with two arrowheads, and identi¬ed by a scalar, rather than a

vector symbol, denotes a moment vector following the right-handed or corkscrew

rule. For example the moment vectors drawn in Fig. 2.23 and identi¬ed by the

scalars M1 , M2 and M3 , respectively, correspond to the moment vectors:

33 Exercises

ez

M3

M2 ey

M1

ex

Figure 2.23

Moment vectors identi¬ed by means of scalars.

ez

M3

M2 ey

M1

ex

Figure 2.24

Moment vectors identi¬ed by means of vectors and having a single arrowhead.

M 1 = M1 e x , M 2 = ’M2 ey , M 3 = M3 e z . (2.70)

Notice that the moment vector M 2 is pointing into the negative y-direction. Alter-

natively, if in the ¬gure the moment vectors are drawn with a single arrowhead, as

in Fig. 2.24, they denote actual vectors and are identi¬ed with vector symbols.

Exercises

The vector bases {e1 , e2 , e3 } and { 1 , 2 , 3 } are orthonormal. The following

2.1

relations exist:

1√ 1√

1= 2e1 + 2e2

2 2

1√ 1√

2=’ 2e1 + 2e2

2 2

3 = e3 .

The force vector F is de¬ned with respect to basis {e1 , e2 , e3 } according to:

F = 2e1 + 3e2 ’ 4e3 .

(a) Determine the decomposition of F with respect to basis { 1 , 2 , 3 }.

The concepts of force and moment

34

(b) Determine the length of F, using the speci¬cations of F expressed in

{e1 , e2 , e3 } and in { 1 , 2 , 3 }.

2.2 For the points P, Q and R the following location vectors are given,

respectively:

xP = ex + 2ey

xQ = 4ex + 2ey

xR = 3ex + ey .

The force vector F = 2ex acts on point Q.

Q

P

F

xQ

ey

xP R

xR

ez

ex

(a) Calculate the moment of the force F with respect to point P and with

respect to point R.

(b) Write the vectors, mentioned above, in column notation according

to the right-handed orthonormal basis {ex , ey , ez } and calculate the

moment of the force with respect to the points P and R by using

Eq. (1.35).

2.3 Calculate for each of the situations given below the resulting moment with

respect to point P.

F

F

F F

45°

F P

P P 2

P

P

F 2

F

2

F

F

F

(a) (b) (c) (d) (e)

2.4 On an axis a wheel with radius R is ¬xed to a smaller wheel with radius r.

The forces F and f are tangentially applied to the contours of both wheels

(as shown in the ¬gure). Calculate the ratio between the forces F and f in

the case where the total moment with respect to the centroid P is zero.

35 Exercises

F

r

P

f

R

On the body in the drawing the moments M 1 = ’ 3ex and M 2 = 4ex and

2.5

the forces F 1 = ’ 2ez and F 2 = ez are exerted. The forces are acting at the

points x1 = ’2ex + 3ez and x2 = 3ex + ey + 3ez . Calculate the resulting

moment vector with respect to point P, given by the position vector xP =

2ex + 4ey + 2ez .

F2

M2

M1

x1 x2

F1 P

ez

xP

ey

ex

2.6 A person is pulling a rope. This results in a force F acting on the hand

at point H as depicted in the ¬gure. Calculate the moment in the shoulder

(point S) and the elbow (point E) as a result of the force F. The following

vectors are given: