<<

. 8
( 67 .)



>>

Figure 2.20
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:

<<

. 8
( 67 .)



>>