Page 39 - Dynamics of Mechanical Systems

0593_C02_fm Page 20 Monday, May 6, 2002 1:46 PM

20 Dynamics of Mechanical Systems

Then, the magnitude of R is:

/
2
R = (a 2 + b 2 + ) 12 (2.4.5)
c
A question that arises is what if A, B, and C are not mutually perpendicular? How then
can we find the magnitude of the resultant? A powerful feature of the vector method is
that the same general procedure can be used regardless of the directions of the components.
All that is required is to express the components as sums of vectors parallel to the X-, Y-,
and Z-axes (or other convenient mutually perpendicular directions). For example, suppose
a vector A is inclined relative to the X-, Y-, and Z-axes, as in Figure 2.4.3. Let θ , θ , and
x
y
θ be the angles that A makes with the axes. Next, let us express A in the desired form:
z
A = A + A + A (2.4.6)
x y z
where A , A , and A are vector components of A parallel to X, Y, and Z. A , A , and A z
y
y
z
x
x
may be considered as “projections” of A along the X-, Y-, and Z-axes. Their magnitudes
are proportional to the magnitude of A and the cosines of the angles θ , θ , and θ . That is,
z
x
y
A = A cosθ = a
x x x
A = A cosθ = a (2.4.7)
y y y
A = A cosθ = a
z z z
where a , a , and a are defined as given in the equations. As before, let n , n , and n be
x
y
z
y
x
z
unit vectors parallel to X, Y, and Z. Then, A , A , and A can be expressed as:
z
x
y
A = a n = A cosθ n
x x x x x
A = a n = A cosθ n (2.4.8)
y y y y y
A = a n = A cosθ n
z z z z z
Finally, by substituting into Eq. (2.4.6), A may be expressed in the form:
A = a n + a n + a n = A cosθ n + A cosθ n + A cosθ n (2.4.9)
x x y y z z x x y y z z
Z
n
z θ
z
n
θ x θ y
Y
n y y
FIGURE 2.4.3
A vector A inclined relative to the n
coordinate axes. X x