Question 2.6: Express the force F shown in Fig. 2-28 as a Cartesian vector......

Since only two coordinate direction angles are specified. the third angle α must be determined from Eq. 2-8; i.e.,

cos² α + cos² β + cos² γ = 1                (2-8)
cos² α + cos² 60° + cos² 45° = 1

cos\ \alpha =\sqrt{1-(0.5)^2-(0.707)^2}=\pm 0.5

Hence, two possibilities exist, namely,

α = cos^{-1}(0.5) = 60°       or       α = cos^{-1}(-0.5) = 120°

By inspection it is necessary that α = 60°, since F_x must be in the +x direction.

Using Eq. 2-9, with F = 200 N, we have

A = Au_A
= A cos αi + A cos βj+ A cos γk                        (2-9)
= A_xi + A_yj + A_zk

F = F cos αi + F cos βj + F cos γk
= (200 cos 60° N)i + (200 cos 60° N)j – (200 cos 45° N)k
= { 1oo.oi + 100.0j + 141.4k } N

Show that indeed the magnitude of F = 200 N.