Question 2.7: Determine the magnitude and the coordinate direction angles ......
Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring in Fig. 2-29a.
Since each force is represented in Cartesian vector form, the resultant force, shown in Fig. 2-29b , is
F_R = ΣF = F_1 + F_2 = { 6oj – 8ok } lb + { 5oi — 1ooj + 100k } lb
= { 50i – 40j + 18ok } lb
The magnitude of F_R is
F_R=\sqrt{(50\ lb)^2+(-40\ lb)^2 + (180\ lb)^2} = 191.0\ lb \\ \quad \ \ = 191\ lb
The coordinate direction angles α, β, γ are determined from the components of the unit vector acting in the direction of F_R.
u_{F_R} =\frac{F_R}{F_R} =\frac{50}{191.0}i – \frac{40}{191.0}j +\frac{180}{191.0}k
= 0.2617i – 0.2094j + 0.9422k
so that
cos α = 0.2617 α = 74.8°
cos β = -0.2094 β = 102°
cos γ = 0.9422 γ = 19.6°
These angles are shown in Fig. 2-29b.
NOTE: In particular, notice that β > 90° since the j component of u_{F_R} is negative. This seems reasonable considering how F_1 and F_2 add according to the parallelogram law.