Question

problem 2
replace the loading system with an equivalent force and couple moment located at point a. show the resultant force and couple moment on a sketch of the object.
$f_x = 15 \cos\left(\frac{4}{5}\
ight) = 14.999$
$f_y = 15 \sin\left(\frac{3}{5}\
ight) = 0.157$
$f_r = \sqrt{(f_x)^2 + (f_y)^2}$
$f_r = \sqrt{(14.999)^2 + (0.157)^2}$
$\boxed{f_r = 15.00\\ \text{kn}}$

Explanation:

Answer:

$F_R = 18.2$ kN, $\theta = 109.2^\circ$ (or $19.2^\circ$ left of vertical); $M_{RA} = 95$ kN$\cdot$m (Counter-clockwise)