Answer to Question #309567 in Mechanics | Relativity for Man

Question

 John, in going to his Physics class, walks 30m, 450 N of E, then 40 m, 600 W of S and finally 50 m, straight south. Use the polygon method to determine the resultant vector 𝑅⃑ and check your result using the component method.

Answer 1

Let us determine the distance OB:

$OB^2 = OA^2+AB^2 - 2OA\cdot AB\cos(60^\circ-45^\circ),\\ OB^2 = 30^2+40^2 - 2\cdot30\cdot40\cos15^\circ,\\ OB = 13.5\,\mathrm{m}.$

$\dfrac{\sin\angle AOB}{AB} = \dfrac{\sin OAB}{OB}, \\ \angle AOB = 130^\circ. \\ \angle BOW = 180^\circ - 130^\circ - 45^\circ = 5^\circ.$

$\angle OBW = 90^\circ -\angle BOW = 85^\circ.$

$|\vec{R}|=OC -?\\ OC^2= BC^2 + OB^2 - 2BC\cdot OB\cos\angle OBW,\\ OC = 50.6\,\mathrm{m}.$

$\dfrac{\sin\angle BOC}{BC} = \dfrac{\sin\angle OBW}{OC}, \; \angle BOC = 80^\circ.\\ \angle COS = 90^\circ - (\angle BOC - \angle BOW) = 15^\circ.$

Let us determine the parameters of R using the coordinate method.

$\vec{R}_x = \vec{OA}_x + \vec{AB}_x + \vec{BC}_x, \\ \vec{R}_x =30\cos45^\circ + 40\cos(180^\circ + 30^\circ) + 50\cos270^\circ, \\ \vec{R}_x = -13.4. \\ \vec{R}_y = \vec{OA}_y + \vec{AB}_y + \vec{BC}_y, \\ \vec{R}_y =30\sin45^\circ + 40\sin(180^\circ + 30^\circ) + 50\sin270^\circ, \\ \vec{R}_y = -48.8.$