Answer to Question #254628 in Classical Mechanics for aza

Question

Find the resultant of the following two displacements: 2 m at 40o and 4 m at 120o. Angles taken relative to + x-axis.

Answer 1

$S_1 = 2m,\angle\alpha = 40\degree$

$S_{1x}= S_1*\cos \alpha=2*\cos40\degree= 1.53m$

$S_{1y}= S_1*\sin \alpha = 2*\sin40\degree=1.29 m$

$S_2 = 4m,\angle\beta= 120\degree$

$S_{2x}= S_2*\cos \beta=4*\cos120\degree= -2m$

$S_{2y}= S_2*\sin \beta=4*\sin120\degree= 3.46m$

$S_x = S_{1x}+ S_{2x}= 1.53-2=-0.47m$

$S_y = S_{1y}+ S_{2y}= 1.29+3.46=4.75m$

$S = \sqrt{S_x^2+S_y^2}=\sqrt{-0.47^2+4.75^2}=4.77m$

$\cos\gamma=\frac{Sx}{S}=\frac{-0.47}{4.77}= -0.09853$

$\gamma=\arccos(-0.09853)\approx95\degree$

$\text{Answer: }4.77m ;\text{angle }95\degree$