Question #254628
Find the resultant of the following two displacements: 2 m at 40o and 4 m at 120o. Angles taken relative to + x-axis.
1
Expert's answer
2021-10-21T16:33:36-0400

S1=2m,α=40°S_1 = 2m,\angle\alpha = 40\degree

S1x=S1cosα=2cos40°=1.53mS_{1x}= S_1*\cos \alpha=2*\cos40\degree= 1.53m

S1y=S1sinα=2sin40°=1.29mS_{1y}= S_1*\sin \alpha = 2*\sin40\degree=1.29 m

S2=4m,β=120°S_2 = 4m,\angle\beta= 120\degree

S2x=S2cosβ=4cos120°=2mS_{2x}= S_2*\cos \beta=4*\cos120\degree= -2m

S2y=S2sinβ=4sin120°=3.46mS_{2y}= S_2*\sin \beta=4*\sin120\degree= 3.46m

Sx=S1x+S2x=1.532=0.47mS_x = S_{1x}+ S_{2x}= 1.53-2=-0.47m

Sy=S1y+S2y=1.29+3.46=4.75mS_y = S_{1y}+ S_{2y}= 1.29+3.46=4.75m

S=Sx2+Sy2=0.472+4.752=4.77mS = \sqrt{S_x^2+S_y^2}=\sqrt{-0.47^2+4.75^2}=4.77m

cosγ=SxS=0.474.77=0.09853\cos\gamma=\frac{Sx}{S}=\frac{-0.47}{4.77}= -0.09853

γ=arccos(0.09853)95°\gamma=\arccos(-0.09853)\approx95\degree


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


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