Given:
A= 10.0 m, 30º
B= 5.0 m, north
C= 12.0 m, 40º east of north
Required:
Find A+B+C graphically and analytically
Solution (analytical):
First, find the components of the vectors:
"A_x=10\\text{ cos}30\u00b0=8.66\\text{ m},\\\\\nA_y=10\\text{ sin}30\u00b0=5\\text{ m},\\\\\nA=8.66i+5j.\\\\\nB_x=5\\text{ cos90}\u00b0=0,\\\\\nB_y=5\\text{ sin}90\u00b0=5\\text{ m},\\\\\nB=0i+5j.\\\\\nC_x=12\\text{ sin}40\u00b0=7.71\\text{ m},\\\\\nC_y=12\\text{ cos}40\u00b0=9.19\\text{ m},\\\\\nC=7.71i+9.19j." The resultant vector is
"D=D_xi+D_yj=A+B+C=\\\\=8.66i+5j+0i+5j+7.71i+9.19j=\\\\=(16.37i+19.19j)\\text{ m},\\\\\n|D|=\\sqrt{D_x^2+D_y^2}=25.2\\text{ m},\\\\\n\\theta=\\text{atan}\\frac{19.19}{16.37}=49.5\u00b0." Solution (graphical):
If we measure the angle and length of the resultant vector, we get approximately 25 m and 50°.
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