Posted by marissa on Friday, September 14, 2012 at 3:50pm.
Fn = 31.4N @ 12o,CCW + 12.7N @ 192o,CCW.
X=Hor.=31.4*cos12+12.7*cos192=18.3 N.
Y=Ver.=31.4*sin12+12.7*sin192=3.89 N.
a. Mag.=sqrt((18.3)^2+(3.89)^2=18.7 N
b. tanA = Y/X = 3.89/18.3 = 0.21257
A = 12o, CCW = 12o North of East.
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