posted by mike on .
Two forces act on a 4.00-kg object in a manner that the object has an
acceleration 3.00 m/s2 in a direction 20.0° north of east. The first force is
15.00 N in a direction 10.0° west of north. What is the second force?
F1 = 15N@100o = -2.60 + 14.8i.
a = 3m/s^2@20o. = 2.82+1.03i.
F1 + F2 = ma.
(-2.60+14.8i)+F2 = 4.0*(2.82+1.03i)
-2.60+14.8i+F2 = 11.28+4.12i
F2 = 11.28+2.60 + 4.12i-14.8i
F2 = 13.88 -10.68i = 17.5N@ -37.6o.
F2 = 17.5N@ 37.6o South of East.