find the resultant of two forces, one 3N due north and 4N due east?
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Explanation:
X = 4N.
Y = 3N.
X + Yi + F3 = 0
4 + 3i + F3 = 0
F3 = -4 - 3i
Tan Ar = Y/X = -3/-4 = 0.75000
r = 36.87o(Q1). = Reference angle.
A = 36.87 + 180 = 216.87o(Q3).
F3 = X/Cos A = -4/Cos216.87 = 5.0 N. @
216.87o.
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