Homework Answers
here,
Force F1 = 47 cos θ 1 i + 47 sin θ 1 j
F1 = 47 cos 63 i + 47 sin 63 j
F1 = 21.33 i + 41.87 j
F2 = 59 cos 27 i + 59 sin 27 (-j)
also θ1 + θ2 = 90
F2 = 52.56 i - 26.78 j
thus F3 = F 3 cos θ3 ( -i) + F 3 sin θ3 ( -j)
The particle is in equilibrium.
So, F1+ F2+ F3 = 0
21.33 i + 41.87 j + 52.56 i -26.78 j = F3 cos θ3 - F3 sin θ3
F3 cos θ3 = 74.89-----1
F3 sin θ3 = 15.09 -----2
eq 2/eq1 = tan θ3 = (15.09/74.89) = 0.201
θ3 = 11.39 deg
so
F3 = 74.89/(cos 11.39)
F3 = 76.39 N (answer)
angle θ3 = 11.39 deg (answer)
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