Ask a New Question
Search
Questions and answers by
Philip Almeria
Answers (1)
F= 270 2 +300 2 +2(270)(300)cos60 =494 N \theta=\arctan{\frac{(300)\sin{60}}{270+(300)\cos{60}}}=31.7\degreeθ=arctan 270+(300)cos60 (300)sin60 =31.7°
