Let's first find x x x - and y y y -components of resultant displacement:
R x = 2000 l b ⋅ c o s 9 0 ∘ + 4000 l b ⋅ c o s 15 0 ∘ + 5000 l b ⋅ c o s 23 0 ∘ + 1000 l b ⋅ c o s 27 0 ∘ + 3000 l b ⋅ c o s 33 0 ∘ = − 4080 l b , R_x=2000\ lb\cdot cos90^{\circ}+4000\ lb\cdot cos150^{\circ}+5000\ lb\cdot cos230^{\circ}+1000\ lb\cdot cos270^{\circ}+3000\ lb\cdot cos330^{\circ}=-4080\ lb, R x = 2000 l b ⋅ cos 9 0 ∘ + 4000 l b ⋅ cos 15 0 ∘ + 5000 l b ⋅ cos 23 0 ∘ + 1000 l b ⋅ cos 27 0 ∘ + 3000 l b ⋅ cos 33 0 ∘ = − 4080 l b ,
R y = 2000 l b ⋅ s i n 9 0 ∘ + 4000 l b ⋅ s i n 15 0 ∘ + 5000 l b ⋅ s i n 23 0 ∘ + 1000 l b ⋅ s i n 27 0 ∘ + 3000 l b ⋅ s i n 33 0 ∘ = − 2330 l b . R_y=2000\ lb\cdot sin90^{\circ}+4000\ lb\cdot sin150^{\circ}+5000\ lb\cdot sin230^{\circ}+1000\ lb\cdot sin270^{\circ}+3000\ lb\cdot sin330^{\circ}=-2330\ lb. R y = 2000 l b ⋅ s in 9 0 ∘ + 4000 l b ⋅ s in 15 0 ∘ + 5000 l b ⋅ s in 23 0 ∘ + 1000 l b ⋅ s in 27 0 ∘ + 3000 l b ⋅ s in 33 0 ∘ = − 2330 l b .
We can find the magnitude of the resultant displacement from the Pythagorean theorem:
R = R x 2 + R y 2 = ( − 4080 l b ) 2 + ( − 2330 l b ) 2 = 4698.5 l b . R=\sqrt{R_x^2+R_y^2}=\sqrt{(-4080\ lb)^2+(-2330\ lb)^2}=4698.5\ lb. R = R x 2 + R y 2 = ( − 4080 l b ) 2 + ( − 2330 l b ) 2 = 4698.5 l b . We can find the direction of the resultant displacement from the geometry:
θ = s i n − 1 ( R y R ) , \theta=sin^{-1}(\dfrac{R_y}{R}), θ = s i n − 1 ( R R y ) , θ = s i n − 1 ( − 2330 l b 4698.5 l b ) = 3 0 ∘ W o f S . \theta=sin^{-1}(\dfrac{-2330\ lb}{4698.5\ lb})=30^{\circ}\ W\ of\ S. θ = s i n − 1 ( 4698.5 l b − 2330 l b ) = 3 0 ∘ W o f S .
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