Let's determine the resultant of the following vectors using the component method:
R x = A x + B x + C x + D x + E x , R_x=A_x+B_x+C_x+D_x+E_x, R x = A x + B x + C x + D x + E x , R x = 250 N ⋅ c o s ( 18 0 ∘ + 3 7 ∘ ) + 150 N ⋅ c o s ( − 27 0 ∘ ) + 300 N ⋅ c o s ( 6 0 ∘ ) + 200 N ⋅ c o s ( 9 0 ∘ + 5 3 ∘ ) + 100 N ⋅ c o s ( 27 0 ∘ + 3 0 ∘ ) = − 159.38 N . R_x=250\ N\cdot cos(180^{\circ}+37^{\circ})+150\ N\cdot\ cos(-270^{\circ})+300\ N\cdot cos(60^{\circ})+200\ N\cdot cos(90^{\circ}+53^{\circ})+100\ N\cdot cos(270^{\circ}+30^{\circ})=-159.38\ N. R x = 250 N ⋅ cos ( 18 0 ∘ + 3 7 ∘ ) + 150 N ⋅ cos ( − 27 0 ∘ ) + 300 N ⋅ cos ( 6 0 ∘ ) + 200 N ⋅ cos ( 9 0 ∘ + 5 3 ∘ ) + 100 N ⋅ cos ( 27 0 ∘ + 3 0 ∘ ) = − 159.38 N .
R y = A y + B y + C y + D y + E y , R_y=A_y+B_y+C_y+D_y+E_y, R y = A y + B y + C y + D y + E y , R x = 250 N ⋅ s i n ( 18 0 ∘ + 3 7 ∘ ) + 150 N ⋅ s i n ( − 27 0 ∘ ) + 300 N ⋅ s i n ( 6 0 ∘ ) + 200 N ⋅ s i n ( 9 0 ∘ + 5 3 ∘ ) + 100 N ⋅ s i n ( 27 0 ∘ + 3 0 ∘ ) = 293.11 N . R_x=250\ N\cdot sin(180^{\circ}+37^{\circ})+150\ N\cdot\ sin(-270^{\circ})+300\ N\cdot sin(60^{\circ})+200\ N\cdot sin(90^{\circ}+53^{\circ})+100\ N\cdot sin(270^{\circ}+30^{\circ})=293.11\ N. R x = 250 N ⋅ s in ( 18 0 ∘ + 3 7 ∘ ) + 150 N ⋅ s in ( − 27 0 ∘ ) + 300 N ⋅ s in ( 6 0 ∘ ) + 200 N ⋅ s in ( 9 0 ∘ + 5 3 ∘ ) + 100 N ⋅ s in ( 27 0 ∘ + 3 0 ∘ ) = 293.11 N .
We can find the resultant magnitude from the Pythagorean theorem:
R = R x 2 + R y 2 = ( − 159.38 N ) 2 + ( 293.11 N ) 2 = 333.64 N . R=\sqrt{R_x^2+R_y^2}=\sqrt{(-159.38\ N)^2+(293.11\ N)^2}=333.64\ N. R = R x 2 + R y 2 = ( − 159.38 N ) 2 + ( 293.11 N ) 2 = 333.64 N . We can find the direction of the resultant from the geometry:
θ = s i n − 1 ( R y R ) = s i n − 1 ( 293.11 N 333.64 ) = 61. 5 ∘ N o f W . \theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{293.11\ N}{333.64})=61.5^{\circ}\ N\ of\ W. θ = s i n − 1 ( R R y ) = s i n − 1 ( 333.64 293.11 N ) = 61. 5 ∘ N o f W .
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