Question #167143

1.       Using the component method, determine the resultant of the following vectors:


                                      A = 250N     37o S of W

                                      B = 150N      due N

                                      C = 300N      60o N of E

                                      D =  200N      53o N of W

                                      E = 100N      30o S of E



1
Expert's answer
2021-02-28T07:37:03-0500

Let's determine the resultant of the following vectors using the component method:


Rx=Ax+Bx+Cx+Dx+Ex,R_x=A_x+B_x+C_x+D_x+E_x,

Rx=250 Ncos(180+37)+150 N cos(270)+300 Ncos(60)+200 Ncos(90+53)+100 Ncos(270+30)=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.


Ry=Ay+By+Cy+Dy+Ey,R_y=A_y+B_y+C_y+D_y+E_y,

Rx=250 Nsin(180+37)+150 N sin(270)+300 Nsin(60)+200 Nsin(90+53)+100 Nsin(270+30)=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.

We can find the resultant magnitude from the Pythagorean theorem:


R=Rx2+Ry2=(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.

We can find the direction of the resultant from the geometry:


θ=sin1(RyR)=sin1(293.11 N333.64)=61.5 N of W.\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{293.11\ N}{333.64})=61.5^{\circ}\ N\ of\ W.


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