Answer to Question #167143 in Physics for rhea

Question

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

                                      A = 250N     37^o S of W

                                      B = 150N      due N

                                      C = 300N      60^o N of E

                                      D =  200N      53^o N of W

                                      E = 100N      30^o S of E

Answer 1

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

$$R_x=A_x+B_x+C_x+D_x+E_x,$$

$R_x=250\ N\cdot cos(180^{\circ}+37^{\circ})+150\ N\cdot\ cos(-270^{\circ})+300\ N\cdot cos(60^{\circ})+200\ N\cdot cos(90^{\circ}+53^{\circ})+100\ N\cdot cos(270^{\circ}+30^{\circ})=-159.38\ N.$

$$R_y=A_y+B_y+C_y+D_y+E_y,$$

$R_x=250\ N\cdot sin(180^{\circ}+37^{\circ})+150\ N\cdot\ sin(-270^{\circ})+300\ N\cdot sin(60^{\circ})+200\ N\cdot sin(90^{\circ}+53^{\circ})+100\ N\cdot sin(270^{\circ}+30^{\circ})=293.11\ N.$

We can find the resultant magnitude from the Pythagorean theorem:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-159.38\ N)^2+(293.11\ N)^2}=333.64\ N.$$

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

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{293.11\ N}{333.64})=61.5^{\circ}\ N\ of\ W.$$