The three displacement vectors in the drawing have magnitudes of A = 5.30 m, B =

Question

The three displacement vectors in the drawing have magnitudes of A = 5.30 m, B = 4.50 m, and C = 4.15 m. Find the resultant (magnitude and directional angle) of the three vectors by means of the component method. Express the directional angle as an angle above or below the positive or negative x axis. (Assume alpha = 21 degree and beta = 53 degree. Give an answer between 0 and 90 degree.) Two forces are applied to a tree stump to pull it out of the ground. Force F_A has a magnitude of 2251 newtons and points 32.0 degree south of east, while force F_a has a magnitude of 3102 newtons and points due south. Using the component method, find the magnitude and direction of the resultant force F_A + F_B that is applied to the stump. Specify the direction with respect to due east.

Explanation / Answer

1)
let R is the resultant vector.

Rx = Ax + Bx + Cx

= -5.3*cos(21) + 4.5*cos(53) + 0

= -2.24

Ry = Ax + Bx + Cx

= 5.3*sin(21) + 4.5*sin(53) - 4.15

= 1.34

so,

|R| = sqrt(Rx^2 + Ry^2)

= sqrt(2.24^2 + 1.34^2)

= 2.61 m <<<<<<<<-----------------Answer

direction : theta = tan^-1(Ry/Rx)

= tan^-1(1.34/(-2.24))

= 30.9 degrees below -x axis

2) let East be +x axis

Fx = FAx + FBx

= 2251*cos(32) + 0

= 1909 N

Fy = FAy + FBy

= 2251*sin(32) + 3102

= 4295 N

|F| = sqrt(Fx^2 + Fy^2)

= sqrt(1909^2 + 4295^2)

= 4700 N

direction : theta = tan^-1(Fy.Fx)

= tan^-1(4295/1909)

= 66.0 degrees south of east

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