Asked by Megan
How can one find the magnitude and the directions of a resultant if Vector A is 35N @ 60' NE, and Vector B is 50N @ 20' NW?
Answers
Answered by
Henry
A = 35N @ 60deg,NE = 35N. @ 60deg,CCW.
B = 50N. @ 20deg,NW = 50N @ 160deg,CCW.
X=hor. = 35cos60 + 50cos160 = -29.48N.
Y = ver. = 35sin60 + 50sin160 = 47.4N.
R^2 = X^2 + Y^2,
R^2 = (-29.48)^2 + (47.4)^2 = 3117,
R = 55.8N.
tanA = Y/X = 47.4 / -29.48 = -1.60787,
A = -58.1 Deg.,CW.
A = -58.1 + 180 = 121.9 Deg,CCW. =
Direction.
R = 55.8N. @ 121.9 Deg.
B = 50N. @ 20deg,NW = 50N @ 160deg,CCW.
X=hor. = 35cos60 + 50cos160 = -29.48N.
Y = ver. = 35sin60 + 50sin160 = 47.4N.
R^2 = X^2 + Y^2,
R^2 = (-29.48)^2 + (47.4)^2 = 3117,
R = 55.8N.
tanA = Y/X = 47.4 / -29.48 = -1.60787,
A = -58.1 Deg.,CW.
A = -58.1 + 180 = 121.9 Deg,CCW. =
Direction.
R = 55.8N. @ 121.9 Deg.