Find the​ magnitude, to the nearest hundredth of a​ pound, and the direction​ angle, to the nearest tenth of a​ degree, of the resultant force.

Question

The magnitude and direction of two forces acting on an object are

120 pounds, S58°​E, and 80 pounds, N69°​E, respectively. Find the​ magnitude, to the nearest hundredth of a​ pound, and the direction​ angle, to the nearest tenth of a​ degree, of the resultant force.

Answer 1

The S58E bearing translates to 32° below positive x-axis, and N69E to 21° above this axis. Then we can determine x and y components of the forces: F_x1 = 120 cos (-32) = 101.766 (I'm keeping 3 decimals to avoid round-off error in the answer), F_x2 = 80 cos 21 = 74.686, total F_x = 176.452; F_y1 = 120 sin (-32) = -63.590, F_y2 = 80 sin 21 = 28.669, total F_y = -34.921. Combining the components quadratically gives a magnitude of 179.87 pounds.

The angle θ of the resultant force, relative to the positive x-axis, comes from tan θ = F_y /F_x = (-34.921) / (176.452) = -0.1979, θ = tan^-1 (-0.1979) = -11.2°. So the bearing is S78.8E.