Question

Part A) If theta= 58 degrees and F=500 N, determine the magnitude of the resultant force. Part B) If theta= 58 degrees and F=500 N, determine the resultant force direction, measured counterclockwise from the positive x axis.

Answer 1

in x axis
500cos(58)-700cos(15) = -411.188 i
in y axis
500sin(58) -700sin(15) = 242.85 j
magnitude = 477.54 N
direction = 30.56 degree from positive x axis in counter clock wise dierction

Answer 2

Resultant force = √(F1²+F2²+2F1F2 cos θ )= √( 2500 + 4900 + 2*500*700* cos 137 = 477.55 N

OR

horizontal force = 700 cos 15 - 500 cos 58 = 411.19 N

also net vertical force = 700 sin 15 - 500 sin 58 = -242.85 N

Resultant = 477.55 N

Angle = inverse tan (242.85 / 411.19) = 30.57 degrees

so required angle = 360 - 30.57 = 329.43 degrees.