Question

what third force must be added to the two forces 30 Newton at 30 degree and 80 Newton at 135 degree the resolution for equal to zero​

Answer 1

Resolve the vectors:

F1 = 30 cos 30 i + 30 sin 30 j = 26 i + 15 j

F2 = 80 cos 135 i + 80 sin 135 j = -56.6 i + 56,6 j

F1 + F2 = (26 - 56.6) i + (15 + 56.6) j = -30.6 i + 71.6 j

The offsetting force would be

F3 = 30.6 i - 71.6 j

The magnitude of this vector is (30.6^2 + 71.6^2)^1/2 = 77.9

cos theta 3 = 30.6 / 77.9 = 66.9 deg

Check:

77.9 * sin 66.9 = 71.6 agreeing in magnitude with F3

Now we need the quadrant of the offsetting vector

sin theta 3 = - cos theta 3 = +

This occurs in the fourth quadrant and theta 3 = 360 - 66.9 = 293.1 deg