The magnitude of a force (xi + 15j) is 17N. Find the :
(a) possible values of x ;
(b) directions of the forces, correct to the nearest degree.
Explanation
(a) (F = xi + 15j)
(|F| = sqrt{x^{2} + 15^{2}} = 17)
(x^{2} = 17^{2} – 15^{2} = 64)
(x = pm sqrt{64} = pm 8)
(x = 8 ; x = -8)
(b)
(tan alpha = frac{15}{8} = 1.875)
(alpha = tan^{-1} (1.875) = 61.9° approxeq 62°)
(tan beta = frac{15}{-8} = -1.875)
(beta = tan^{-1} (-1.875) = -61.9°)
= (180° – 61.9° = 118.1° approxeq 118°)
Directions of forces are 62° and 118° each with the positive x- axis.