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.

Answer Details

The force is \(x\mathbf{i}+15\mathbf{j}\) with magnitude 17 N.

(a) Possible values of x. The magnitude is \(\sqrt{x^2+15^2}:\)

\[\sqrt{x^2+225}=17\Rightarrow x^2+225=289\Rightarrow x^2=64\Rightarrow x=\pm8.\]

So \(x=8\) or \(x=-8.\)

(b) Directions (to the nearest degree), measured from the positive x-axis.

When \(x=8\): the force \(8\mathbf{i}+15\mathbf{j}\) lies in the first quadrant.

\[\theta=\tan^{-1}\!\left(\frac{15}{8}\right)\approx61.9^\circ\approx62^\circ.\]

When \(x=-8\): the force \(-8\mathbf{i}+15\mathbf{j}\) lies in the second quadrant.

\[\theta=180^\circ-\tan^{-1}\!\left(\frac{15}{8}\right)\approx180^\circ-61.9^\circ\approx118^\circ.\]

So the two forces act at approximately \(62^\circ\) and \(118^\circ\) to the positive x-axis.