Question 1 Report
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.
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.
Answer Details
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