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.
(a) possible values of x ;
(b) directions of the forces, correct to the nearest degree.
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Correct Answer: Option n
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.
(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.