Find the angle between forces of magnitude 7N and 4N if their resultant has a magnitude of 9N.
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Correct Answer: Option B
Explanation:
\(F_{1} = 7i + 0j\)
\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)
\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)
\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)
\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)
Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)
\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)
\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)
\(\theta = \cos^{-1} 0.2857 = 73.40°\)
\(F_{1} = 7i + 0j\)
\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)
\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)
\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)
\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)
Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)
\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)
\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)
\(\theta = \cos^{-1} 0.2857 = 73.40°\)