Two forces \( \vec{F}_1 \) and \( \vec{F}_2 \) are acting on a body. One force has magnitude thrice that of the other force, and the resultant of the two forces is equal to the force of larger magnitude. The angle between \( \vec{F}_1 \) and \( \vec{F}_2 \) is \( \cos^{-1}\left(\frac{1}{n}\right) \). The value of \( |n| \) is _____.
Correct Answer: 6
Solution and Explanation
Given:
\[ |\vec{F}_1| = F, \quad |\vec{F}_2| = 3F \]
The resultant force \( |\vec{F}_R| \) is equal to the larger force:
\[ |\vec{F}_R| = 3F \]
The magnitude of the resultant is given by:
\[ |\vec{F}_R|^2 = |\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2| \cos \theta \]
Substituting the values:
\[ (3F)^2 = F^2 + (3F)^2 + 2 \times F \times 3F \cos \theta \]
Simplifying:
\[ 9F^2 = F^2 + 9F^2 + 6F^2 \cos \theta \]
Rearranging terms:
\[ 0 = 6F^2 \cos \theta \]
\[ \cos \theta = -\frac{1}{6} \]
Therefore:
\[ \theta = \cos^{-1} \left( -\frac{1}{6} \right) \]
Given that \( \cos \theta = \frac{1}{n} \),
we have:
\[ n = -6 \implies |n| = 6 \]
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