Question

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 _____.

Answer 1

Correct Answer: 6

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:
\[ n = -6 \implies |n| = 6 \]