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

Answer 2

Step 1: Given information.
Let the two forces be \( \vec{F}_1 \) and \( \vec{F}_2 \).
Given that one force has a magnitude three times the other, we can assume:
\[ F_1 = 3F_2 \] and the magnitude of their resultant is equal to the larger force, i.e.
\[ R = F_1 \]
Step 2: Relation for the magnitude of the resultant.
For two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \), the resultant is:
\[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \] Substitute \( R = F_1 \):
\[ F_1 = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \]
Step 3: Simplify the equation.
Square both sides:
\[ F_1^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \] Cancel \( F_1^2 \) from both sides:
\[ 0 = F_2^2 + 2F_1F_2\cos\theta \] \[ \Rightarrow \cos\theta = -\frac{F_2}{2F_1} \]
Step 4: Substitute \( F_1 = 3F_2 \).
\[ \cos\theta = -\frac{F_2}{2 \times 3F_2} = -\frac{1}{6} \] Thus, the angle between the two forces is:
\[ \theta = \cos^{-1}\left(-\frac{1}{6}\right) \]
Since the question gives the angle as \( \cos^{-1}\left(\frac{1}{n}\right) \), we have:
\[ \frac{1}{n} = -\frac{1}{6} \quad \Rightarrow \quad |n| = 6 \]
Step 5: Final Answer.
\[ \boxed{|n| = 6} \]