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

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:
$n = -6 \implies |n| = 6$