The resultant of two equal forces inclined at an angle α is half of resultant when they are inclined at the angle β, then

Concept:

Law of Parallelogram of forces: This law is used to determine the resultant of two coplanar forces acting at a point.

It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through that common point.”

If F1 and F2 are two vectors represented by the adjacent sides of a parallelogram having angle θ between them, then the resultant vector R is given by:

\({\rm{R}} = \sqrt {{\rm{F}}_1^2 + {\rm{F}}_2^2 + 2{{\rm{F}}_1}{{\rm{F}}_2}\cos {\rm{\theta }}}\)

Calculation:

Let F be the force.

When angle between them is β 

∴ Resultant  \(\rm {R_2} = \sqrt {F^2 + F^2 + 2F^2\cos β}\)

According to the question:

\(R_2=2R_1\)

\(⇒R_2^2=4R_1^2\)

\(⇒ \rm 2F^2 + 2F^2\cos \beta = \rm 8F^2 + 8F^2\cos \alpha\)

\(⇒ \rm 2F^2(1 + \cos \beta) = 4\times 2F^2(1 + \cos \alpha)\)

\(⇒ \rm (1 + \cos \beta) = 4(1 + \cos \alpha)\)

\(⇒ 2\cos^2 (\beta/2) = 4 \times 2\cos^2 (\alpha/2)\)

\(⇒ \cos (\beta/2) = 2 \cos (\alpha/2)\)