Question

two force act on a particle in the ratio 1:2 resu;tant of these forces is three times the frist force angle between then is

Answer 1

Since the magnitude of the two forces acting on the particle is 1:2, let the forces be $\vec{\sf{F}}$ and $\vec{\sf{2F}}.$ Let the angle between them be $\theta.$

Given that the magnitude of their resultant force is three times the first force, so the resultant force will be $\vec{\sf{3F}}.$

Then the magnitude of the resultant force will be,

$\longrightarrow\sf{3F=\sqrt{F^2+(2F)^2+2\cdot F\cdot 2F\cos\theta}}$

$\longrightarrow\sf{3F=\sqrt{F^2+4F^2+4F^2\cos\theta}}$

$\longrightarrow\sf{9F^2=5F^2+4F^2\cos\theta}$

$\longrightarrow\sf{4F^2=4F^2\cos\theta}$

$\longrightarrow\sf{4F^2\cos\theta-4F^2=0}$

$\longrightarrow\sf{4F^2\left(\cos\theta-1\right)=0}$

Since $\vec{\sf{F}}$ is non - zero,

$\longrightarrow\sf{\cos\theta-1=0}$

$\longrightarrow\sf{\cos\theta=1}$

Since $\theta$ is assumed to be between $\sf{0^o}$ and $\sf{180^o,}$

$\Longrightarrow\sf{\underline{\underline{\theta=0^o}}}$

Hence $\bf{0^o}$ is the answer.