Question

When two forces (equal) are inclined at an angle $2\alpha$ their resultant is twice as great as when they act on angle $\beta$ . Show that $\cos \alpha =2\cos \beta /2$

Answer 1

When angle is $2\alpha$, the resultant force is,

$R_1=(F^2+F^2+2\times F\times F\times \cos 2\alpha {)}^{1/2}$

$R_1={\left(2F^2\left(1+\cos 2\alpha \right)\right)}^{1/2}$

Similarly, When angle is $\beta$, the resultant force is,

$R_2={\left(2F^2\left(1+\cos \beta \right)\right)}^{1/2}$

Now, according to the given condition,

$R_1=2R_2$

${R_1}^2=4\times {R_2}^2$

$2F^2\left(1+\cos 2\alpha \right)=4\times 2F^2\left(1+\cos \beta \right)$

$\left(1+\cos 2\alpha \right)=4\times \left(1+\cos \beta \right)$

$\cos \alpha =2\cos \frac{\beta }{2}$