Question

A particle is subjected to two equal forces along two different directions. If one of them is halved, the angle which the resultant makes with the other is also halved. The angle between the forces is

Answer 1

using the cosine formula for both conditions and dividing them

Answer 2

The “angle between the forces” is $120^{\circ}$

To find:

The ‘angle between’ the given forces.

Solution:

Let F be the two forces acting on a particle and θ be the angle between them.

Suppose R be the resultant of the two forces.

$R^{2}=F^{2}+F^{2}+2 \times F \times F \times \cos \theta$

$\Rightarrow R^{2}=4 F^{2} \cos ^{2} \frac{\theta}{2}$

When one of the force is halved, the angle which the new “resultant force “makes with the “other is halved”.

$\frac{F}{R}=\frac{F}{R}$

(Using internal angle bisector theorem)

$\Rightarrow \mathrm{R}=\mathrm{F}$

$\Rightarrow 2 F \cos \theta$

$=\frac{1}{2}$

$=\frac{\left(\cos 60^{\circ}\right)}{2}$        

$\Rightarrow \frac{\theta}{2}=60^{\circ}$  

$\Rightarrow \theta=120^{\circ}$