Question

Consider two displacements, one of magnitude 6 m and another of magnitude 8 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 14 m, (b) 2 m, and (c) 10 m ?

Answer 1

Given first displacement `d_2=6` m

and 2nd displacement `d_2=8` m.

(a) when their resultant `R =14` m ,let the angle between the displacement be `alpha`

So

`R^2=d_1^2+d_2^2+2d_1d_2cosalpha`

`=>14^2=6^2+8^2+2*6*8*cosalpha`

`=>cosalpha=(14^2-6^2-8^2)/(2*6*8)=96/96=1`

So `alpha=0^@`

(b) when their resultant `R =2` m ,let the angle between the displacement be `beta`

So

`R^2=d_1^2+d_2^2+2d_1d_2cosbeta`

`=>2^2=6^2+8^2+2*6*8*cosbeta`

`=>cosbeta=(2^2-6^2-8^2)/(2*6*8)=-96/96=-1`

So `beta=180^@`

(c) when their resultant `R =10` m ,let the angle between the displacement be `gamma`

So

`R^2=d_1^2+d_2^2+2d_1d_2*cosgamma`

`=>10^2=6^2+8^2+2*6*8*cosgamma`

`=>cosgamma=(10^2-6^2-8^2)/(2*6*8)=0/96=0`

So `gamma=90^@`