Question

The resultant of the two forces `3N` and`2N` at an angle `theta` is doubled when first force in increased to `6N`. Find `theta`?

Answer 1

`theta=(2pi)/3` or `120^@`

Explanation:

Formula for resultant force of two forces `P` and `Q` is given by `R=sqrt(P^2+Q^2+2PxxQxxcostheta)`.

When two forces `3N` and `2N` are at an angle `theta`, value of the first resultant is `R_1` is given by

`R_1=Nsqrt(3^2+2^2+2xx3xx2xxcostheta)=sqrt(13+12costheta)`

When first force is increased to `6N`, the resultant force `R_2` is given by

`R_2=Nsqrt(6^2+2^2+2xx6xx2xxcostheta)=sqrt(40+24costheta)`

As `R_2` is doubled

`Nsqrt(40+24costheta)=2xxNsqrt(13+12costheta)` or

`40+24costheta=4xx(13+12costheta)=52+48costheta`

Hence, `24costheta=-12` or `costheta=-1/2`

and `theta=(2pi)/3` or `120^@`