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

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Answer 1 · 2016-03-07T05:29:30

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

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 $120^@$

The resultant of the two forces 3N3N and2N2N at an angle thetatheta is doubled when first force in increased to 6N6N. Find thetatheta?