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

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

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Answer 1 · 2016-03-07T05:29:30

$θ = \frac{2 π}{3}$ $theta=(2pi)/3$ or $120^{\circ}$ $120^@$

Explanation:

Formula for resultant force of two forces $P$ $P$ and $Q$ $Q$ is given by $R = \sqrt{P^{2} + Q^{2} + 2 P \times Q \times cos θ}$ $R=sqrt(P^2+Q^2+2PxxQxxcostheta)$ .

When two forces $3 N$ $3N$ and $2 N$ $2N$ are at an angle $θ$ $theta$ , value of the first resultant is $R_{1}$ $R_1$ is given by

$R_{1} = N \sqrt{3^{2} + 2^{2} + 2 \times 3 \times 2 \times cos θ} = \sqrt{13 + 12 cos θ}$ $R_1=Nsqrt(3^2+2^2+2xx3xx2xxcostheta)=sqrt(13+12costheta)$

When first force is increased to $6 N$ $6N$ , the resultant force $R_{2}$ $R_2$ is given by

$R_{2} = N \sqrt{6^{2} + 2^{2} + 2 \times 6 \times 2 \times cos θ} = \sqrt{40 + 24 cos θ}$ $R_2=Nsqrt(6^2+2^2+2xx6xx2xxcostheta)=sqrt(40+24costheta)$

As $R_{2}$ $R_2$ is doubled

$N \sqrt{40 + 24 cos θ} = 2 \times N \sqrt{13 + 12 cos θ}$ $Nsqrt(40+24costheta)=2xxNsqrt(13+12costheta)$ or

$40 + 24 cos θ = 4 \times (13 + 12 cos θ) = 52 + 48 cos θ$ $40+24costheta=4xx(13+12costheta)=52+48costheta$

Hence, $24 cos θ = - 12$ $24costheta=-12$ or $cos θ = - \frac{1}{2}$ $costheta=-1/2$

and $θ = \frac{2 π}{3}$ $theta=(2pi)/3$ or $120^{\circ}$ $120^@$

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

1 Answer

Explanation:

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