Question #15905

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Question #15905

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Answer 1 · 2017-07-04T01:48:23

The resultant force will be $6000 \sqrt{3} N$ $6000sqrt3N$ .

Explanation:

Suppose, $- - \to A B$ $vec (AB)$ and $- - \to A C$ $vec (AC )$ are the two boats and $- - \to A D$ $vec( AD)$ is the direction of ship.
We know the formula of resultant is

$A D = \sqrt{A C^{2} + C D^{2} + 2 \times A B \times C D \times cos 60}$ $AD=sqrt(AC^2+CD^2+2×AB×CD×cos 60$

$A D = \sqrt{2 {(6000)}^{2} + 2 \times 6000 \times 6000 \times (\frac{1}{2})}$ $AD=sqrt(2(6000)^2+2×6000×6000×(1/2)$

$A D = \sqrt{3 \times 6000^{2}}$ $AD=sqrt(3×6000^2)$

$A D = 6000 \sqrt{3}$ $AD=6000sqrt3$ final answer

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Question #15905

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