The wind exerts a force of 452 N north on a sailboat, while the water exerts a force of 325 N west on the sailboat. What is the magnitude and direction of the net force on the sailboat?

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The wind exerts a force of 452 N north on a sailboat, while the water exerts a force of 325 N west on the sailboat. What is the magnitude and direction of the net force on the sailboat?

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Answer 1 · 2015-11-08T00:50:40

Direction = $45^{o}$ $45^o$ North of West (Due North-West)
Magnitude = 557 N

Explanation:

In this question you have a force (452N) due North and a force (325N) due West. Since these directions are $90^{o}$ $90^o$ apart from one another the angle of the net force is $45^{o}$ $45^o$ .

To calculate the magnitude you must reach WAAAAAY back in your math classes to the pythagorean theorem; since you can view these vectors as a triangle with the net force being the hypotenuse.
$a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$
${(325 N)}^{2} + {(452 N)}^{2} = 309929 N^{2}$ $(325N)^2 + (452N)^2 =309929N^2$

You can then take the square root of your answer to find the magnitude in Newtons.

$\sqrt{309929 N^{2}} = 557 N$ $sqrt(309929N^2) = 557N$

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The wind exerts a force of 452 N north on a sailboat, while the water exerts a force of 325 N west on the sailboat. What is the magnitude and direction of the net force on the sailboat?

1 Answer

Explanation:

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