The force exerted by the wind on the sails of a sailboat is fsail = 410 n north. the water exerts a force of fkeel = 200 n east. if the boat (including its crew) has a mass of 300 kg, what are the magnitude and direction of its acceleration?

Question

Caesar

Physics

21 April, 22:23

The force exerted by the wind on the sails of a sailboat is fsail = 410 n north. the water exerts a force of fkeel = 200 n east. if the boat (including its crew) has a mass of 300 kg, what are the magnitude and direction of its acceleration?

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Aurora Harrell · Answer 1

The North and East forces are 90° apart, making a right triangle with the resultant as the hypotenuse.

(200) ² + (410) ² = F²

40,000 + 168,100 = F²

√ (208,100) = F

456.18 N = F

F = ma

456.18 N = (300 kg) a

456.18 / 300kg = a

1.52 m/s² = a

The sailboat will be heading North East. To find the angle of the boats trajectory use inverse tangent function.

tan (Ф) = opposite/adjacent

arctan (opposite/adjacent) = Ф

arctan (410/200) = 64° North East