The force exerted by the wind on the sails of a sailboat is Fsail = 445 N north.

Question

The force exerted by the wind on the sails of a sailboat is Fsail = 445 N north. The water exerts a force of Fkeel = 185 N east. If the boat (including its crew) has a mass of 305 kg, what are the magnitude and direction of its acceleration?

We choose east to be the positive x-direction and north to be the positive y-direction. The horizontal forces acting on the 305 kg boat (including crew) are shown in the diagram.





One method of finding the desired acceleration is to compute the resultant force R. For the magnitude of the force, where
?Fx = and ?Fy =

We have
FR = (?Fx)2 + (?Fy)2



= ? 105 N2 = N.

Explanation / Answer

magnitude is sqrt(185^2 + 445^2) = 481.92 N F = ma, so a is 481.92/305 = 1.58 m/s^2 direction is arctan(445/185) = 67.43 degrees. :)

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