The force exerted by the wind on the sails of a sailboat is Fsail = 425 N north.

Question

The force exerted by the wind on the sails of a sailboat is Fsail = 425 N north. The water exerts a force of Fkeel = 255 N east. If the boat (including its crew) has a mass of 255 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 255 kg boat (including crew) are shown in the diagram.

One method of finding the desired acceleration is to compute the resultant force FarrowboldR. For the magnitude of the force, where ?Fx = N and ?Fy = N, we have FR = (?Fx)2 + (?Fy)2 = cross 105 N2 = N.

we have

FR = (?Fx)2 + (?Fy)2 = 105 N2 = N. The force exerted by the wind on the sails of a sailboat is Fsail = 425 N north. The water exerts a force of Fkeel = 255 N east. If the boat (including its crew) has a mass of 255 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 255 kg boat (including crew) are shown in the diagram. One method of finding the desired acceleration is to compute the resultant force FarrowboldR. For the magnitude of the force, where ?Fx = N and ?Fy = N, we have FR = (?Fx)2 + (?Fy)2 = cross 105 N2 = N.

Explanation / Answer

Fx = 255i....Fy = 425j

Fnet = 255i+425j = Fxi+Fyj

F net = sqrt(Fx^2+Fy^2) = 495N

a = Fnet /m = (255i+425j)/255= 1i+1.67j

lal =sqrt(ax^2+ay^2) =1.9436m/s^2

direction = theta = tan^-1(ay/ax) = 59.086 with east

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