The force exerted by the wind on the sails of a sailboat is Fsail = 440 N north. The water exerts a force of Fkeel = 160 N east. If the boat (including its crew) has a mass of 250 kg, what are the magnitude and direction of its acceleration?

Use the method applied here:

http://www.jiskha.com/display.cgi?id=1304278772

To find the magnitude and direction of the boat's acceleration, we need to consider the net force acting on the boat.

First, let's break down the forces into their components. The force exerted by the wind, Fsail = 440 N, is directed north. The force exerted by the water, Fkeel = 160 N, is directed east. We can represent these forces as vectors:

Fsail = 440 N north = (0 N east, 440 N north)
Fkeel = 160 N east = (160 N east, 0 N north)

Since we are interested in the net force, we can add these vectors component-wise:

Fnet = Fsail + Fkeel
= (0 N east, 440 N north) + (160 N east, 0 N north)
= (160 N east, 440 N north)

Now, we can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

Fnet = m * a

Plugging in the values, we have:

(160 N east, 440 N north) = 250 kg * a

To find the magnitude of the acceleration, we can use the Pythagorean theorem:

|a| = sqrt((160 N)^2 + (440 N)^2)

|a| = sqrt(25,600 N^2 + 193,600 N^2)
= sqrt(219,200 N^2)
≈ 468.20 N

The direction of the acceleration can be found by taking the inverse tangent of the y-component over the x-component of the acceleration vector:

θ = arctan((440 N)/(160 N))

θ ≈ 69.30°

Therefore, the magnitude of the boat's acceleration is approximately 468.20 N and it is directed at an angle of approximately 69.30° north of east.