A 358-kg boat is sailing 15.9° north of east at a speed of 1.80 m/s. Thirty seconds later, it is sailing 36.6° north of east at a speed of 3.68 m/s. During this time, three forces act on the boat: a 29.7-N force directed 15.9° north of east (due to an auxiliary engine), a 23.9-N force directed 15.9° south of west (resistance due to the water), and W (due to the wind). Find the magnitude and direction of the force W. Express the direction as an angle with respect to due east.

To solve this problem, we can start by breaking down the given information and finding the resultant force acting on the boat.

Given:
Mass of the boat (m) = 358 kg
Initial velocity of the boat (v1) = 1.80 m/s
Final velocity of the boat (v2) = 3.68 m/s

Forces acting on the boat:
Force due to auxiliary engine (FA) = 29.7 N, 15.9° N of E
Force due to water resistance (FR) = 23.9 N, 15.9° S of W
Force due to the wind (W) = ?

We need to find the magnitude and direction of the force W due to the wind.

Step 1: Find the change in momentum (Δp) of the boat.
Δp = mv2 - mv1

Step 2: Use the definition of impulse (J) to find the net impulse acting on the boat.
J = Δp

Step 3: Calculate the components of the net impulse along the horizontal and vertical directions.

Horizontal component:
Jx = J * cos(θ)
Jx = J * cos(θ) = (mv2x - mv1x)

Vertical component:
Jy = J * sin(θ)
Jy = J * sin(θ) = (mv2y - mv1y)

where θ is the angle of the net impulse with respect to the horizontal direction.

We can solve for Jx and Jy using the given information.

Step 4: Find the net horizontal force (Fx) and net vertical force (Fy) acting on the boat.
Fx = Jx / Δt
Fy = Jy / Δt

where Δt is the time interval of 30 seconds.

Step 5: Find the magnitude and direction of the net force acting on the boat.
F = √(Fx^2 + Fy^2)
θF = tan^(-1)(Fy / Fx)

Finally, we can solve for the magnitude and direction of the force W by subtracting the other known forces from the net force.

Magnitude of force W = √(Fx^2 + Fy^2) - √(FAx^2 + FAy^2 + FRx^2 + FRy^2)
Direction (angle with respect to due east) = θF - θA

Let's plug in the values to find the solution.

To solve this problem, we can break down the forces into their horizontal (x-axis) and vertical (y-axis) components. Let's consider the horizontal components first.

The boat is sailing on a 15.9° north of east direction. To find the horizontal component of this movement, we can use trigonometry. The cosine of the angle gives us the horizontal component:

Horizontal component = speed * cos(angle)

Using the first set of information, we can calculate the initial horizontal component:

Horizontal component1 = 1.80 m/s * cos(15.9°)

Similarly, for the second set of information:

Horizontal component2 = 3.68 m/s * cos(36.6°)

Next, let's consider the vertical components. Using the same logic, we can find the vertical component of the boat's movement using the sine of the angle:

Vertical component = speed * sin(angle)

Using the first set of information:

Vertical component1 = 1.80 m/s * sin(15.9°)

And for the second set of information:

Vertical component2 = 3.68 m/s * sin(36.6°)

Now, let's calculate the net horizontal and vertical components of the forces acting on the boat. We know that both components add up to give the net force.

Net horizontal force = Force1 (horizontal component) + Force2 (horizontal component) + Horizontal component1 + Horizontal component2

Similarly,

Net vertical force = Force1 (vertical component) - Force2 (vertical component) + Vertical component1 + Vertical component2

Now, let's substitute the known values:

Net horizontal force = (29.7 N * cos(15.9°)) + (23.9 N * cos(-15.9°)) + Horizontal component1 + Horizontal component2

Net vertical force = (29.7 N * sin(15.9°)) - (23.9 N * sin(-15.9°)) + Vertical component1 + Vertical component2

Simplifying these equations will give us the net force components. Then, we can calculate the magnitude and direction of the force W using the Pythagorean theorem and trigonometry.