A 301-kg boat is sailing 16.5° north of east at a speed of 2.08 m/s. Thirty seconds later, it is sailing 37.2° north of east at a speed of 3.60 m/s. During this time, three forces act on the boat: a 32.7-N force directed 16.5° north of east (due to an auxiliary engine), a 21.7-N force directed 16.5° 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. magnitude 1 N

direction 2 °

To find the magnitude and direction of the force W, we need to consider the forces acting on the boat and apply Newton's second law of motion.

Let's break down the given information:

1. Boat's initial speed and direction:
- Speed: 2.08 m/s
- Direction: 16.5° north of east

2. Boat's speed and direction after 30 seconds:
- Speed: 3.60 m/s
- Direction: 37.2° north of east

3. Forces acting on the boat:
- Auxiliary engine force: 32.7 N, directed 16.5° north of east
- Water resistance force: 21.7 N, directed 16.5° south of west
- Force W (due to the wind): To be determined

Now, let's calculate the net force acting on the boat:

1. Decompose the forces into their x and y components:
- Auxiliary engine force: 32.7 N x cos(16.5°) = 31.52 N in the x-direction
32.7 N x sin(16.5°) = 9.00 N in the y-direction
- Water resistance force: 21.7 N x cos(-16.5°) = 20.66 N in the x-direction
21.7 N x sin(-16.5°) = -6.30 N in the y-direction
- Force W: Wx in the x-direction and Wy in the y-direction

2. Calculate the total x and y components of the net force:
- Net force in the x-direction:
Fnet,x = 31.52 N + 20.66 N + Wx
- Net force in the y-direction:
Fnet,y = 9.00 N - 6.30 N + Wy

3. Apply Newton's second law: The net force equals the mass of the boat times its acceleration:
- Fnet,x = m * ax
- Fnet,y = m * ay

Since the boat is moving at constant velocity, its acceleration is zero. Therefore, the net force must be zero as well:

Fnet,x = 0
Fnet,y = 0

Now, we can solve for Wx and Wy:

31.52 N + 20.66 N + Wx = 0
9.00 N - 6.30 N + Wy = 0

Simplifying the equations:

Wx = -52.18 N
Wy = -2.70 N

Finally, we can find the magnitude and direction of force W:

Magnitude of force W: |W| = sqrt(Wx^2 + Wy^2) = sqrt((-52.18 N)^2 + (-2.70 N)^2) ≈ 52.3 N

Direction of force W: tan(θ) = Wy / Wx
θ = atan(Wy / Wx)
θ = atan((-2.70 N) / (-52.18 N))
θ ≈ 2°

Therefore, the magnitude of the force W is approximately 52.3 N, and its direction is approximately 2° with respect to due east.