A 316-kg boat is sailing 12.0° north of east at a speed of 1.60 m/s. 39.0 s later, it is sailing 36.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.4-N force directed 12.0° north of east (due to an auxiliary engine), a 20.7-N force directed 12.0° south of west (resistance due to the water), and (due to the wind). Find the (a) the magnitude and (b) direction of the force . Express the direction as an angle with respect to due east.

Fn = 31.4N @ 12o,CCW + 12.7N @ 192o,CCW.

X=Hor.=31.4*cos12+12.7*cos192=18.3 N.
Y=Ver.=31.4*sin12+12.7*sin192=3.89 N.

a. Mag.=sqrt((18.3)^2+(3.89)^2=18.7 N

b. tanA = Y/X = 3.89/18.3 = 0.21257
A = 12o, CCW = 12o North of East.

To find the magnitude and direction of the net force acting on the boat, we need to analyze the forces acting on the boat separately and then combine them using vector addition.

First, let's break down the given forces:

1. The force due to the auxiliary engine is 31.4 N and directed 12.0° north of east. We can express this force as a vector:

Fa = 31.4 N (12.0° north of east)

2. The resistance force due to the water is 20.7 N and directed 12.0° south of west. To calculate the vector form, we change the direction to its opposite:

Fw = -20.7 N (12.0° north of east)

3. The force due to the wind is unknown and is directed 36.0° north of east. We need to find this force.

Now, let's find the force due to the wind:

The net force acting on the boat is equal to the mass of the boat multiplied by its acceleration. In this case, the acceleration is due to the change in the boat's velocity over the given time interval.

We know the initial and final velocities of the boat, so we can calculate the change in velocity:

Δv = vf - vi
= (4.00 m/s) - (1.60 m/s)
= 2.40 m/s

The change in the velocity vector gives us the direction of the force due to the wind:

Δv = Δxî + Δyĵ

Since we are given that the boat is sailing north of east, the change in velocity has both an x and y component:

Δx = Δv * cos(36.0°)
= (2.40 m/s) * cos(36.0°)
≈ 1.945 m/s

Δy = Δv * sin(36.0°)
= (2.40 m/s) * sin(36.0°)
≈ 1.459 m/s

The force due to the wind can be expressed as a vector:

Fwind = Δxî + Δyĵ
= (1.945 m/s)î + (1.459 m/s)ĵ

Now, let's combine all the forces acting on the boat:

Fnet = Fa + Fw + Fwind
= (31.4 N) (12.0° north of east) + (-20.7 N) (12.0° north of east) + (1.945 m/s)î + (1.459 m/s)ĵ

To find the magnitude and direction of the net force, we can calculate its components:

Fnet,x = (31.4 N * cos(12.0°)) + (-20.7 N * cos(12.0°)) + 1.945 N
Fnet,y = (31.4 N * sin(12.0°)) + (-20.7 N * sin(12.0°)) + 1.459 N

The magnitude of the net force can be calculated using the Pythagorean theorem:

|Fnet| = sqrt((Fnet,x)^2 + (Fnet,y)^2)

The direction of the net force can be found using trigonometry:

θ = atan(Fnet,y / Fnet,x)

Plug in the values above to find the magnitude and direction of the net force acting on the boat.