A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.0-N force directed 15.0° north of east (due to an auxiliary engine), a 23.0-N force directed 15.0° south of west (resistance due to the water), and (due to the wind). Find the magnitude and direction of the force . Express the direction as an angle with respect to due east.

13 N 59.3 degrees NofE

To find the magnitude and direction of the force, we need to first calculate the net force acting on the boat.

Step 1: Calculate the net force in the x-direction (east).

The force due to the auxiliary engine is given by F_aux = 31.0 N at an angle of 15.0° north of east. We can break this force into its x and y components:

F_aux_x = F_aux * cos(15.0°)
F_aux_y = F_aux * sin(15.0°)

F_aux_x = 31.0 N * cos(15.0°)
F_aux_x = 29.787 N

F_aux_y = 31.0 N * sin(15.0°)
F_aux_y = 8.139 N

The force due to the water resistance is given by F_water = 23.0 N at an angle of 15.0° south of west. We also need to break this force into its x and y components:

F_water_x = -F_water * cos(15.0°)
F_water_y = -F_water * sin(15.0°)

F_water_x = -23.0 N * cos(15.0°)
F_water_x = -22.021 N

F_water_y = -23.0 N * sin(15.0°)
F_water_y = -6.267 N

The force due to the wind is denoted as F_wind.

The net force in the x-direction is given by:

F_net_x = F_aux_x + F_water_x + F_wind_x
F_net_x = 29.787 N - 22.021 N + F_wind_x
F_net_x = 7.766 N + F_wind_x

Step 2: Calculate the net force in the y-direction (north).

The net force in the y-direction is given by:

F_net_y = F_aux_y + F_water_y + F_wind_y
F_net_y = 8.139 N - 6.267 N + F_wind_y
F_net_y = 1.872 N + F_wind_y

Step 3: Calculate the magnitude and direction of the net force.

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

|F_net| = sqrt(F_net_x^2 + F_net_y^2)
|F_net| = sqrt((7.766 N + F_wind_x)^2 + (1.872 N + F_wind_y)^2)

Since the magnitude of the wind force (F_wind) is unknown, we can't directly calculate |F_net|. However, we can solve for F_wind by using information from the boat's motion.

The change in momentum of the boat is given by:

Δp = m * (v_f - v_i)

Where:
m = mass of the boat = 325 kg
v_f = final velocity of the boat = 4.00 m/s
v_i = initial velocity of the boat = 2.00 m/s

Δp = 325 kg * (4.00 m/s - 2.00 m/s)
Δp = 650 kg m/s

Since the net force is responsible for producing this change in momentum, we can write:

F_net * Δt = Δp

Where:
Δt = change in time = 30 s

(7.766 N + F_wind_x) * 30 s = 650 kg m/s

Now we need to solve this equation to find F_wind_x.

Step 4: Solve the equation to find F_wind_x.

7.766 N * 30 s + 30 s * F_wind_x = 650 kg m/s

232.98 N + 30 s * F_wind_x = 650 kg m/s

30 s * F_wind_x = 650 kg m/s - 232.98 N

F_wind_x = (650 kg m/s - 232.98 N) / 30 s

F_wind_x = 13.24 N

Now we can substitute this value back into the expression for F_net_x:

F_net_x = 7.766 N + F_wind_x
F_net_x = 7.766 N + 13.24 N
F_net_x = 20.00 N

Step 5: Substitute F_wind_x into the equation for F_net_y.

Knowing F_wind_x = 13.24 N, we can calculate F_net_y:

F_net_y = 1.872 N + F_wind_y
F_net_y = 1.872 N + F_wind_y

Since F_wind_y is unknown, we can't directly calculate F_net_y.

Step 6: Calculate the magnitude and direction of the net force.

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

|F_net| = sqrt(F_net_x^2 + F_net_y^2)
|F_net| = sqrt((20.00 N)^2 + F_net_y^2)

Again, since F_net_y is unknown, we can't directly calculate |F_net|.

To find the direction of the net force, we can use trigonometry. The angle θ can be calculated as follows:

θ = arctan(F_net_y / F_net_x)
θ = arctan(F_net_y / 20.00 N)

Since we don't have the value of F_net_y, we can't directly calculate the angle θ.