Magnitude and direction of Wind Force

[tryton]

Homework Statement

1. Homework Statement
A 373-kg boat is sailing 13.0 ° north of east at a speed of 2.00 m/s. 32.0 s later, it is sailing 33.0 ° north of east at a speed of 3.90 m/s. During this time, three forces act on the boat: a 33.1-N force directed 13.0 ° north of east (due to an auxiliary engine), a 22.9-N force directed 13.0 ° south of west (resistance due to the water), and Fw(due to the wind). Find the (a) the magnitude and (b) direction of the force Fw. Express the direction as an angle with respect to due east.

Homework Equations

Force=ma
Kinematic equations

The Attempt at a Solution

33.1N-22.9N=10.2N Narrowed down the forces as these two are facing the same direction
M=373Kg
V0=2.0m/s
vx0=1.95m/s
vy0=.450m/s
v=3.9m/s
vx=3.27m/s
vy=2.13m/s
theta1=13degrees
theta2=33degrees

I am not sure where to go from here. What is missing is x,y,a, and Force of wind.
Can get x,y,a from breaking Force down into the x and y components and then use that to find F=ma for x=v0t+1/2at^2 but then I don't know if that is right and still don't know how to get the force of the wind...

Is that the right direction to go in at least?
Thanks for any help on this one.

 

[Rayquesto]

I think you are supposed to find the resultant force that the boat makes with the auxiliary engine and the water. Do you know how to add components and find the resultant vector?

 

[tryton]

I realized that part of the question didn't show up. I need to find the Fw, the Force of the wind.

I can add the two forces but I don't know how to separate out Fw.

so
Fa=Force auxiliary engine
Fax=22.1cos13=34.25
Fay=33.1sin13=7.45

Fwc=Force of watercurrent and wind
Fwcx=-29.9cos13=-29.1
Fwcy=-29.9sin13=-6.73

Added the Net Force is
Fx=5.12
Fy=.720

Then
F=ma
ax=5.12/373=.0137
ay=.720/373=.00193

a= square root of (.0137 ^2+.00193^2)=.0139

arctan(.00193/.0137)=8.006degrees

But then that is both water AND wind ...
and they just want the force of the wind.

x=V0t+1/2at^2
x= 2.0m/s * 32.0s + 2 * .0139m/s^2 * 32.0s^2
x=92.47m

Do I need to find Y as well? and then add x and y?

 

[Rayquesto]

oh I am sorry. I didnt see the Fw part. Well, the next question is, what do you think, conceptually, the net force is going to be, if there is a net velocity of 3.9m/s during that instentaneous time at 32 seconds?

 

[rwisz]

I would approach this problem by first identifying the known and unknown variables. From the given information, we know the mass of the boat (373 kg), its initial velocity (2.00 m/s) and final velocity (3.90 m/s), and the three forces acting on it (33.1 N, 22.9 N, and Fw). We also know the directions of the boat's motion (13.0 ° north of east and 33.0 ° north of east).

To find the magnitude and direction of the force Fw, we can use the equation F=ma, where F is the net force acting on the boat, m is the mass of the boat, and a is the acceleration of the boat. We can also use the kinematic equations to relate the initial and final velocities, time, and acceleration.

First, let's find the acceleration of the boat. From the given information, we can calculate the change in velocity in the x and y directions: Δvx = 3.27 m/s - 1.95 m/s = 1.32 m/s, and Δvy = 2.13 m/s - 0.45 m/s = 1.68 m/s. Using the kinematic equation vf = vi + at, we can find the acceleration in each direction: ax = Δvx/t = 1.32 m/s / 32 s = 0.04125 m/s^2, and ay = Δvy/t = 1.68 m/s / 32 s = 0.0525 m/s^2.

Next, we can use the Pythagorean theorem to find the magnitude of the net force acting on the boat: F = √(33.1 N)^2 + (22.9 N)^2 = 39.8 N. Since we know the mass of the boat, we can now use the equation F=ma to find the magnitude of the force Fw: 39.8 N = (373 kg)(a). Solving for a, we get a = 0.107 m/s^2.

Now, we can use the equations for projectile motion to find the components of the force Fw in the x and y directions. In the x direction, the only force acting on the boat is the force Fw, so we can use the equation Fx = max to find Fwx = (373