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Magnitude and direction of Wind Force

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    Force=ma
    Kinematic equations


    3. 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.
     
    Last edited: Sep 23, 2011
  2. jcsd
  3. Sep 23, 2011 #2
    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?
     
  4. Sep 23, 2011 #3
    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?
     
  5. Sep 24, 2011 #4
    oh im 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?
     
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