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Finding the angle of the traveling direction of a sailboat

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    As a sailboat sails 55 m due north, a breeze exerts a constant force F1 on the boat's sails. This force is directed at an angle west of due north. A force F2 of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just 42 m. What is the angle between the direction of F1 and due north?

    2. Relevant equations
    Don't know if this pertains to adding the forces together.

    3. The attempt at a solution
    I assume that trigonometry is involved here to solve this problem. Other than that, that's how far I have gotten into the problem.
  2. jcsd
  3. Oct 3, 2011 #2


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    Well W=Fd, where the force F is in the same direction as the distance d.

    You know that F1=F2 from the problem.
    And W1=W2

    And, from the second case the work W2=F2 x 42m

    But as the sailboat is travelling due north, and a breeze is acting on the sails with a force F1 at a certain angle, west of due north. Here, the work W1 on the sailboat would be the component of F1 in the north direction and the distance 55m.
  4. Oct 3, 2011 #3
    So with this, W=F1(cosθ)d

    And plugging in the equations,
    F1(cos[itex]\Theta[/itex])55m = F2 X 42m (being that cos0°=1)

    divide F on both sides,
    cosθ55m = 42m

    divide 55m with 42m, and I assume that in the end,
    cosθ = .7(63 repeating)°
    cosθ = .76°

    Am I close?
  5. Oct 4, 2011 #4


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    0.76° --> the little dot goes on the angle but not on the cos of the angle. Sine, cosine, tangent come from the ratios of the sides of a traingle and so dimensionless.

    What angle θ will give cosθ=0.76.
    They ask for the angle, and, I presume, you have to state whether or not it is east of north or west of north angle.
    If you draw a diagram of the sailboat and forces you should be able to figure that out visually.

    to check if your answer is correct or not, put the values back into the equation.
    ie W2 = F2 x 42

    W1 = F1 cosθ x 55 = F1 0.76 x 55 = F1 x 42

    Since F1 = F2, then W1 = W2 and correct.

    You could do that for all problems you encounter as a check.
  6. Oct 4, 2011 #5
    I get the impression that the author of this problem doesn't understand how sailboats work.
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