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Boat in river: vectors

  1. Jul 26, 2004 #1
    Let me know if the following statement is wrong :
    If a boat is going to cross a river directly ( i.e perpendicularly ) the x, horizontal, component of the boat velocity must be cancelled, that's be equal, by the water stream vector.

    I followed this idea in a problem but my answer is not true according to the solution manual included in the textbook.
  2. jcsd
  3. Jul 26, 2004 #2
    The statement looks right to me (per my understanding). Maybe you have an error somewhere in your calculations. What is the solutions manual's explanation?
  4. Jul 26, 2004 #3
    Here is both the problem and explanation :
    If the water flows at 20 mi/h, and the boat at 30 mi/h in what angle would you head the boat to cut directly across the river?

    v_BE cuts directly across the river. Head at v_BW somewhat upstream at an angle Theta such that v_BE = v_BW + v_WE, sin (Theta) = (20 mi/h)/(30 mi/h) = 0.667 so Theta = 42 degrees.

    I assumed the horizontal component of v_boat to be 20 mi/h, so the vertical component became 22 mi/h then the sin(Theta) = (22 mi/h)/(30 mi/h) which leads to a Theta equal to 48.
    What is wrong ?
  5. Jul 26, 2004 #4

    Doc Al

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    Staff: Mentor

    Since the water moves east (I presume) at 20 mi/h, the boat must move at an angle so that its east-west component is 20 mi/h west. Thus [itex]30 cos\theta = 20[/itex], which gives the boat's needed direction as [itex]\theta = 48\deg[/itex] north of west.
  6. Jul 28, 2004 #5

    This is my answer too, but the answer in the solutions manual is 42 deg, so you mean that's wrong ?
  7. Jul 28, 2004 #6

    Doc Al

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    Staff: Mentor

    42 deg with respect to what? Realize that 48 deg N of W = 42 deg W of N. :smile:
  8. Jul 28, 2004 #7

    :biggrin: :biggrin: :biggrin: :biggrin: :redface:
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