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Laws of Motion

  1. Jun 26, 2011 #1
    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 31.0N force directed 15.0° north of east, a 23.0N force directed 15.0° south of west and F
    Find the force F
    Answer : [18.4N, 68.0° north of east]

    Fr = F1 + F2 + F3 (due to 3 forces acting on it)

    F3 = Fr - F1 - F2 (as question wanted to look for the 3rd force)

    F1 = 31cos15° = 30.0N (to the right)

    F2 = -(23cos15) = -22.2N (to the right. Or +ve to the left)

    a = (v - u)/t

    Do I need to search for the vertical, y and the horizontal, x component of the velocity to turn it into resultant Force? Through square root of (Fy^2 +Fx^2)? I do not understand the first 2 sentences of the question. So in the end they will turn into the resultant force acted on the boat? I am confused.

    It is a relatively new topic in mechanics to me. Im sorry I could not even make a single line of solution.. >< i have no idea where to start calculating actually. Though I did forces of equilibrium before, this doesn't seems like anything from it.
     
    Last edited: Jun 26, 2011
  2. jcsd
  3. Jun 26, 2011 #2

    Doc Al

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    Staff: Mentor

    But what about the change in velocity? What might that have to do with force?
     
  4. Jun 26, 2011 #3
    Oh so I would use the change in velocity to find the acceleration and thus make it into my resultant force? Hmm which means :

    Uy = 2sin15° = 0.52m/s Vy = 4sin35° = 2.29m/s
    Ux = 2cos15° = 1.93m.s Vx = 4cos35° = 3.28m/s

    In which I have t=35s

    Which means my a for x and y-component would be :

    x : a = (3.28-1.93m/s)/35 = 0.04m/s^2
    y : a = (2.29-0.52m/s)/35 = 0.05m/s^2

    and from there I find my Fr which is

    x : Fr = 13.0N
    y : Fr = 16.3N

    Then Fr = 20.8N?

    Edit : Just tried solving it, I din get the amount as given by the answer. Im not sure where have I gone wrong.
     
  5. Jun 26, 2011 #4

    Doc Al

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    Staff: Mentor

    Recheck this value for time.
     
  6. Jun 26, 2011 #5
    Hey I got my answer! Thanks for the help Doc! I understood some part wrongly but I fixed my understanding now.
    I think I should have no problem with the other questions now. Appreciate your real direct-to-the-point replies. Have a good day sir!
     
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