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Model the forces as constant over a short interval of time

  1. Oct 10, 2007 #1
    The figure below shows a worker poling a boat-a very efficient mode of transportation-across a shallow lake. He pushes parallel to the length of the light pole, exerting on the bottom of the lake a force of 225 N. Assume the pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of 30.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 360 kg.

    (a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force.

    (b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.400 s after the moment described.



    I got the answer for (a), 3.34e3 N. I calculate:
    [Force of gravity] = [mass][gravity] = [360kg][9.81m/s^2] = 3531.6 N
    [Normal force] = [Force exerted on bottom of the lake]*cos30 = 225N*cos30 = 194.9 N
    [Magnitude of the force] = [Force of gravity]-[Normal force] = 3531.6 N-194.9 N = 3337N



    But I couldn't find the answer for (b) What I got so far is to calculate the:

    (Net force) = (Horizontal drag force) + [(Force exerted on the bottom of the lake)*(cos
    60)] = -47.5N + (225*cos 60) = 65.0N

    then find accerleration, a

    a = F/m = 65.0N/360kg = 0.180556m/s^2

    then find v(0.400)

    v = at = 0.180556m/s^2 * 0.400s = 0.07222m/s

    then add both velocities

    0.007222m/s + 0.857m/s = 0.9292m/s = 9.29e-1m/s

    and I don't understand why it's a wrong answer. Can you figure out why?
     
    Last edited: Oct 10, 2007
  2. jcsd
  3. Oct 10, 2007 #2

    Doc Al

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    Staff: Mentor

    Why cos(70)? You want the horizontal component of the pole force.
     
  4. Oct 10, 2007 #3
    yes, I want the horizontal component of the pole force
     
  5. Oct 10, 2007 #4

    Doc Al

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    But where did the 70 degrees come from? The only angle given was 30 degrees.
     
  6. Oct 10, 2007 #5
    oh...my bad...the angle given is 30 degrees and to look for the horizontal component should be 60 degrees. sorry

    ***i recalculate on the first reply of this topic
     
    Last edited: Oct 10, 2007
  7. Oct 10, 2007 #6

    Doc Al

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    Good. That's better.

    Your revised speed looks OK to me. Why do you say its wrong?
     
    Last edited: Oct 10, 2007
  8. Oct 10, 2007 #7
    well when I submit the answer, 9.29e-1, it comes out like this

    "Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

    but I never rounded any values while I'm calculating. So I was wondering did I do calculations right or it is completely wrong?
     
  9. Oct 10, 2007 #8

    Doc Al

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    I don't see anything wrong with your method or calculation. I get the same numbers. (Sometimes those online systems can be flaky.)
     
  10. Oct 10, 2007 #9
    So even your own work and calculations come out with a same answer?
     
  11. Oct 10, 2007 #10

    Doc Al

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    Yes.
     
  12. Oct 10, 2007 #11
    Do you know any way to make this question correct so it doesn't show up this reason below:

    "Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error."

    I only got two submissions left
     
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