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Finding the couple vector of this ice boat

  1. Aug 30, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and its solution are attached as TheProblemAndSolution.jpg.

    2. Relevant equations
    Vector addition. Cross product.

    3. The attempt at a solution
    Given that I am just supposed to take Fig. 1 for what it is based on the way the problem is phrased, my biggest confusion is in understand how Fig. 1 was translated to Fig. 2 as well as in interpreting Fig. 2.

    Firstly, I feel it shouldn't matter what the orientation of the boat is since the forces we are considering in the diagram will have the same magnitudes and they will all be pointing in the same directions with respect to each other. More specifically, defining the axes to be aligned with the orientation of the boat, the 20 lb force on the sail seems to me like it should be 5 feet upward in the same direction as the i_3 vector while directed in the direction of the i_2 vector but it should not be 6 feet from the tail of the i_2 vector. The other 20 lb force vector's tail seems to me like it should be placed on the top-right corner of the bottom face of the rectangular prism directed toward the opposite direction of the i_2 vector instead of it being translated 6 feet in the direction opposite the i_2 vector from where I feel it shoul dbe. As for the 10 lb force pointing downward, I agree with where it is. For the upward pointing 10 lb force vector, I also feel it should be on the top-right of the bottom face of the rectangular prism but, unlike 20 lb force vector that is not applied to the sail, it should be pointing in the direction of the i_3 vector.

    Looking online, I found: (1) “In mechanics, a couple is a system of forces with a resultant (a.k.a. net, or sum) moment but no resultant force.” and (2) “Another term for a couple is a pure moment”.

    What I found online makes sense to me so I get what a couple vector is and I also get the algebra in computing it. That is, it's the non-zero vector sum of all the torques (applied per force) where the net force is zero.

    To me, the r_n and f_n numbers obtained for the 10 lb and 20 lb forces do not seem consistent with the information provided by Fig. 1.2.

    If you need me to say more, just ask.

    Any help in solving this problem would be greatly appreciated!
    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2012 #2
    The only way to arrive at the solution, in my opinion, is to completely forget about Fig.2. It is, as you pointed out, inconsistent with Fig.1. This problem is so badly written the author ought to be ashamed of him/herself.

    Whenever you take the moment of a system of forces, it is always with respect to a specific point in space, we'll call it P. If you have the freedom to choose both P and the origin, then you can choose to make them the same point, P = (0,0,0). This is helpful because if you place P = (0,0,0) on the origin of a force vector, you know that its moment about that point P is zero, because r = (0,0,0).

    We seemingly have that freedom in this problem (although if the solution hadn't been provided, there would have been no indication of that). Looking at Fig.1, we can place our origin (0,0,0) and the point P, about which we calculate the moment, at the end of the left-runner, where the 10 and 20-lb forces originate. If we do that, we don't have to worry about their individual moments, because once again r = (0,0,0).

    Again, we're going to let our orthonormal basis originate from the end of the left-runner where the 10 and 20-lb forces originate (so that we only have to calculate two individual moments instead of four). Let i1 point toward the end of the iceboat (that is, in the direction of the 20-lb force crossed with the 10-lb force), let i2 point toward the body of the iceboat (that is, in the direction of the 20-lb force on the left-runner), and let i3 point directly upward.

    Now, using the basis described above, you should be able to calculate the individual moments for the remaining two forces. Unlike the author, you should get r20-lb = (3,6,5) crossed with f20-lb = (0,-20,0) for the second individual moment. Add the two individual moments together, and you should arrive at the author's answer.

    It wasn't at all clear to me why the author chose to take the moment about the end of the left-runner, other than for the sake of reducing the amount of math involved. Perhaps there's another way of reconciling the two figures, or of choosing a basis such that you obtain the same individual moments as the author. But I wouldn't spend too much time trying to decipher the author's cryptic (and likely flawed) thought processes. If this problem is out of a book you're required to use for class, I'd recommend getting a supplementary book. If you'd like suggestions on some good statics texts, just send me a PM. Or if you have more questions about the problem, feel free to ask.
     
    Last edited: Aug 30, 2012
  4. Aug 31, 2012 #3
    The iceboat is a rigid body. O force acting on a rigid body at some particular point may be moved to another point at that same body along the line of its direction. That's what the author has done. The 20 lb force acting on the runner was moved to the zero point to simplify the computations. That's a useful technique, master it.
     
  5. Sep 7, 2012 #4

    s3a

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    Sorry for the late reply. After having read what you guys said and thought about the problem for a long time, I can now say that I fully understand it :) . Thank you.

    To add, I can now see for myself that the second figure is technically correct but inconsistent with the first figure. In other words, it seems that steps were skipped in going from the first figure to the second figure.
     
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