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Statics: 3D Equilibrium Problem r X F or Fd?

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data

    The 25-kg rectangular access door is held in the 90° open position by the single prop CD. Determine the force F in the prop and the magnitude of the force normal to the hinge axis AB in each of the small hinges A and B.

    w0361.jpg

    2. Relevant equations
    ΣF=0
    ΣM=0



    3. The attempt at a solution

    I actually solved this problem entirely already. I used point A as the origin and I solved by first taking the sum of the moments about the y-axis (since A and B have zero moment arms there, and this just left the weight of the door and the force in the prop CD). I then used the same methods for the other axes until all the forces were found.

    My question for everyone is: what works better and is more easier in this situation? Using the cross product or taking (F)(d)? I am having a debate with some of my classmates who think the cross product is a better approach. I think (F)(d) works better because all of the distances are already given in the problem and their would be no need to find position vectors and do lengthy cross products(I think that F(d) is much easier, since each cross product has a total of six separate products to compute). Let me know what you all think and how you might have solved it.

    Josh
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2016 #2

    Simon Bridge

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    The cross product will always be right - but the one that works better (in the sense you are more likely to get the marks) is the one you can use most reliably. That will be different for different people.
    Bottom line: define "better".
     
  4. Oct 30, 2016 #3
    In this particular case, by "better" I mean, which of the two approaches would be the most expedient in finding the solution to this problem? In problems like these, there is usually a number of approaches that can be taken to solve them. Sometimes I have used the cross product when it was probably easier to use force times distance to calculate the moment. Also, I have used force times distance when the cross product would have worked more efficiently. I think what you said makes a lot of sense. That different people will use different approaches, and the one that is most comfortable to them. Thanks!
     
  5. Nov 4, 2016 #4

    Simon Bridge

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    expedient = convenient and practical although possibly improper or immoral? Maybe you mean "easier for you to use"?
    ... I know you already aid: I think what you said makes a lot of sense. That different people will use different approaches, and the one that is most comfortable to them. Thanks!" ... and thank you.
    What I am illustrating here is how carefully asking the question leads you to the answer... this is most of what the scientific method is about.
    As soon as you realise you want to know the method you, personally, find easiest, you realise that nobody else can answer that question :)
     
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