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Classical mechanics: motion and acceleration

  1. Dec 30, 2017 #1
    1. The problem statement, all variables and given/known data

    Referring to the attached diagram, the object ABCD is kept in an isolated system where it is under no external influence of any kind.

    A· An instantaneous force F acts along the edge BC of the block ABCD, where M is its mass, with its center of mass located at O.

    B· X-Y is the coordinate system that has been used to resolve the aforementioned force

    C· The component of force towards O (F cosθ) is further resolved using another coordinate system X'-Y'.

    Here are the questions:

    1. Is choosing these kinds of coordinate systems legit in mechanics (meaning, one would intuitively take AB and BC as the X and the Y coordinates respectively. Would it be wrong if someone doesn't)?

    If yes, then:

    2. Will ABCD have an instantaneous acceleration in the X' direction which is equal to (F cosθ(cosα))/M, when F is applied?

    3. Will ABCD have an instantaneous acceleration in the Y' direction which is equal to (F cosθ(sinα))/M, when F is applied?

    4. What will the F sinθ do (i.e will it have any effect on the motion of ABCD)?

    5. Which component of the force F will cause rotation in ABCD and about which point?

    6. If the edges BC and AD are reduced to 0, then will F cause pure rotational motion and no translation motion?

    2. Relevant equations
    No such equations

    3. The attempt at a solution
    I intentionally framed this question in my head so that the concept of choosing coordinates becomes clearer. So, as such I have not reached a solution as yet.
  2. jcsd
  3. Dec 30, 2017 #2


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    You can choose any coordinate system that you please and it would not be wrong to do so. The object's subsequent motion after the instantaneous force is applied will not be different because you chose your coordinates one way and not another. Only the mathematical description of the motion will differ but not in any significant way. The magnitudes of all vectors will be the same.

    Personally, I prefer to choose a system that makes the writing of vectors as simple as possible. In this case, I would pick my axes along BA and BC. Note that in this system, the force has only a vertical component which is equal to its magnitude. Also note that you must have made a mistake from going through your XY system to your X'Y' system: The X'Y' system is parallel to the BA-BC system. How can force F have both a vertical and a horizontal component in X'Y' when it points straight up along BA? Can you spot your mistake?
  4. Dec 31, 2017 #3
    But choosing the axes, like i have done here, is showing that the vertical force F is indeed giving a horizontal component. And i haven't yet spotted the mistake. Please help me solve this.
  5. Dec 31, 2017 #4


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    This has already been answered with yes.



    Yes. It is part of the overall force acting on the system. The centre of mass acceleration is given by the total force acting on the system and that includes all components of all forces. Now, that component will also lead to a non-zero torque relative to the centre of mass, also leading to an angular acceleration, but this does not mean you can ignore it for computing the linear acceleration.

    I suggest studying this in the non-inertial frame where the centre of mass is at rest. (Note that, in that frame, the centre of mass is a fixed point since the total force on the object, including inertial forces, is zero.)

    Will reducing those sides to zero lead to the total force on the object being zero?

    Edit: I suggest you do the following experiment. Take a pen and place it on the floor. Swipe your finger fast to hit the pen at the end orthogonal to the pen itself. How does the pen move? Does it just rotate or does it also fly away from you?
  6. Dec 31, 2017 #5


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    You have drawn the ##F\cos\theta## component in the X'Y' frame. Fair enough. What happened to the ##F\sin\theta## component? Why was that omitted? Draw it in and see what happens. It will be instructive. Also, angle ##\alpha## is related to angle ##\theta##. How? The answer will help you simplify the final result.
  7. Jan 1, 2018 #6
    Thank you very much @kuruman and @Orodruin
    I can now progress with solving it :smile: now I can mark it solved!
  8. Jan 9, 2018 #7
    Can we ask questions on here?
  9. Jan 9, 2018 #8


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    You can ask questions and post replies here only if they are directly related to the original posting or to one of the replies. Otherwise, you have to start a new thread. If your question is homework-related, be sure to read the guidelines and use the template.
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