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mechanics trouble

Discussion in 'Mathematics' started by fudgesweets, Feb 26, 2012.

  1. fudgesweets

    fudgesweets New commenter

    I have read the following example:

    Forces F1=2i+4j F2=-5i+4j and F3=6i-5j act on a particle of mass 3kg. Find the acceleration of the particle,

    Resultant force= F1+F2+F3
    =3i+3j
    3i+3j=3a thus a=(i+j) ms-2

    When finding the resultant force why is not the weight of the particle considered? i.e. -3gj N is not included.....what have i misunderstood??

    Thankful to any responses
     
  2. fieldextension

    fieldextension New commenter

    If the weight was in the direction of the k vector then it certainly WOULD have an effect on the acceleration.
     
  3. fieldextension

    fieldextension New commenter

    The question means "Only forces F1=2i+4j F2=-5i+4j and F3=6i-5j act on a particle of mass 3kg."
    So either one of those forces is the weight (presumably in a gravitational field other than earth's, since none of them has magnitude of 3x9.8) or else the process is happening somewhere there is no (significant) gravitational field.
     
  4. This is not true. If you think about a connected particle on a smooth table connected to a particle hanging off the edge of the table (with a pulley on the edge), the weight of the particle on the table is not involved in F=ma. The weight is equal to the Normal Reaction force.
    Similarly, for this question the table is the i-j plane and the weight is acting down (in the k direction) so not causing the acceleration.
     
  5. stewarty

    stewarty New commenter

    Please explain why it WOULD have an effect.
    It is perpendicular to the other two so surely it would have NO effect.
     
  6. Some of you seem to be imagining a table.
     
  7. arsinh

    arsinh New commenter

    Not me. I'm imagining a concrete floor.
     
  8. The original question was very clear, albeit highly artificial.
    Why complicate things with gravity, smooth tables, not to
    mention spoon benders and other extraneous influences.
    BTW It is a very heavy particle, or perhaps I should have
    said a very massive particle.
     
  9. fieldextension

    fieldextension New commenter

    What I wrote is correct. The reason that weight is not involved in the F=ma example you mention is that there is a force of equal magnitude acting in the opposite direction to the weight (the normal reaction). All forces should be included, but those two cancel each other out so in practice you can leave them out.
    An object which has horizontal forces acting on it, but is in freefall in a gravitational field certainly does have its acceleration influenced by its weight (otherwise it would not be falling!). The resultant force would be the vector sum of all the forces acting on it (including the weight) and the acceleration would be in the same direction. By your way of thinking, a rocket fired horizontally from a cliff would never fall to the ground!
     
  10. But the question doesn't say it is in freefall, you've imagined this, so why can't I imagine that it is on a table (or a concrete floor)?

     
  11. fieldextension

    fieldextension New commenter

    I nowhere said that the question involves the object being in freefall. I brought that issue up to make another point entirely. I said that the question entails that either:
    (a) The object in in deep space, far from any gravitational fields; or
    (b) The object is in some gravitational field other than earth's (since none of the stated forces have magnitude 3x9.8).

    You say that the object is on a table. I do not see how that might work. Which of the three forces might be the normal reaction? Which might be the weight?
     
  12. They are two completely different forces not given in the question because they are not in the plane/direction of motion (i-j plane).
    I'm not saying that this is what is happening but it is one way of looking at it. IME many pupils try to include the weight (as the OP did) and if they think of it as forces on a horizontal plane (e.g. table) they can see that the weight is not involved as it is balancing the reaction force. They then solve these type of questions correctly and don't include extra forces.
    You are over-complicating things, in Mechanics modelling is used to simplify situations.

     
  13. fieldextension

    fieldextension New commenter

    I suppose you might interpret the question that way. It would mean that in addition to the three forces mentioned in the question there are at least another two forces which they failed to mention. Maybe there are even more. On my understanding, there are just the three forces stated in the question. You are over-complicating what is relatively simple by adding numerous assumptions and extra forces nowhere mentioned in the question. Three forces are simpler than five or more. Other people here are similarly mystified about the assumption that there must be a table or something similar involved in the question.


     
  14. Ahh, but we are ignoring two of them (as many things in Mechanics are ignored!)
    I'm just using it to explain to a pupil a possible reason why weight is not included, rather than just say "ignore the weight"!

    I think we'll leave it here shall we...

     
  15. fieldextension

    fieldextension New commenter

    You are inventing two of them, then telling your students to ignore the two additional forces you invented. That does not seem simpler to me, but maybe it is a matter of taste.

    I wouldn't say "Ignore the weight." I would say "There are only three forces acting so presumably there is no weight." I don't want my pupils to have the misconception that everything has a weight and that all mechanics problems happen on a table or something similar on planet earth. I also would not want to encourage the idea that you can import unjustified assumptions into a question.
     
  16. Very well put. IMHO the vast majority of maths teachers would agree with fieldextension on this matter.
     
  17. GoldMaths

    GoldMaths New commenter

    Why would you not assume the particle was in equilibrium before the three forces were applied? This means you can ignore all other forces as you are only looking at the resultant of the "new" forces.
     
  18. Me neither. I can SEE a table quite clearly.
     
  19. fieldextension

    fieldextension New commenter

    Because the question does not mention any forces other than the three it mentions, not even ones which cancel each other out. The idea that there were other forces, which still act in addition to the three mentioned, and cancel each other out, is a needless and unjustified complication.
     
  20. fieldextension

    fieldextension New commenter

    I agree. I would never bring these issues up unprompted, but it is not unreasonable or suprising for students to ask questions such as "Do we need to include weight as one of the forces? How can there possibly be no weight? How might a problem such as this exist in the real world even in principle?" Then I think we ought to be able to give them a good answer.

     

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