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Why weight on an incline is different?

  1. Oct 19, 2011 #1
    Hey guys, first post here...
    So, I'm having an issue comprehending this topic:
    Let's say you have a box which weighs 3 N, and you place a 7 kg block on top of it. So, since the block is laying on a horizontal plane, the weight is mg, or 6*10=60 N. Total weight=60+3=63

    Let's say you have an inclined box 45 degrees to the ground (held by a stick of negligible mass, or whatever... just pretend that it is inclined) that weighs 3 N and you place a 6 kg block on top of it, and the friction causes it to remain at rest. My teacher said that the weight here is mgsin45, because Normal=mgsin45 at this case. So, (6)(10)(2....)=120ish N. Total weight=120+3=123.

    What I don't understand, is why isn't the weight in BOTH cases mg? Because, if you put both systems each in a separate balance, the weight will be 63 N for both, amirite? The masses certainly don't change, you have the SAME box and the SAME block, except they are at different angles.

    Por favor clarify :D
     
  2. jcsd
  3. Oct 19, 2011 #2
    grams are a unit of weight. Edit (opps a unit of MASS, not weight)

    (mass)(gravitational acceleration) is measured in newtons which is a unit of force. F=ma

    The force along the plane is mgsin45 since the plane is at a 45 degree angle to the direction of force.

    So the force in both cases is mg, unless your measuring the force only along the plane.
     
    Last edited: Oct 19, 2011
  4. Oct 19, 2011 #3
    I guess I understand that, but if a question asks, what is the weight, then it is mg for both cases right? I just have a point of contention on a test which I want to argue, and I put mg but my teacher said it was mgcostheta

    EDIT: My bad on the first post, my teacher says that it is mgcostheta (the y component of F which also equals the Normal force.), not mgsintheta.
     
  5. Oct 19, 2011 #4
    Oh yes, if the question asks only "What is the weight" it is mg. That is almost like a trick question. However if the question is asking what the force is along the plane then it is not simply mg.

    The word weight is usually only used for something pressing straight down due to being in a gravitational force. Like measuring something on a scale, a rocket on a platform, or standing on a scale.
     
  6. Oct 19, 2011 #5
    Hopefully my teacher will understand our logic :P Thanks a bunch
     
  7. Oct 19, 2011 #6

    rcgldr

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    The component of gravitational force normal to the plane is m g cos(θ), the component of gravitational force parallel to the plane is m g sin(θ), the vector sum of these forces is m g. If the block and the box are not sliding (no vertical component of acceleration), then the total force is m g, regardless of the angle.
     
  8. Oct 19, 2011 #7
    I would approach it as a simple misunderstanding, and that you thought it might be a trick question. Let he/she come to the conclusion that your brighter then the average, don't go in holding the attitude that you are.
     
  9. Oct 20, 2011 #8
    If you put a scale on the incline then the weight shown would indeed be different. However, the scale would not be showing the weight of the box, but just a component of that weight.

    So...yeah, I think your teacher needs to reword his questions in the future. Perhaps he
    may argue "that's not what I meant", but there should be no question as to what he meant.

    By the way, there are alot of errors in your first post...
     
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