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Clarification for inclined plane object, force components

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    there is a frictinless slope. And there is a box sliding down the slope.
    let's say that the slope slides from right, towards the down and left.

    object's mass is unknown value m.

    angle of the slope is 35deg.

    calculate the absolute value of the acceleration that the box has when it slides along the slope.


    2. Relevant equations

    F=ma
    trigonometric functions

    3. The attempt at a solution


    I will first draw some pictures and then ask about clarification, because I had some unresolved questions that I didn't have time to ask the teacher. I think my teacher thought those issues were maybe obvious to others in the class, but not to me, so I was also little bit embarrassed to ask.

    for me, the physics- based rationale was a little bit hazy. I know how to calculate angles using trigonometry and geometry, though.

    Using my earlier physics knowledge. I think that gravity somehow pulls the box sliding across the ramp. (gut feeling). I suppose, the more technical requirement in that case would be to calculate the acceleration and Fnet of the box, in the direction of being parallel to the ramp.

    sin35= Fnet/ G
    G*sin35= Fnet
    G*sin35=ma
    mg*sin35=ma | /m
    g*sin35 = a
    a = 5.62 m/s^2



    My understanding was basically that:

    1.)In the same "axis", in which the Normal force resides. There is no accelerations and the net force in this axis is zero. Because, the object stays flat upon the slope, whilst it slides down across the surface of the slope. This is a common sense understanding I suppose.


    2,) Does this mean that S as noted in the picture, is esseentially the counter vector to the Normal force ? [ one could calculate (-1)*(normal force vector) ]

    3.) Is S itself a vector and a force?
    4.) What is the definition of the S length in the picture? The Mystery length S is still a little bit of a mystery to me.

    5.) I'm not sure I understand the rationale why the length of Fnet, must by necessity, be one side of the triangle, and also such that the second side of the triangle must be the length of normal force. The hypotenuse would be the gravity, which I do understand,however. Because I suppose in this model the gravity is always downward.

    6.) can any force vector like gravity, always be broken up into components?
    I think one confusing bit for me was that the components of the gravity force, are written out into the same picture together with regular force like Normal force or gravity ....



    physics incline plane.jpg free body diagram box.jpg physics incline plane.jpg angles box.jpg
     
  2. jcsd
  3. Sep 22, 2016 #2

    haruspex

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    Your answer is correct.
    With your S and Fnet, you have resolved mg into its components in the coordinate system consisting of an axis parallel to the slope and an axis normal to the slope. The component normal to the slope must be equal and opposite to the normal force sincethere is no acceleration in that direction.
    Based on that, what do you think the magnitude of S should be?
     
  4. Sep 22, 2016 #3

    I already know standard convention for drawing vectors into free body diagram.

    What is the convention (notation) for drawing components of a force (like components of gravity in this problem)

    To my mind. For S length it is such that the S is equal size and opposite direction to the Normal Force. They cancel out.
    This is required to fulfill the idea that the box does NOT accelerate perpendicular to the ramp's surface. I could be wrong though. But this assumption is supported by the model of the sliding action across the ramp's surface.

    Are you allowed to add component normal (whatever that is?) Into the normal force vector, therefore calculating the sum of forces in that particular axis (perpendicular to the ramp's surface?)


    Does S equal a force and what is the name of that force strictly speaking?
    (I know that G in the picture means essentially "earth pulls the box")
     
  5. Sep 22, 2016 #4

    haruspex

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    That's what I wrote, isn't it?

    If by that you mean the normal component of the gravitational force, yes.

    the normal component of the gravitational force
     
  6. Sep 22, 2016 #5
    I thought that was what you asked.

    Well... anyway.

    I dont immediately see how you are supposed to calculate the normal force values because we lack the mass values and the force values.

    More about summing the normal component of gravity into the normal force.

    I thought essentially that you simply have components only in the y-axis and x-axis. Especially in twodimensional image like this.
    And with those components you can represent vectors in component form representation.

    I can ask my teacher more about the normal component of gravity tomorrow...
    That explanation I didnt quite understand how that could be the case.
     
  7. Sep 23, 2016 #6

    haruspex

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    Sure, but in terms of mass m and angle of the plane to the horizontal, what is the component of the gravitational force normal to the plane?
    Yes, but you can define x and y to be in any pair of orthogonal directions you like.

    Sorry if I'm not answering your questions. I am having difficulty understanding what it is that bothers you.
     
  8. Sep 23, 2016 #7

    David Lewis

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    My suggestion is to depict the components of a force as a dashed (or different color) line than the resultant force to emphasize that you may have either the components or the resultant, but not both.
     
  9. Sep 23, 2016 #8

    I had trouble internalizing the concept that

    F=ma

    Essentially the new definition according to my teacher was that Fnet is the vector sum of the forces on that object. And yes... the vectors of the gravity and normal force cause resultant force. The resultant force is the net force which gives the box its acceleration (I think)

    I forgot about that definition abd I was wonderring what purpose was all this vector calculus for. I.e. what was the physicsbased rationale for vector calculus.

    But that is essentially what my teacher said today when I asked about the issue.

    I did not do much vector calculation in either math or physics in high school and my previous university. And for that matter I did not really study very advanced high school courses in physics.

    My earlier physics homework problems didnt have much of the vector calculation stuff.
     
  10. Sep 23, 2016 #9

    David Lewis

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    You have two forces acting on the box: the force of gravity, and the force the ramp imposes on the box. When you add these two forces together (using vector addition) you will get the net force.
     
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