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Basic concept on vector

  1. May 13, 2012 #1
    Guys i really need your zealous help! It is a simple concept but i have some doubt on it. A vector can be resolve into its component vectors like unit vectors i,j & k. A single vector can be expressed into its i-components and j-components if it is a 2-dimensional vector. Is it when 2 vectors are equivalent, they must give the same consequences ? And when vectors are used to represent forces, a single force can be resolved into its x and y components. However if the force is acting perpendicular to the x-axis, supposedly it should have no y-components. But for inclined plane where the only force is the vertically downward gravitational pull on the object, we tend to resolve this vertical force into mg.cosθ or mg.sinθ which both these components vector having a horizontal components that cancel out each other! But in inclined plane, it do impart effect to the motion of the object. Why this could happens?? Lets say if we have a quantity 5 and we say this 5 is equal to quantity 10-5. But can that quantity 10 impart any effects on the current situation??
     

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  3. May 13, 2012 #2
    what do you mean by saying that if the force is perpendicular to x axis, it will not have any y component and you can resolve any vector in 2D along any two perpendicular axes lying in the plane also since the force is vertical,it's horizontal component would be zero.
     
  4. May 13, 2012 #3

    Stephen Tashi

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    This is good question and it is not well explained in some textbooks that use the diagram that you show.

    The proper general procedure to analyze the motion of a body is to first diagram the forces that act upon the body, not the forces exerted by the body upon other objects. So the proper diagram for an object on an inclined plane should show the forces exerted upon it. These forces are the force of gravity downward (mg) and the force of the incline plane, which pushes on the body perpendicular to the surface of the plane (N). Having the other forces in the diagram causes confusion.

    The next step is find the combined effect of the forces acting upon the body. The vector sum of the force mg and the force N is the force labeled mg sin(theta). This is the force that is relevant to analyzing the motion of the body. It is the net force acting upon the body.

    The other forces in the diagram (f and mg cos(theta)) are not relevant to the motion of the body. They are relevant to other questions. The force f is the force necessary to keep the body from sliding. The force mg cos(theta) is the force exerted by the body upon the inclined plane.

    You are correct that the x-components of the vectors labeled mg cos(theta) and mg sin(theta) would cancel out. However, the vector mg cos(theta) is not relevant to the motion of the body. The relevant forces upon the body are N = - mg cos(theta) and g. The x components of the forces N and g do not cancel out.
     
  5. May 13, 2012 #4

    sophiecentaur

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    A bit more 'basics' here. A force can be 'resolved into contributions from any number of appropriate forces in any number of directions. It is a useful but totally artificial thing to do and the choice is totally arbitrary and made to suit your convenience. One of the early problems students have with vectors is actually 'knowing' which choice is best; the worked examples in books just choose one way and don't often say why.
    The reason that we resolve into two orthogonal components (or three, when we're considering three dimensional situations) is that you can treat the two components independently. If you can identify a force such as weight, then it is often convenient to choose one of the components as Vertical. If you are dealing with friction on a slope, then it may be convenient to do your resolution parallel and normal to the slope. What will make it convenient is the fact that the equations that you will draw up about the situation will be easier to solve because they will each have fewer variables in them.

    Vector calculations are, effectively, a set of simultaneous equations and, when we are given a set of simultaneous equations to solve, the first thing we have to do is to try an 'eliminate' one of the variables by the familiar jiggery-pokery of multiplying one equation and then subtracting etc. Choosing the 'best' direction for resolving produces the simplest set of equations to start with and just gives you less manipulation to do in order to solve them.
    If you are happy to solve complicated sets of simultaneous equations then you don't need to be so smart about the choice of where to resolve.
    It just makes sense to choose orthogonal vectors for your resolution - or you are really making life hard for yourself.
     
  6. May 13, 2012 #5
    They don't cancel each other, only in special case of 45 degrees. But even so, that doesn't mean the body won't have horisontal motion.
    You seem to be confused by the apparent "emergence" of new forces as we decompose one force. Well, we can decompose into any number of forces but they still act on same body at the same time. Therefore you can't completely analyse motion taking into account only one component of force.
    Yes, but only simultaneously with -5. Don't forget that we replace one force with two forces.
    So it's not
    but "we tend to resolve this vertical force into mg.cosθ and mg.sinθ
     
  7. May 15, 2012 #6
    Your diagram is wrong.
    You either have a single force = mg Or 2 forces, mgSinθ(parallel to plane) and mgCosθ(tangent to plane).

    It just a replacement from single force to 2 forces. Once you resolved to sum of other vectors it does not exist anymore by itself.

    10=6+4
    If you take existence of 6 or/and 4, 10 does not exist until you add 4 to 6.
     
    Last edited: May 15, 2012
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