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Resolution of components

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Revered Members,
    Kindly see my attachments.

    2. Relevant equations

    This is Atwood machine dealing with motion on a frictionless inclined plane.

    3. The attempt at a solution
    To better visualize the forces involved, i isolated m1 and m2 and drawn free body diagrams for both. For mass m1, there are three concurrent forces, that is, T,m1g,and N, where T is the force in the string because of Tension and N is the normal force of the table on the block. The assumption is m1 accelerates up the plane, which is taken in x direction. Now, the weight m1g is broken down into components. The x component is in the assumed direction of acceleration , and the y component acts perpendicularly to the plane and is balanced by the normal force N.
    I have shown the components in the attachment1. My question is
    1) why m1gcosθ and m1gsinθ can't be interchanged, as given in my second attachment?
    2) what is wrong in changing the angle θ, that lies between m1g and m1gcosθ( first attachment) to between m1gsinθ and m1g?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2011 #2
    You can either work with θ or with (90 - θ), the latter angle you can call β. But, of course, you cannot call them both θ in the same problem.
     
  4. Oct 3, 2011 #3
    But invariably in all the books i referred , the Tension is always balanced by the mgcosθ and not by mgsinθ. My teacher said, that 90 - θ should be the sine component always and θ should be cos component. Is that so?
     
  5. Oct 3, 2011 #4
    See the attachment.
     

    Attached Files:

  6. Oct 4, 2011 #5
    Oh! Beautifully explained grzz. Thanks a lot. I got it now.
     
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