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The Component of Forces

  1. Mar 4, 2012 #1
    I have two questions:

    Pic 1: An object is placed on top of a spring scale which strapped to a horizontal surface. They slide down on an incline of 30 degrees. Friction is ignored. What fraction of the weight does the spring scale read of the top object? (Only incline is given, and friction is ignored)

    Would the weight of the topmost object change at all? Would the topmost objects normal force be on an angle or still straight up because it is on a horizontal surface. Does acceleration affect weight at all?

    I drew what I thought the FBD would look like on the topmost object only including Normal force and weight, by this I think the weight is not affected.


    Pic 2: If there is an horizontal force, does the horizontal force go to the middle of the larger circle on an angle or does the force continue horizontally and only apply at the bottom of the larger circle? Also is it possible to lift the larger circle vertically with a horizontal force?

    I drew what I thought would happen to the applied force horizontally.
     

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    Last edited: Mar 4, 2012
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  3. Mar 4, 2012 #2

    SammyS

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    attachment.php?attachmentid=44735&d=1330910812.png

    I assume that there are three objects stacked together on an incline. If so, your FBD for the top object is incorrect.

    How about stating the whole problem ? I presume there is some friction involved as well as other details.
     
  4. Mar 4, 2012 #3
    An object is placed on top of a spring scale which strapped to a horizontal surface. They slide down on an incline of 30 degrees. Friction is ignored. What fraction of the weight does the spring scale read of the top object? (Only incline is given, and friction is ignored)

    The problem is I do not know how the forces will react on each other, so I am confused with where to start. That is what I assumed the FBD to be because an incline is given.

    I know I have to use the formula Fnormal = mgcos30, but not to sure how this will affect the object on the top.

    I was looking up about spring scales and I found somewhere which said that a spring scale detects the Normal force. Does that mean I have to find the Normal force acting on the spring scale?

    I did not post the problem because I just wanted an explanation of what is happening rather than any calculations.
     
    Last edited: Mar 4, 2012
  5. Mar 4, 2012 #4

    SammyS

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    I'll reply to your last comment, first.

    This is fine, but without letting us know that you're doing this, we have a difficult time giving much help, because the problem is incomplete.
    _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

    Each of the objects should have its own FBD .

    The top object sits on a horizontal surface. By definition, the normal force is perpendicular to the surface.

    What is the vertical component of the acceleration of the combination of the three objects ?
     
  6. Mar 4, 2012 #5
    Vertical force should be Fg = mg, but how is there vertical acceleration? Should it not be zero since the Normal force and Force of gravity balance eachother?

    Or would it be a = mtotalgcos30 - mtotalg / mtotal
     
    Last edited: Mar 4, 2012
  7. Mar 4, 2012 #6

    SammyS

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    These object slide down the incline with some acceleration, right?

    That acceleration is parallel to the incline. What component of that acceleration is in the vertical direction?
     
  8. Mar 4, 2012 #7
    Fparallel = mgsin30

    ma = mgsin30
    a = gsin30

    Fperpendicular = mgcos30

    ma = mgcos30(Fn) - mgcos30(Fg in perpendicular direction)
    a = gcos30 - gcos30
    a = 0

    In the vertical direction alone:
    Fg = mg
    ma = mg - Fnormal
    a = 0
     
    Last edited: Mar 4, 2012
  9. Mar 4, 2012 #8

    SammyS

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    If the magnitude of the acceleration is a = g sin30°, then the vertical component is a cos(30°) = g sin(30°) cos(30°)

    Multiply by m to get the vertical component of the net force.

    After this, an accurate FBD for the top object will help complete the task.
     
  10. Mar 4, 2012 #9
    I am sorry but I do not understand, how does "a cos(30°) = g sin(30°) cos(30°)" get vertical component, it gives the same answer as "g sin(30°)" which is parallel to the incline.

    The good news is I found my FBD problem.
     
  11. Mar 5, 2012 #10

    SammyS

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    How is it possible that g sin(30°) cos(30°) is the same as g sin(30°) ?

    a cos(30°) is the vertical component of the acceleration, a.
     
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