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Newton's 2nd Law(Dynamics)

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A series of small packages, each with a mass of 0.5 kg, are discharged from a conveyor belt as shown. Knowing that the coefficient of static friction between each package and the conveyor belt is 0.4, determine (a) the force exerted by the belt on the package just after it has passed point A, (b) the angle θ defining the point B where the packages first slip relative to the belt. (The speed of belt is constant.)

    25i0rrd.png

    2. Relevant equations

    sum of the forces in x = 0
    sum of the forces in y = 0

    3. The attempt at a solution

    I am stuck on the first part, I'm not sure how to incorporate the velocity given into the FBD of the package. Also, for part A would friction be omitted because the package won't be slipping relative to the belt?
     
  2. jcsd
  3. Oct 9, 2009 #2

    rock.freak667

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    For part A:

    What are the two forces acting on the belt at the point A? (At A it begins to move in a circle, so what is the resultant of these two forces equal to?)
     
  4. Oct 9, 2009 #3

    rl.bhat

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    Before A, the force on the belt is mg. Same force is exerted by the belt on the packet. But when it crosses A, it moves in the circular motion. So the belt will have the additional reaction. What is that? What are the forces acting on the packet at B?
     
  5. Oct 9, 2009 #4
    The normal force and the weight are acting on the package. But when it starts to move following the circle you don't know the angle of the normal force anymore. Would normal and tangential components of acceleration come into play when the package begins to move in a circle? Since you are given the radius of curvature and you know the velocity you can find the normal acceleration by an = v2/r
     
    Last edited: Oct 9, 2009
  6. Oct 10, 2009 #5

    rock.freak667

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    Well they told you that the velocity is constant, so what does that say about the component of tangential acceleration?

    At point A, the mass is just now starting to move in a circle, meaning that normal reaction - weight = ?
     
  7. Oct 10, 2009 #6
    Thanks for the help, I figured out part A. For part B would there still be no component of tangential acceleration?
     
    Last edited: Oct 10, 2009
  8. Oct 11, 2009 #7
    I can't seem to figure out part B. I tried using this FBD of the block but it isn't working. Where did I go wrong?

    w82phj.png
     
  9. Oct 11, 2009 #8

    rock.freak667

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    So at point B, the equation governing is

    N-Wy=mv2/r


    Can you find the velocity at point B?
     
  10. Oct 11, 2009 #9
    The velocity at point B wouldn't be 1m/s?
     
  11. Oct 11, 2009 #10

    rock.freak667

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    Nope, you can find the velocity at point B using conservation of energy.
     
  12. Oct 11, 2009 #11
    This question is from a chapter before learning conservation of energy, is there any other way to find the velocity?
     
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