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zaphod

  • 3 years ago

A conveyor belt traveling at 3.0ms and at an angle of 20 to the horizontal has 18kg of sugar dropped on to it each second.

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  1. zaphod
    • 3 years ago
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    |dw:1350219671114:dw|

  2. zaphod
    • 3 years ago
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    the force which the belt must exert on the sugar to accelerate it to the speed of the belt?

  3. zaphod
    • 3 years ago
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    @Callisto @Coolsector @ParthKohli

  4. Coolsector
    • 3 years ago
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    ill give you a hint .. the sugar momentum changes when i gets on the belt use F = dp/dt and find out the force that the belt exert's on the sugar

  5. zaphod
    • 3 years ago
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    now how do we demonstrate this in the diagram, becoz the weight down the slope should also be considered..

  6. Coolsector
    • 3 years ago
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    i think we should add the weight (affective part due to the angle) to the dp/dt term

  7. zaphod
    • 3 years ago
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    can u show these forces in a diagram?

  8. zaphod
    • 3 years ago
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    @Vincent-Lyon.Fr

  9. Coolsector
    • 3 years ago
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    |dw:1350220276938:dw|

  10. Coolsector
    • 3 years ago
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    the dp/dt part acts for a short time

  11. zaphod
    • 3 years ago
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    so where is force exerted on the sugar by belt?

  12. Coolsector
    • 3 years ago
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    dp/dt + mgx

  13. Coolsector
    • 3 years ago
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    the dp/dt is just to give it a speed of 3ms and the mgx is always there

  14. Coolsector
    • 3 years ago
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    no mgx is mgsin20

  15. Vincent-Lyon.Fr
    • 3 years ago
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    You need to consider total momentum before and after an elementary mass dm of sugar joins the conveyor belt.

  16. Coolsector
    • 3 years ago
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    dp/dt is 18 * 3

  17. Coolsector
    • 3 years ago
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    so you have 18 * 9.8 * sin(20) + 18 * 3 = 114.33

  18. zaphod
    • 3 years ago
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    cool sector, i just need a system showing all this forces so tht i can resolve each of them and find force? is i possible?

  19. Coolsector
    • 3 years ago
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    this is the system that shows the forces on the sugar at the moment it touches the belt!

  20. zaphod
    • 3 years ago
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    okay il do it thanks :)

  21. Coolsector
    • 3 years ago
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    you can give the dp/dt a different name maybe it will look better for you.. the thing to understand is that at the moment of touch the belt exerts dp/dt + mgx on the sugar

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