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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.

Physics
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|dw:1350219671114:dw|
the force which the belt must exert on the sugar to accelerate it to the speed of the belt?

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Other answers:

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
now how do we demonstrate this in the diagram, becoz the weight down the slope should also be considered..
i think we should add the weight (affective part due to the angle) to the dp/dt term
can u show these forces in a diagram?
@Vincent-Lyon.Fr
|dw:1350220276938:dw|
the dp/dt part acts for a short time
so where is force exerted on the sugar by belt?
dp/dt + mgx
the dp/dt is just to give it a speed of 3ms and the mgx is always there
no mgx is mgsin20
You need to consider total momentum before and after an elementary mass dm of sugar joins the conveyor belt.
dp/dt is 18 * 3
so you have 18 * 9.8 * sin(20) + 18 * 3 = 114.33
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?
this is the system that shows the forces on the sugar at the moment it touches the belt!
okay il do it thanks :)
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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