zaphod
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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zaphod
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|dw:1350219671114:dw|
zaphod
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the force which the belt must exert on the sugar to accelerate it to the speed of the belt?
zaphod
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@Callisto @Coolsector @ParthKohli
Coolsector
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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
zaphod
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now how do we demonstrate this in the diagram, becoz the weight down the slope should also be considered..
Coolsector
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i think we should add the weight (affective part due to the angle) to the dp/dt term
zaphod
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can u show these forces in a diagram?
zaphod
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@Vincent-Lyon.Fr
Coolsector
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|dw:1350220276938:dw|
Coolsector
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the dp/dt part acts for a short time
zaphod
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so where is force exerted on the sugar by belt?
Coolsector
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dp/dt + mgx
Coolsector
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the dp/dt is just to give it a speed of 3ms and the mgx is always there
Coolsector
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no mgx is mgsin20
Vincent-Lyon.Fr
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You need to consider total momentum before and after an elementary mass dm of sugar joins the conveyor belt.
Coolsector
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dp/dt is 18 * 3
Coolsector
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so you have
18 * 9.8 * sin(20) + 18 * 3 = 114.33
zaphod
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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?
Coolsector
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this is the system that shows the forces on the sugar at the moment it touches the belt!
zaphod
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okay il do it thanks :)
Coolsector
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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