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zaphod
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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.
 2 years ago
 2 years ago
zaphod Group Title
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.
 2 years ago
 2 years ago

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zaphod Group TitleBest ResponseYou've already chosen the best response.0
dw:1350219671114:dw
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
the force which the belt must exert on the sugar to accelerate it to the speed of the belt?
 2 years ago

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@Callisto @Coolsector @ParthKohli
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
now how do we demonstrate this in the diagram, becoz the weight down the slope should also be considered..
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
i think we should add the weight (affective part due to the angle) to the dp/dt term
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
can u show these forces in a diagram?
 2 years ago

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@VincentLyon.Fr
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
dw:1350220276938:dw
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
the dp/dt part acts for a short time
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
so where is force exerted on the sugar by belt?
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
dp/dt + mgx
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
the dp/dt is just to give it a speed of 3ms and the mgx is always there
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
no mgx is mgsin20
 2 years ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
You need to consider total momentum before and after an elementary mass dm of sugar joins the conveyor belt.
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
dp/dt is 18 * 3
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
so you have 18 * 9.8 * sin(20) + 18 * 3 = 114.33
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
this is the system that shows the forces on the sugar at the moment it touches the belt!
 2 years ago

zaphod Group TitleBest ResponseYou've already chosen the best response.0
okay il do it thanks :)
 2 years ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago
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