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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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zaphod Group TitleBest ResponseYou've already chosen the best response.0
dw:1350219671114:dw
 one year 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?
 one year ago

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@Callisto @Coolsector @ParthKohli
 one year 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
 one year 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..
 one year 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
 one year ago

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

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@VincentLyon.Fr
 one year ago

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

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

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

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

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

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

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

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