Force in belt

  • Thread starter Antu Ghosh
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  • #1
1. Problem Statement:
Sand is falling on a conveyor at a steady rate of 2kgs-¹.The motor of conveyor belt is rotating so that the belt is moving at a constant speed of 3.5m/s.what is force required to maintain the speed of the speed of the belt?

2. Relevant Equation:
F=v×(dm/dt)


The Attempt at a Solution


F=(3.5)×2
=7N
 

Answers and Replies

  • #2
tnich
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1. Problem Statement:
Sand is falling on a conveyor at a steady rate of 2kgs-¹.The motor of conveyor belt is rotating so that the belt is moving at a constant speed of 3.5m/s.what is force required to maintain the speed of the speed of the belt?

2. Relevant Equation:
F=v×(dm/dt)


The Attempt at a Solution


F=(3.5)×2
=7N
Your numerical answer is correct. However I think you should start with ##F = \dot p = \dot mv + m\dot v##, and work from there.
 
  • #3
haruspex
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Your numerical answer is correct. However I think you should start with ##F = \dot p = \dot mv + m\dot v##, and work from there.
Many, me included, object to that equation. In principle, it treats mass as something that can be created and destroyed, with momentum somehow being conserved. In reality, of course, it is entering or leaving the system to which p refers, and it may do so with momentum. The equation only works when it enters or leaves without momentum (as it does in this case).
 
  • #4
tnich
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Many, me included, object to that equation. In principle, it treats mass as something that can be created and destroyed, with momentum somehow being conserved. In reality, of course, it is entering or leaving the system to which p refers, and it may do so with momentum. The equation only works when it enters or leaves without momentum (as it does in this case).
You raise a good point. What is your preferred way to deal with situations like this?
 
  • #5
haruspex
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You raise a good point. What is your preferred way to deal with situations like this?
One way is to consider the mass change as exerting a force. If the departing/entering mass does so with velocity vΔm then the force that exerts is ##\dot mv_{\Delta m}##. So ##\dot{(mv)} = F_{applied}-\dot mv_{\Delta m}##, or ##m\dot v = F_{applied}-\dot m(v-v_{\Delta m})##.
Not sure that is the conceptually clearest, though.
 

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