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Find the force on a conveyor belt

  1. Apr 14, 2014 #1

    PhysicoRaj

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    1. The problem statement, all variables and given/known data
    Sand is being poured onto a conveyor belt at the rate of 2kgs-1. Find the force required to move the belt at an acceleration of 3ms-2. Given the sand is not dumped off the belt at any point and the mass of the conveyor is negligible.

    2. Relevant equations
    F=ma=dP/dt
    F-ma+v(dm/dt)=0
    Conservation of momentum (:uhh:)

    3. The attempt at a solution
    I'm really stuck. Have I considered the correct equations? dm/dt and dv/dt are given. With only these two I can't find a place to start. Or maybe I'm missing something simple and common which makes this opaque for me. Thanks for the help!
     
  2. jcsd
  3. Apr 14, 2014 #2

    maajdl

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    F=ma only if m is constant.
    The general equation is F=dp/dt .
     
  4. Apr 14, 2014 #3

    PhysicoRaj

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    I think this is a varying force, determined by the mass on the belt at the instant and the instantaneous velocity, as is obvious from F=m[dv/dt]+v[dm/dt]. How can there be a specific value as an answer for this question?
    [Edit:]I'm starting to think the problem is wrong. Here are the options given:
    A]6N
    B]38N
    c]zero
    D]1.5N
     
    Last edited: Apr 14, 2014
  5. Apr 14, 2014 #4

    rude man

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    I agree with your thoughts. The force is not constant and the given answers are in error.

    However, there is still an answer to the question.
    Hint: how about dF/dt?
     
  6. Apr 15, 2014 #5

    PhysicoRaj

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    Thanks. That would mean d2P/dt2 which gives me.. 2(dm/dt)(dv/dt) [because acceleration is constant and dm/dt is constant]that would be 12N. But it's not given in the options. I think they were meaning velocity of 3ms-1 instead of accel. If it was so, then I would get F=m(dv/dt)+v(dm/dt)=v(dm/dt)=3*2=6N. Right?
     
  7. Apr 15, 2014 #6

    rude man

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    Well, the problem clearly stated an "acceleration of 3m/s^2" so I don't think they meant 3 m/s velocity. (If they did mean 3 m/s the answer would be F=6N as you say.)

    In your post 3 you speculated that a time-varying force was needed. So how can you wind up with a constant force (of 12N)? Had you checked dimensions you would have caught the error immediately.

    It would give you dF/dt = what you derived above, i.e. 2(dm/dt)(dv/dt) which is not F.
    So solve dF/dt = 2(dm/dt)(dv/dt) = 2a (dm/dt), what do you get for F?

    Hint: you need an initial condition for the complete solution. You can for example assume that at t=0 the mass just starts to pile up on the conveyor belt.
     
  8. Apr 15, 2014 #7

    PhysicoRaj

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    Ah.. I get it.
    Considering [itex]\frac{dF}{dt}=2a\frac{dm}{dt}[/itex]
    [itex]dF=2a.dm[/itex]

    [itex]F=2am[/itex]
    Am I right?

    At t=0, m=0 so F=0.
    At t=t(say), m=2t (bcz dm=2dt), so [itex]F(t)=4at[/itex]
    That would fetch me [itex]F(t)=12t[/itex]
    (At the end of one second, the force would be 12N.)
     
    Last edited: Apr 16, 2014
  9. Apr 16, 2014 #8

    rude man

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    Bingo! Good work.
     
  10. Apr 16, 2014 #9

    PhysicoRaj

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    Thanks a lot rude man!:smile:
    (and now for some LaTex practice..)
     
    Last edited: Apr 16, 2014
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