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Find the force on a conveyor belt

  • Thread starter PhysicoRaj
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PhysicoRaj

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1. Homework Statement
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. Homework 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!
 

maajdl

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F=ma only if m is constant.
The general equation is F=dp/dt .
 

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:

rude man

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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
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?
 

PhysicoRaj

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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?
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?
 

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.
 

PhysicoRaj

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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?
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.)
 
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rude man

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Ah.. I get it.
Considering dF/dt = 2a(dm/dt)
dF=2a(dm)
F=2am
Am I right?
At t=0, m=0 so F=0.
At t=t(say), m=2t (bcz dm=2dt), so F(t)=4at
That would fetch me F(t)=12t
(At the end of one second, the force would be 12N.)
Bingo! Good work.
 

PhysicoRaj

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Thanks a lot rude man!:smile:
(and now for some LaTex practice..)
 
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