Find the force on a conveyor belt

[PhysicoRaj]

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.

Homework Equations

F=ma=dP/dt
F-ma+v(dm/dt)=0
Conservation of momentum ()

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]

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

 

[PhysicoRaj]

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

 

[rude man]

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]

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]

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]

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 \frac{dF}{dt}=2a\frac{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.)

 

[rude man]

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]

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