Mass falling to a scale from a funnel

[Eitan Levy]

Homework Statement

A closed funnel is 10 meters above a weighing scale. At t=0 we open the funnel. and 15kg of sand are coming out of it every second. The velocity of the sand when it comes out is 13m/s, directed downwards (towards the scale). At a certain moment the scale displays a value of 700N. What is the mass of sand on the scale at this moment?

Assume the sand loses its velocity the moment it hits the scale.

Homework Equations

F=dP/dt
Kinematics

The Attempt at a Solution

I am having trouble understanding why the mass won't simply be 700/g. If the sand is on the weight it has no velocity, no acceleration - it has a constant velocity of 0, doesn't it?

Any help would be appreciated.

 

[BvU]

The scale does not measure mass; it measures force and divides by g . Is that a good hint ?

 

[Eitan Levy]

The scale does not measure mass; it measures force and divides by g . Is that a good hint ?

Still struggling to understand why the equation will not be mg=700. Is there something that applies force on the scale but the sand on it? If it's only the sand, the force on all of it would be the mass of it * g, and the normal applied by the scale will be equal to this.

 

[gneill]

What's the difference between the sand coming from the funnel and the sand already on the scale?

 

[Eitan Levy]

What's the difference between the sand coming from the funnel and the sand already on the scale?

The sand coming from the funnel has a velocity different then 0 (and a momentum different then it). How does that affect the scale though?

 

[jbriggs444]

The sand coming from the funnel has a velocity different then 0 (and a momentum different then it). How does that affect the scale though?

Is a force required to slow it down?

 

[CWatters]

So the velocity of the sand changes. That means there is acceleration going on. What does Newton day about that?

Edit: cross posted with jbriggs.

 

[Eitan Levy]

Is a force required to slow it down?

I thought about it, but in the question it says that the sand loses its velocity the moment it hits the scale. I think that means that a very large force, perhaps infinite, is applied on the sand for a very short time. Can I use an infinite F in dP/dt=F? Is that the way to go?

 

[jbriggs444]

In an infinitesimal interval, dt, the incremental mass is infinitesimal and undergoes a finite change in velocity. F=dp/dt is finite.

 

[Eitan Levy]

In an infinitesimal interval, dt, the incremental mass is infinitesimal and undergoes a finite change in velocity. F=dp/dt is finite.

If I get this correctly, I am given the force the scale applies on the sand on it at a certain moment (700N). 700=F+Mg, where F is the force it applies on sand hitting it at that moment, and M is the mass of sand on it with no velocity at that moment.

I am struggling to calculate F because I have never seen something like that before.

F=dP/dt=m'(t)*v(t)+m(t)*v'(t). I think m'(t)=0 because the change in momentum is due to the force applied by the scale and not by a change of mass. F=dP/dt=m(t)*v'(t)=m(t)*dv/dt. I know the velocity of the sand right when it hits the scale, by I can't figure how to continue to calculate F. Could you provide some guidance?

 

[CWatters]

If I get this correctly, I am given the force the scale applies on the sand on it at a certain moment (700N). 700=F+Mg, where F is the force it applies on sand hitting it at that moment, and M is the mass of sand on it with no velocity at that moment.

I am struggling to calculate F because I have never seen something like that before.

F=dP/dt...

The flow rate of sand is given in kg per second and you are given the before and after velocity. Try using that equation and think about the right hand side as "change in momentum per second". I mean literally what is the change in momentum that occurs in one second.

 

[Eitan Levy]

The flow rate of sand is given in kg per second and you are given the before and after velocity. Try using that equation and think about the right hand side as "change in momentum per second". I mean literally what is the change in momentum that occurs in one second.

Another question, is the 700=F+Mg even correct? Do I need to add mg, where m is the mass of sand that hits the scale at the moment.

I think the mass of sand that hits the scale every second is 15/t, where t is the time it takes the sand to reach the to the scale once it starts to fall. The sand also hits the scale with a velocity equals to V, that we can calculate. If it's correct, the momentum lost every second is 15V/t. How can I reach the force though? Mathematically speaking, I can't realize how to continue with F= F=dP/dt=m(t)*v'(t)=m(t)*dv/dt from this point. As I mentioned before I have no experience with this kind of problems.

 

[CWatters]

Another question, is the 700=F+Mg even correct? Do I need to add mg, where m is the mass of sand that hits the scale at the moment.

Yes its correct. There are only two forces that contribute to the 700N. 1) The force due to the sand just sitting on the pan and 2) the force due to the changing momentum of sand hitting the pan.

I think the mass of sand that hits the scale every second is 15/t, where t is the time it takes the sand to reach the to the scale once it starts to fall.

No. The mass flow rate of the sand hitting the pan is the same as that leaving the funnel = 15 kg per second. It all goes in the pan. None misses the pan. None accumulates in mid air.

The _velocity_ with which the sand hits the pan might be different to the _velocity_ with which it leaves the funnel because gravity causes it to accelerate.

[strike]
However i think we will have to ignore any increase as we don't know the height it fall. Assume its a short distance so the difference is negligible.
[/strike]

Edit: Oops the height is given see posts below.

It's less complicated than you are making it...

In one second 15kg changes velocity by ?? m/s so dP/dt= ?

 

[gneill]

However i think we will have to ignore any increase as we don't know the height it fall.

A closed funnel is 10 meters above a weighing scale.

We do know the height of fall.

 

[CWatters]

Oh bother. How did I miss that. OK well its easy work out how much faster its going when it hits the pan. Just apply equations of motion.