Total time wagon needs to fully pass the hopper

In summary, the train wagon passes the sand hopper in a straight line and it takes T=M/C*(exp(CL/MV)-1) for the wagon to completely pass.
  • #1
freutel
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Homework Statement


A train wagon of mass M moves on a rail with constant velocity V (without friction). It passes a sand hopper which pours sand in the wagon at constant rate C [kg/s]. The sand falls vertically so it does not transfer any horizontal momentum to the wagon. The length of the wagon is L [m].
Question 1: Determine the velocity of the train wagon as a function of time V(t) while it is under the sand hopper.
Question 2: Show that the total time T that it takes for the wagon to fully pass the hopper machine is T=M/C*(exp(CL/MV)-1)

Homework Equations


  • Conservation of momentum MwagonV0=(Mwagon+msand)*V1
  • F=ma=dP/dt (Newton's second law)
  • t=msand/C
  • Straight-line motion equations
vx=v0x+axt (Equation 1)

x=x0+v0xt+½axt2 (Equation 2)

vx2=v0x2+2a(x-x0) (Equation 3)

x-x0=½(v0x+vx)t (Equation 4)


The Attempt at a Solution



The first question was pretty easy. Because I need the answer from question 1 to solve question 2 I want to make sure that I have the right answer for question 1.
Question 1:
Using conservation of momentum I got MwagonV0=(Mwagon+msand)*V1 and the mass of the sand can be written as Ct. Subbing that in you get MwagonV0=(Mwagon+Ct)*Vt. This gives an equation for Vt -> Vt=Mwagon*V0/(Ct+Mwagon)

Question 2:
This one is tricky and it definitely needs integrating if you look at the equation. I have a feeling there are multiple ways to approach this but every approach I did ended up in chaotic equations that did not help. First I used Newton's second law --> F=dP/dt
Pt=Mtvt with the mass being dependent of time because of Mwagon+Ct

dP/dt=Mtdv/dt+vtdm/dt (dm/dt is the change in mass per unit time which is C)

F=Mtdv/dt+Cvt

When isolating dV/dt and integrating that I got

V=(F-CV)ln(M2+C2T2)

M2+C2T2=exp(V/(F-CV)

This is definitely going in the wrong direction and I got no L in the equations.

Then I started with equation 2 with x being L --> L=V0t+½at2.

Acceleration is dV/dt which is CMV/((Ct+M)2) but subbing this in gives again nothing I can work with.

Then I used equation 3 which resulted in

(MV)2/((M+Ct)2)=V02+2(dV/dt)*L

Isolating dV/dt gives

-C2t2V2/(2(M2L+C2T2L)=dV/dt

This is also a pain to integrate but I used an integration calculator and I ended up with an arctan in my equation.

This is going nowhere and I would really appreciate some help.
 
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  • #2
Hi, freutel.

The constant acceleration equations that you wrote as relevant equations are not applicable in this problem.

You're making it more difficult than necessary. You found from question 1 that Vt=Mwagon*V0/(Ct+Mwagon).
Can you see what to do if you write Vt = dx/dt?
 
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  • #3
I always fail to see the easiest way to solve the problem..
Ok so I did Vt=dx/dt which gives x=((MwagonV)/C)*ln(Mwagon+Ct)
The x is of course L so I isolated everything from the natural log which gives LC/(MwagonV)=ln(Mwagon+Ct)
This results in Mwagon+Ct=exp(CL/MwagonV).
Now I get for t --> t=(1/C)*exp((CL/MwagonV)-M/C.

It appears I am missing a factor M at the exponential function... I'm looking for that missing M but I cannot find it. Did I oversee something or could it be that the given equation has a factor M too many?
 
  • #4
TSny said:
Hi, freutel.

The constant acceleration equations that you wrote as relevant equations are not applicable in this problem.

You're making it more difficult than necessary. You found from question 1 that Vt=Mwagon*V0/(Ct+Mwagon).
Can you see what to do if you write Vt = dx/dt?

I know you are not supposed to give full answers but I really cannot find that missing M. Can you please help me?
 
  • #5
When you integrate you can either do indefinite integrals or definite integrals.

If you use indefinite integrals, you will have an arbitrary constant of integration that you will need to determine.

If you use definite integrals, then make sure you use the appropriate limits of integration for x and t and evaluate the integrals at both the upper and lower limits.
 
  • #6
Yes! I made a little mistake, I forgot to set the boundaries from 0 to T but now I got it!
Thank you very much, TSny!
 
  • #7
OK. Good work!
 

Related to Total time wagon needs to fully pass the hopper

1. What factors affect the total time for a wagon to fully pass the hopper?

The total time for a wagon to fully pass the hopper is affected by several factors, including the length and weight of the wagon, the speed of the wagon, the size and capacity of the hopper, and any external forces such as wind or friction.

2. How does the speed of the wagon impact the total time for it to pass the hopper?

The speed of the wagon has a direct impact on the total time for it to pass the hopper. The faster the wagon is moving, the shorter the total time will be. This is because the wagon will cover more distance in less time, reducing the overall time it takes to pass the hopper.

3. Is the size of the hopper a significant factor in the total time for a wagon to pass?

Yes, the size and capacity of the hopper can greatly affect the total time for a wagon to pass. A larger hopper will take longer to fill up, resulting in a longer overall time for the wagon to pass. Conversely, a smaller hopper will fill up more quickly, reducing the total time for the wagon to pass.

4. How does the weight of the wagon impact the total time for it to pass the hopper?

The weight of the wagon can impact the total time for it to pass the hopper in two ways. First, a heavier wagon will take longer to accelerate and decelerate, slowing down the overall process. Second, a heavier wagon will require more force to move, which can affect the speed and efficiency of the process.

5. Is there a way to calculate or estimate the total time for a wagon to fully pass the hopper?

Yes, the total time for a wagon to fully pass the hopper can be calculated or estimated using various mathematical formulas that take into account the factors mentioned above. These calculations can provide an approximate time, but actual results may vary depending on external factors and variables.

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