Leaking cart being accelerated

[PhysicsRock]

Homework Statement

A cart with mass $m_0 = 10 \, \text{t}$ is filled with $15 \, \text{t}$ of sand. It is then accelerated with a Force of $F = 5000 \, \text{N}$. However, a hatch was left open and sand is leaking from the cart at a constant rate of $\Phi = 0.5 \, \frac{\text{t}}{\text{s}}$. Give the speed of the cart after the sand is all gone as a function of $F$ and $\Phi$. You may neglect friction in this problem.

Relevant Equations

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My approach is to use the definition of the Force with $\displaystyle F = \frac{dp}{dt} = \dot{m} v + m \dot{v}$. Since $m(t)$ decreases linearly, I should be able to set $m(t) = M - \Phi t$, thus $F = - \Phi v + (M - \Phi t) \dot{v}$, which gives $\displaystyle v = -\frac{ F - (M - \Phi t) \dot{v} }{\Phi}$. Here's the problem. I don't know how I am supposed to use this, as $v$ is unknown and thus $\dot{v}$ is unknown too. I tried solving this like an ODE, however, with that I got speeds of $15,000 \frac{\text{m}}{\text{s}}$, which is obviously quite unrealistic. Anything I did wrong in the approach itself? Or is my ODE solution wrong? I calculated it to be $\displaystyle v(t) = \frac{ F t }{M - \Phi t}$.

Thank you in advance.

 

[pasmith]

Pay attention to units. A force of $5000\,\mathrm{N} = 5000\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-2}$ is only $5\,\mathrm{t}\,\mathrm{m}\,\mathrm{s}^{-2}$.

 

[PhysicsRock]

Pay attention to units. A force of $5000\,\mathrm{N} = 5000\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-2}$ is only $5\,\mathrm{t}\,\mathrm{m}\,\mathrm{s}^{-2}$.

Oh my god, yes. I converted everything to kilogramms but forgot to do so for $\Phi$, so I was basically doubling everything rather than dividing by $500 \frac{\text{kg}}{\text{s}}$. Thank you for the reminder, that really helped!

 

[erobz]

So, is the mass leaking from the cart giving an impulse to the cart in the direction of the applied force?

 

[PhysicsRock]

So, is the mass leaking from the cart giving an impulse to the cart in the direction of the applied force?

No, I forgot to include that. It doesn't contribute to momentum.

 

[erobz]

No, I forgot to include that. It doesn't contribute to momentum.

Then:

$$ F = M(t) \dot v$$

Where:
$$ M(t) = M_c + M_s(t)$$

$$M_s(t) = M_{s_o} - kt$$

The equation you are trying to apply here (incorrectly) the ejecta is contributing to the forward momentum of the cart.

 

[PhysicsRock]

Then:

$$ F = M(t) \dot v$$

Where:
$M(t) = M_c + M_s(t)$

$M_s(t) = M_{s_o} - kt$

But since all the remaining sand is in motion with the cart, the change of mass has to be regarded as well, right? Force is defined as change in momentum over time and mass is decreasing at a non-zero rate, so the product rule has to be applied.

 

[erobz]

But since all the remaining sand is in motion with the cart, the change of mass has to be regarded as well, right? Force is defined as change in momentum over time and mass is decreasing at a non-zero rate, so the product rule has to be applied.

The ejected momentum is not contributing to the forward momentum of the cart. In your bastardization of Newtons Second ( as @PeroK would refer to it ) it clearly is. Notice $\dot M_s$ is negative so when it's moved to the LHS:

$$ F + \dot M_s v = M \dot v$$

You can see that term $\dot M_s v$ is adding to the net force in the direction of motion.

If it's not a rocket, don't treat it as such.

 

[PhysicsRock]

The ejected momentum is not contributing to the forward momentum of the cart. In your bastardization of Newtons Second ( as @PeroK would refer to it ) it clearly is. Notice $\dot M_s$ is negative so when it's moved to the LHS:

$$ F + \dot M_s v = M \dot v$$

You can see that term $\dot M_s v$ is adding to the net force in the direction of motion.

If it's not a rocket, don't treat it as such.

That's good reasoning. However, when I use that approach, solve for $v(t)$ and then plug in all the values, I get that $|v| \approx 92 \, \frac{\text{m}}{\text{s}}$. That seems to be a bit too much.

Edit: With my original idea I get a speed of exactly $15 \, \frac{\text{m}}{\text{s}}$. That's more realistic to me, considering an object with ten tons of mass is being moved by only five Kilonewtons.

 

[erobz]

That's good reasoning. However, when I use that approach, solve for $v(t)$ and then plug in all the values, I get that $|v| \approx 92 \, \frac{\text{m}}{\text{s}}$. That seems to be a bit too much.

Edit: With my original idea I get a speed of exactly $15 \, \frac{\text{m}}{\text{s}}$. That's more realistic to me, considering an object with ten tons of mass is being moved by only five Kilonewtons.

That doesn't make any sense. In "your" method you are adding to the net force on the cart, while accounting for the same change in mass in both scenarios...BUT ending up with a lower velocity?!? Brake Check!

When I solve the problem, I get about $9~ \rm\frac{m}{s}$

 

[PhysicsRock]

That doesn't make any sense. In "your" method you are adding to the net force on the cart, while accounting for the same change in mass in both scenarios...BUT ending up with a lower velocity?!? Brake Check!

When I solve the problem, I get about $9~ \rm\frac{m}{s}$

Could you share your solution? Maybe I messed something up. For my way, I end up with the expression given in the question. For yours, I get $v = - \frac{F}{\Phi} \ln(M - \Phi t)$.

 

[erobz]

Could you share your solution? Maybe I messed something up. For my way, I end up with the expression given in the question. For yours, I get $v = - \frac{F}{\Phi} \ln(M - \Phi t)$.

$$ F = M \frac{dv}{dt} \tag{1}$$

Where

$M = M_c + M_s \tag{2}$

It follows that:

$$ \frac{dM}{dt} = \frac{dM_s}{dt} = -k $$

Then change variables from time to mass:

$$ F = M \frac{dv}{dM}\frac{dM}{dt} = M \frac{dv}{dM} (-k) \tag{3}$$

Solving $(3)$:

$$ v_f = \int_{0}^{v_f} dv = -\frac{F}{k} \int_{M_o}^{M_f} \frac{dM}{M} = \frac{F}{k} \ln \left( \frac{M_c + M_{s_o} }{M_c}\right) $$

 

[Steve4Physics]

I get $v = - \frac{F}{\Phi} \ln(M - \Phi t)$.

Hi @PhysicsRock. Can I add to what @erobz has just said? When deriving the above formula, you forgot the constant of integration (the constant is not zero in this situation). If you do it correctly (and use the appropriate value of t) you will get the same as @erobz.

 

[PhysicsRock]

Hi @PhysicsRock. Can I add to what @erobz has just said? When deriving the above formula, you forgot the constant of integration (the constant is not zero in this situation). If you do it correctly (and use the appropriate value of t) you will get the same as @erobz.

Oh, right. The good old $+c$. Thank you.

 

[kuruman]

Belatedly, I add this observation. Draw a free body diagram of the cart and its contents after some sand has already leaked out. Newton's second law says $M(t)\dfrac{dv}{dt}=F$. We don't consider the mass that has leaked out because as it drops it retains its horizontal momentum without imparting anything to the cart. Same situation as the standard projectile problem in which a plane releases a package at rest with respect to it.

 

[haruspex]

$\displaystyle F = \frac{dp}{dt} = \dot{m} v + m \dot{v}$.

This is one of my pet gripes.
Mass is neither created nor destroyed, so if $\dot m\neq 0$ it is not a closed system. Any mass entering or leaving may be bringing in or taking away momentum. To make that equation work, this momentum transfer has to be included as a virtual force.

 

[ChiralSuperfields]

This is one of my pet gripes.
Mass is neither created nor destroyed, so if $\dot m\neq 0$ it is not a closed system. Any mass entering or leaving may be bringing in or taking away momentum. To make that equation work, this momentum transfer has to be included as a virtual force.

Do you please know how do you solve the problem using a virtual force? I am interesting to know how the problem can be solved by using the product rule to take the derivative of momentum with respect to time.

Many thanks!

 

[haruspex]

Do you please know how do you solve the problem using a virtual force? I am interesting to know how the problem can be solved by using the product rule to take the derivative of momentum with respect to time.

Many thanks!

If mass with velocity v is being added to the system at rate $\dot m$ then there is a virtual force $\dot mv$. The equation then becomes $\Sigma F_{real}+\dot mv=\frac{dp}{dt}=\dot mv+m\dot v$, which reduces to $\Sigma F_{real}=m\dot v$, as per master Isaac.

 

[ChiralSuperfields]

If mass with velocity v is being added to the system at rate $\dot m$ then there is a virtual force $\dot mv$. The equation then becomes $\Sigma F_{real}+\dot mv=\frac{dp}{dt}=\dot mv+m\dot v$, which reduces to $\Sigma F_{real}=m\dot v$, as per master Isaac.

Thanks for your reply! Do yo please know the definition of a virtual force? - It's not included in my classical mechanics textbook.

Also isn't the mass leaving the system too, so the virtual force should be negative? Or is the ground surrounding the cart the system?

Many thanks!

 

[haruspex]

Do yo please know the definition of a virtual force?

Anything that has an effect within the equations that is like a force but is not a physical force in the usual sense. Usually it refers to corrections that have to be made to equations if using a non inertial frame. In the present case it is a correction for not using a closed system as far as mass is concerned.

the virtual force should be negative?

In the present case, $\dot m<0$, so yes, the force would be negative.

 

[ChiralSuperfields]

Anything that has an effect within the equations that is like a force but is not a physical force in the usual sense. Usually it refers to corrections that have to be made to equations if using a non inertial frame. In the present case it is a correction for not using a closed system as far as mass is concerned.

In the present case, $\dot m<0$, so yes, the force would be negative.

Thanks for your help haruspex!

 

[PhysicsRock]

I think I should tell you, what turned out to be correct. Thanks to @erobz you were right. Not including the $\dot{M} \vec{v}$ term turned out to be correct with a final speed of ca. $9.2 \, \frac{\text{m}}{\text{s}}$.