What is going on here? - Mechanics Puzzler

[emob2p]

A boat is moving in the x-direction with constant velocity. The frictional force against the boat is constant and independent of its mass. There is a pile of sand on the boat. Someone starts shoveling the sand off the boat strictly in the y-direction with respect to the boat.
Here is my question:
If the frictional force is independent of the boat's mass, then the forces in the x-direction (force of engine + frictional force) should always cancel regardles off the sand being shoveled off. But at the same time, dp/dt in the x-direction should not be zero because the boat's mass is changing even though the mass is being shoveled off in the y-direction. How to resolve this problem?

 

[geometer]

A boat is moving in the x-direction with constant velocity. The frictional force against the boat is constant and independent of its mass. There is a pile of sand on the boat. Someone starts shoveling the sand off the boat strictly in the y-direction with respect to the boat.
Here is my question:
If the frictional force is independent of the boat's mass, then the forces in the x-direction (force of engine + frictional force) should always cancel regardles off the sand being shoveled off. But at the same time, dp/dt in the x-direction should not be zero because the boat's mass is changing even though the mass is being shoveled off in the y-direction. How to resolve this problem?

I think your dilemma lies in your assumption that removing the sand will not affect the friction force. By the way, in this situation drag is the correct term vs. classic friction. Removing the sand will change the boat's mass, which will cause it to ride higher in the water than before which will lower the surface area of the boat in the water which will lower the drag.
(edited to correct editorial mistakes)

 

[emob2p]

I understand that removing the sand would alter the drag force. But then forget the drag force. Just assume there is a constant force opposing the force of the engine. Then regardless of the mass, the boat will still move with constant velocity in the x-direction. Yet there still is a dp/dt in the x-direction. The problem is still not solved.

 

[geometer]

I understand that removing the sand would alter the drag force. But then forget the drag force. Just assume there is a constant force opposing the force of the engine. Then regardless of the mass, the boat will still move with constant velocity in the x-direction. Yet there still is a dp/dt in the x-direction. The problem is still not solved.

My point is I don't think you can assume a constant force opposing the force of the engine - that's not physically possible.

 

[pervect]

If the frictional force is independent of the boat's mass, then the forces in the x-direction (force of engine + frictional force) should always cancel regardles off the sand being shoveled off. But at the same time, dp/dt in the x-direction should not be zero because the boat's mass is changing even though the mass is being shoveled off in the y-direction. How to resolve this problem?

Given that p=mv, and that p,m, and v are functions of time, you can use in the most general case the chain rule for derivatives

d(uv) = udv + vdu

which yields

dp/dt = m dv/dt + v dm/dt

When you shovel the sand off sideways, the velocity as a function of time stays constant, but the momentum changes (because of the v dm/dt term).

 

[geometer]

I don't see how your treatment solves the problem pervect. The force pushing the boat is created by the prop which accelerates a given amount of water in the opposite direction to the boat's motion. The reaction force to this pushes the boat forward. Assuming the propeller accelerates the same amount of water at the same rate regardless of the mass of the boat, as the mass of the boat goes down, the acceleration of the boat, and therefore the velocity of the boat should go up.

 

[geometer]

Another problem with your reasoning emob is that the retarding forces and the acclerating forces do not always have to cancel. If you are in constant unnacelerated motion, they will, but when you change the situation by shovelling the sand over the side, you are no longer in equilibrium and they don't cancel. The boat speeds up. Once you finish shovelling the sand over the side, you settle down to a new equilibrium, with a new drag force.

 

[emob2p]

Let me simplify the problem to isolate the dilemma. Imagine that our ship is in outer space such that no external forces are acting on it. It is also moving with constant velocity in the x-direction with respect to our frame. Now, someone begins shoveling off the sand strictly in the y-direction in our frame. Thus, we calculate that there is a change of momentum in the x-direction (since it has a positive velocity in the x-direction and its mass changes) yet there are still no forces acting on the ship in the x-direction.

 

[Cyrus]

Hmm, I am not sure, but maybe momentum does not change. You will decrease your mass, but increase your velocity proportionaly so that the product of the mass and velocity is exactly the same as when you started. I am not sure though just a guess. Momentum might be changing in the Y direction, though, because you need a force to move that sand, so somehow something must push down in order to push the sand up and out.

 

[emob2p]

Yes, but you only increase your velocity in the y-direction. In the x-direction your velocity is constant yet your momentum is not.

 

[Cyrus]

why would your velocity be constant, the momentu is changing, P=m*v, m is decreasing, but momentum is conserved, so v must increase to compensate. Dont quote me though, lol

 

[cepheid]

$$\sum\vec{F} = \frac{d\vec{p}}{dt}$$

$$\sum{F_x} = 0$$

$$\sum{F_y} = \frac{dp_y}{dt}$$


Guys...whatever force is used to eject matter from the ship vertically is a force imposed on the matter by the ship (launching platform system or whatever). By Newton's third law there is a corresponding force of the matter on the ship (pushing down on the platform). The ship must therefore be gaining a vertical (y) component of momentum equal and opposite to the rate of change of in momentum of the ejected matter. There are no net external forces in the x-direction, so that component of the velocity is unaffected. Would not the resulting path be very similar to that of a body in free fall, since the path would be parabolic (in the frame that emob described). And in a frame moving with the same x velocity as the ship, would the ship not appear to move straight downward?

Maybe I should take a second look:

$$\frac{dp_y}{dt} = \frac{d(mv_{y})}{dt}$$

= $$m\frac{dv_{y}}{dt} + v_y\frac{dm}{dt}$$

What do the two terms represent physically? Both have units of $kg \cdot \frac{m}{s^2}$

I'm not sure actually. The first is just equal to the force that increases the velocity in the y-direction (i.e. the mass at that instant times the y component of acceleration). The second is equal to a different force...I'm not sure what. All I know is that dm/dt is negative. Can anyone explain? And please point out if I have made any mistakes!

 

[gerben]

Let me simplify the problem to isolate the dilemma. Imagine that our ship is in outer space such that no external forces are acting on it. It is also moving with constant velocity in the x-direction with respect to our frame. Now, someone begins shoveling off the sand strictly in the y-direction in our frame. Thus, we calculate that there is a change of momentum in the x-direction (since it has a positive velocity in the x-direction and its mass changes) yet there are still no forces acting on the ship in the x-direction.

The momentum of the ship changes, but the total momentum of the ship and the (now shovelled off) sand its still the same...
(just as the boat the sand keeps moving in the x-direction)

 

[Cyrus]

Shouldent the boat speed up? If its loosing mass, and momentum is conserved independently in the x and y directions, then won't the speed have to increase as the mass decreases so that the momentum is constant?

On second though, could it be possible that the speed of the boat remains constant, but now that the mass is less, the total momentum is less than before. In order to account for the difference in momentum, we have to conside that the sand chucked out the boat is also moving with the same x velocity, so that's the remaining momentum.

But I thought that a rocket moves because of momentum. Since momentum is conserved, as you chuck propellent out the back, you decrease the mass, so the velocity has go to up. (or is it true in that case because the force produced is in the same direction as motion.) In our case, the force is transverse.

 

[sal]

A boat is moving in the x-direction with constant velocity. The frictional force against the boat is constant and independent of its mass. There is a pile of sand on the boat. Someone starts shoveling the sand off the boat strictly in the y-direction with respect to the boat.
Here is my question:
If the frictional force is independent of the boat's mass, then the forces in the x-direction (force of engine + frictional force) should always cancel regardles off the sand being shoveled off. But at the same time, dp/dt in the x-direction should not be zero because the boat's mass is changing even though the mass is being shoveled off in the y-direction. How to resolve this problem?

First of all, for a particular body, such as the boat, in Newtonian mechanics,
$$\Sigma f = \frac{dp}{dt}$$
is only true so long as the mass is constant. You've got somebody shoveling mass overboard, so of course, the momentum in the boat is going to be changing. Mass is decreasing, velocity is constant, so
$$\frac{dp_x}{dt} < 0$$.

The forces from the engine and water are red herrings. You've said v starts out constant, and the forces don't change. So, net force on the boat from the engine and drag is zero.

The only force to worry about, then, is the result of shoveling sand overboard. You've got a rocket. Force on the boat from the sand will be
$$f_{sand} = v_s \cdot \frac{dm}{dt}$$
where v_s is the velocity with which the sand is thrown -- exhaust velocity -- and dm/dt is the "burn rate" of the sand.

If the shoveler always throws the sand perpendicular to the line of motion, the boat will follow a spiral path, with decreasing radius as the mass decreases. If the sand is always thrown along the y-axis then determining the path will take more effort.

 

[pervect]

I don't see how your treatment solves the problem pervect. The force pushing the boat is created by the prop which accelerates a given amount of water in the opposite direction to the boat's motion. The reaction force to this pushes the boat forward. Assuming the propeller accelerates the same amount of water at the same rate regardless of the mass of the boat, as the mass of the boat goes down, the acceleration of the boat, and therefore the velocity of the boat should go up.

I was assuming that there was no friction - similar to the way the problem was later modified. I think the chain rule solution answers that question adaquately.

 

[snbose]

The problem is simple.
The net effect of the boat loosing mass (no matter which direction) is that its momentum tends to decrease. The engine that drives the boat is running at the same power level, so the net change in momentun is 0. Naturally as everyone concluded the velocity in the X direction will INCREASE to compenste the mass lost.

The friction or more correctly VISCOUS drag will also increase and the boat will attain an equilibrium when the forces cancel each other. Why? Recall that viscous drag is proportional tovelocity: Poisoille Theorum:-)

So when the boat looses mass, it hovers from one equilibrium to another (provided you allow sufficient time for a steady state to be achieved). In each equilibrium state progressively the boat moves faster and the viscous drag increases and balances the engines thrust.

 

[Cyrus]

If the shoveler always throws the sand perpendicular to the line of motion, the boat will follow a spiral path, with decreasing radius as the mass decreases. If the sand is always thrown along the y-axis then determining the path will take more effort.[\quote]

I do not follow how this is true. If there is a change in momentum strictly in the y direction, then the boad should drift always down, speeding up, while it goes along the x axis. It should follow the path of a parabola. Its thrown transverse to the line of motion, so there is nothing to curve it. At the first time sand is thrown, the boad moves down and across, but does not "turn" about any axis.

 

[sal]

If the shoveler always throws the sand perpendicular to the line of motion, the boat will follow a spiral path, with decreasing radius as the mass decreases. If the sand is always thrown along the y-axis then determining the path will take more effort.

I do not follow how this is true. If there is a change in momentum strictly in the y direction, then the boad should drift always down, speeding up, while it goes along the x axis. It should follow the path of a parabola. Its thrown transverse to the line of motion, so there is nothing to curve it. At the first time sand is thrown, the boat moves down and across, but does not "turn" about any axis.

You're partially correct, of course, based on the exact statement of the problem. The shoveler shovels sand in the Y direction. Given that, it will certainly not trace out a spiral.

But if you think of a "real" boat, it's not actually going to go through the water sideways -- boats have keels; in this scenario after a while the boat is moving almost directly sideways. So, in a small nod to realism, I took the liberty of assuming the boat turns to point in the direction it's going. In that case, it would also be natural for the shoveler to keep shoveling the sand over the side, rather than shoveling it off the stern as the boat turns to -Y.

In that scenario, the direction of the sand rotates as the boat rotates and the boat traces out a spiral. The boat also maintains a constant speed through the water in this case, which makes the assumption of constant friction slightly more realistic.

Now, to take your assertion that it'll trace out a parabola if the sand is always thrown in the +y direction ... it won't. Force due to shoveling is

$$f_{sand} = v_{sand} \cdot \frac{dm}{dt}$$

where dm/dt = b = "burn rate" of the sand. But the boat's mass is decreasing. So, if M = initial mass of boat + sand, then the acceleration in the Y direction is going to be

$$\frac{dv_y}{dt} = \frac{f}{M - b \cdot t}$$

Now, dx/dt is constant, by assumption, so

$$\frac{dy}{dt} \propto \frac{dy}{dx}$$

and

$$\frac{dv_y}{dt} = \frac{d^2y}{dt^2} \propto \frac{d^2y}{dx^2}$$

For a parabola, d^2y/dx^2 is constant. In this case it certainly isn't. So, I'm not sure what the figure is, but it isn't a parabola. Also note that it has a singularity at

$$t = \frac{M}{b}$$

Off hand it looks more like a tangent than a parabola.

 

[Cyrus]

Your a bit above me right now, lol. Can you help me with this dilema. The mass is thrown off the side. So with time, mass of the boat changes. But all the mass thrown off the side moves forward as well, neglecting air resistance, with the same forward speed as the boat. So the momentum of the sand and the boat in the x direction is always the same. ( I guess this is why it does not speed up as it looses mass?). Now a similar argument could be made about the momentum in the y direction. Initially there is no momentum in the y direction. So afterwards there should be no momentum either, if we consider the whole system, sand and boat. So as sand is chucked out the boat, the sand gains momentum in the Y direction, but the boat has equal and opposite momentum, so that the sum of their momentum remains zero, at all points in time. Does this correctly explain why the boat gains velocity in the Y direction, but not in the x direction? (when I read the problem I thought the Y direction was up. Would this mean that the boat bobs up and down as the sands chucked?)

When I first saw this problem it reminded me of the ideal rocket equation. But I think I mixed it up by assuming that because the mass changes, the speed changes in both the x and y direction. Throwing the mass out the Y axis would be "thrust" along the Y direction. So there is only a change in velocity along the same axis you "chuck" material. Is that a general assumption I can safely make.

Could I also think of the situation as follows: When the sand is thrown in the Y direction, if we ignore the implicatinos of this, and the mass is changing, we could think of this as the boat breaking into pieces. The boats now into several small pieces, but they all move with the same velocity still. ( I hope this is clear). Its not what is actually happening, but isint it equivalent strictly in the x direction.

 

[snbose]

Here is my question:
If the frictional force is independent of the boat's mass, then the forces in the x-direction (force of engine + frictional force) should always cancel regardles off the sand being shoveled off. But at the same time, dp/dt in the x-direction should not be zero because the boat's mass is changing even though the mass is being shoveled off in the y-direction. How to resolve this problem?

The analyses of you guys are great but ...

In the actual problem the detailed challanges invovled shovelling sand are not relevent. The bottomline is the boat looses mass. Say its cargo catches fires and smokes in random directions. The cargo evaporates (sublimates). Now what?

On any event the boat will gain speed and will finally attain steady state. Now the viscous drag stll balances the engines forward thrust.

 

[NoTime]

On a 50 ton boat, if you shovel off a few pounds of sand it just is not going to make any significant or likely any difference to the drag.

True the momentum of the boat will change, but momentum applies to the total system. The sand did not lose any momentum just because you shoveled it off the boat.
Same with the smoke molecules.
Eventually, the boat would be slightly easier to stop.

 

[sal]

Your a bit above me right now, lol. Can you help me with this dilema. The mass is thrown off the side. So with time, mass of the boat changes. But all the mass thrown off the side moves forward as well, neglecting air resistance, with the same forward speed as the boat. So the momentum of the sand and the boat in the x direction is always the same. ( I guess this is why it does not speed up as it looses mass?).

Yes, exactly.

Now a similar argument could be made about the momentum in the y direction. Initially there is no momentum in the y direction. So afterwards there should be no momentum either, if we consider the whole system, sand and boat. So as sand is chucked out the boat, the sand gains momentum in the Y direction, but the boat has equal and opposite momentum, so that the sum of their momentum remains zero, at all points in time. Does this correctly explain why the boat gains velocity in the Y direction, but not in the x direction?

Yes.

(when I read the problem I thought the Y direction was up. Would this mean that the boat bobs up and down as the sands chucked?)

Yes. But what goes up must come down, so wouldn't the sand just fall back to the deck, and back onto the pile afterwards, in that case?

When I first saw this problem it reminded me of the ideal rocket equation.

That's exactly what it is.

But I think I mixed it up by assuming that because the mass changes, the speed changes in both the x and y direction.

Right, speed in the x direction is constant. (Assuming you don't take my "spirial path" approach...)

Throwing the mass out the Y axis would be "thrust" along the Y direction. So there is only a change in velocity along the same axis you "chuck" material. Is that a general assumption I can safely make.

Yes.

Could I also think of the situation as follows: When the sand is thrown in the Y direction, if we ignore the implicatinos of this, and the mass is changing, we could think of this as the boat breaking into pieces. The boats now into several small pieces, but they all move with the same velocity still. ( I hope this is clear).

Yes, when you look at it that way, momentum is constant, and the net force on the whole "system" is zero. All that's happening is the original mass of "boat + sand" is breaking up. The same thing's true of a rocket in space.

Its not what is actually happening, but isint it equivalent strictly in the x direction.

Yes.

 

[Cyrus]

Sal, I think we should just agree to disagree on this one. Just kidding.

 

[arildno]

First of all, I would like to thank "sal" for coming up with a correct interpretation, I would however, like to make a few general points, though:
1. The laws of classical mechanics.
a)Newton's 2.law
Newton's 2.law of motion says that the sum of forces acting upon a system consisting of THE SAME MATERIAL PARTICLES, is equal to the rate of change of the MOMENTUM of those particles.
That is, in integral form:
$$\frac{d}{dt}\int_{V_{m}(t)}\rho\vec{v}dV=\vec{F} (1)$$
i.)Here, $$V_{m}(t)$$ signifies the region in space our particle system occupies at time "t".
(Note that, by Newton's 3.law, the internal force couples disappears, so that only exterior forces occur on the right-hand side)

ii)$$\rho,\vec{v}$$ are the constituent particles' density and velocity; these are often represented through the use of density&velocity FIELDS (The Eulerian formalism).
iii) The mass of the system, M, is given by:
$$M=\int_{V_{m}(t)}\rho{dV}$$
iv)It is easy to rewrite (1) in an alternate form by introducing the velocity of the center of mass as:
$$M\vec{v}_{c.m}=\int_{V_{m}(t)}\rho\vec{v}dV$$
Hence, for a general system, we have:
$$\frac{d}{dt}(M\vec{v}_{c.m})=\vec{F}$$
b) Mass conservation
In classical mechanics, a fundamental law is that a system consisting of the same particles throughout time CONSERVES its mass.
That is:
$$\frac{d}{dt}\int_{V_{m}}\rho{dV}=0$$
Note that this does NOT mean that the shape or volume of the material system cannot change; however, we may reformulate our law of mass conservation in terms of a constraint on the density and velocity functions.
(The resulting partial diff.eq. is called the continuity equation).
(Note that the constancy of mass of a material system is contradicted by relativity)


PARTIAL CONCLUSION:
These laws hold therefore for material systems; they do NOT hold in general for systems which do not contain the same particles over time.
In particular, the "boat"-system is NOT such a system.

I'll get back to how we tweak Newton's 2.law for "non"-material systems (i.e, systems of particles where the set of individual particles might change over time (i.e, the boat system, for instance))

 

[arildno]

2. A first example:
In order to prepare an understanding of why Newton's 2.law must be tweaked for systems which do not contain the same set of particles over time, consider the following example from fluid mechanics:
Let's have a U-shaped tube, and let a fluid enter the tube at one end with velocity $$U\vec{i}$$ and leave the tube with velocity $$-U\vec{i}$$
We assume a steady flow, that is, the velocity measured at a fixed point in space do not change over time (even though it will be DIFFERENT particles occupying that space point at various times).

Now, let's consider the MOMENTUM contained within that tube.
Since the flow is steady, the amount of momentum contained within the tube must be CONSTANT THROUGH TIME!

Does this mean that the tube doesn't exert a net force on the fluid contained in it?
Of course it does!
A given fluid particle experience a momentum change $$-\delta{m}2U\vec{i}$$ by passing through the tube; clearly, this is a result of a force from the tube.
Conversely, the fluid exerts a net force on the tube in the positive $$\vec{i}$$-direction; the tube must be HELD FAST to remain at rest.

This as a first example..

 

[arildno]

3. Derivation of equations in the case of no external forces:
This is basically the same argument used to derive the rocket propulsion equation.
a) We consider a two-body system with constant masses
$$M_{o},m_{f}$$
$$M_{o}$$ is in the rocket case the mass of solid rocket, $$m_{f}$$ is the constant fuel mass, consisting of fuel remaining ($$m_{r}(t)$$+fuel expelled ($$m_{e}(t)$$).
Clearly, we have:
$$m_{r}(t)+m_{e}(t)=m_{f}$$
or, taking the derivatives:
$$\dot{m}_{r}=-\dot{m}_{e}$$
We also define the mass of the ship+remaining fuel:
$$M_{R}(t)=M_{o}+m_{r}(t)$$
Or, taking derivatives:
$$\dot{M}_{R}=\dot{m}_{r}$$
What we want to do, is to find the proper equation for the $$M_{R}(t)$$-system.
b) In an inertial frame in which the ship has velocity $$\vec{v}_{sh}(t)$$
In the time interval (t,t+dt), some a fuel amount $$\dot{m}_{e}dt$$ is expelled with a velocity MEASURED WITH RESPECT TO THE INERTIAL FRAME$$\vec{v}_{fuel}(t)$$.
We define the EJECTION velocity of the fuel as the relative velocity of ejected fuel to the ship:
$$\vec{v}_{rel}=\vec{v}_{fuel}-\vec{v}_{sh}$$
c) Let us consider mass conservation of the material system over the time interval (t,t+dt) (I won't bother with the part of that system consisting of priorly expelled fluid; it is easily included):
$$M_{R}(t)\vec{v}_{sh}(t)=M_{o}\vec{v}_{sh}(t+dt)+(m_{r}(t)+\dot{m}_{r}dt)\vec{v}_{sh}(t+dt)+\dot{m}_{e}dt\vec{v}_{fuel}$$
Or, setting $$\vec{v}_{sh}(t+dt)=\vec{v}_{sh}(t)+\vec{a}_{sh}dt$$
we gain by cancelling some terms and neglecting second order terms (we also divide by dt):
$$\vec{0}={M}_{R}(t)\vec{a}_{sh}+\dot{M}_{R}\vec{v}_{sh}+\dot{m}_{e}\vec{v}_{fuel}$$
d) Rewriting into rocket equation:
By remembering $$\dot{m}_{e}=-\dot{M}_{R}$$
we may regroup our equation into:
$$\dot{M}_{R}\vec{v}_{rel}=M_{R}\vec{a}_{sh}$$
This is the famous rocket equation.
(Since $$\dot{M}_{R}<0$$, the acceleration of the ship is in the opposite direction of the ejection velocity)
e) The momentum and momentum flux equation:
Defining the momentum of the ship+remaining fuel system as:
$$\vec{p}_{R}=M_{R}(t)\vec{sh}(t)$$, we see that we also have the following, equivalent equation:
$$\vec{0}=\frac{d\vec{p}_{R}}{dt}+\dot{m}_{e}\vec{v}_{fuel}$$

The second term is called the MOMENTUM FLUX $$\vec{\mathcal{M}}$$out of the object region, and the proper generalization of Newton's 2.law for a "non"-material system can be written as:
$$\vec{F}_{ext}=\frac{d\vec{p}}{dt}+\vec{\mathcal{M}}$$
Or in words:
The external forces acting on a "non"-material system (contained in some spatial region) equals the rate of change of momentum contained in that region plus the momentum flux out of that region.

Possibly, I'll post some examples later on..