Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is going on here? - Mechanics Puzzler

  1. Oct 18, 2004 #1
    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?
  2. jcsd
  3. Oct 18, 2004 #2
    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)
  4. Oct 18, 2004 #3
    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.
  5. Oct 18, 2004 #4
    My point is I don't think you can assume a constant force opposing the force of the engine - that's not physically possible.
  6. Oct 18, 2004 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    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).
  7. Oct 18, 2004 #6
    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.
  8. Oct 18, 2004 #7
    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.
  9. Oct 18, 2004 #8
    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.
  10. Oct 18, 2004 #9
    Hmm, Im 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. Im 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.
    Last edited: Oct 18, 2004
  11. Oct 19, 2004 #10
    Yes, but you only increase your velocity in the y-direction. In the x-direction your velocity is constant yet your momentum is not.
  12. Oct 19, 2004 #11
    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
  13. Oct 20, 2004 #12


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [tex] \sum\vec{F} = \frac{d\vec{p}}{dt} [/tex]

    [tex] \sum{F_x} = 0 [/tex]

    [tex] \sum{F_y} = \frac{dp_y}{dt} [/tex]

    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:

    [tex] \frac{dp_y}{dt} = \frac{d(mv_{y})}{dt} [/tex]

    = [tex] m\frac{dv_{y}}{dt} + v_y\frac{dm}{dt} [/tex]

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

    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!
  14. Oct 20, 2004 #13
    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)
  15. Oct 20, 2004 #14
    Shouldent the boat speed up? If its loosing mass, and momentum is conserved independently in the x and y directions, then wont 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 thats 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.
    Last edited: Oct 20, 2004
  16. Oct 20, 2004 #15


    User Avatar

    First of all, for a particular body, such as the boat, in Newtonian mechanics,
    [tex]\Sigma f = \frac{dp}{dt}[/tex]
    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
    [tex]\frac{dp_x}{dt} < 0[/tex].

    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
    [tex]f_{sand} = v_s \cdot \frac{dm}{dt}[/tex]
    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.
  17. Oct 20, 2004 #16


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
  18. Oct 20, 2004 #17
    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.
  19. Oct 20, 2004 #18
  20. Oct 20, 2004 #19


    User Avatar

    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

    [tex]f_{sand} = v_{sand} \cdot \frac{dm}{dt}[/tex]

    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

    [tex]\frac{dv_y}{dt} = \frac{f}{M - b \cdot t}[/tex]

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

    [tex]\frac{dy}{dt} \propto \frac{dy}{dx}[/tex]


    [tex]\frac{dv_y}{dt} = \frac{d^2y}{dt^2} \propto \frac{d^2y}{dx^2}[/tex]

    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

    [tex]t = \frac{M}{b}[/tex]

    Off hand it looks more like a tangent than a parabola.
    Last edited: Oct 20, 2004
  21. Oct 20, 2004 #20
    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.
    Last edited: Oct 20, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook