Solving Boat Movement without External Force Applied

[bksree]

Assume there is a boat of length 4 m weighing 40 kg. One man of weight 50 kg is at one end (A) and the other of weight 65 kg is at the other end (B).
Both men now move towards the centre of the boat.
Will the boat move in the water and if so by what distance ? (assume no friction between water & boat)

Taking origin at A, COM originally is 22/15 from A.

Since no external force acts on the system the COM remain stationary. Using this principle, assuming that boat moves by 'x' towards the left of the origin, it is obtained from the eqn for COM that
0 = 50(2-x) + 60(-2-x) + 40(-x) / (150)
Solving this eqn gives x = 2/15 i.e. boat moves 2/15 m towards the left.

THE DOUBT IS WHY DID THE BOAT MOVE WHEN NO EXTERNAL FORCE ACTED ON IT ?

 

[A.T.]

THE DOUBT IS WHY DID THE BOAT MOVE WHEN NO EXTERNAL FORCE ACTED ON IT ?

What is "the boat"?

- Only the boat itself? Then there was an external force from the passengers.

- Boat & passengers? Then it didn't move, because it's COM stayed stationary.

 

[y33t]

Assume there is a boat of length 4 m weighing 40 kg. One man of weight 50 kg is at one end (A) and the other of weight 65 kg is at the other end (B).
Both men now move towards the centre of the boat.
Will the boat move in the water and if so by what distance ? (assume no friction between water & boat)

Taking origin at A, COM originally is 22/15 from A.

Since no external force acts on the system the COM remain stationary. Using this principle, assuming that boat moves by 'x' towards the left of the origin, it is obtained from the eqn for COM that
0 = 50(2-x) + 60(-2-x) + 40(-x) / (150)
Solving this eqn gives x = 2/15 i.e. boat moves 2/15 m towards the left.

THE DOUBT IS WHY DID THE BOAT MOVE WHEN NO EXTERNAL FORCE ACTED ON IT ?

Boat will move even if these two guys don't move (they are positioned at each end) because their weights are different thus center of gravity of whole system is not at exactly in the middle of the boat. Boat will move with a constant acceleration in this case (assuming no sinking). When these guys move toward each other, acceleration will decrease because they will be moving to a single point where in the end center of gravity of the system will be exactly in the middle of the boat and acceleration will decrease to zero so equilibrium.

 

[A.T.]

Boat will move even if these two guys don't move (they are positioned at each end) because their weights are different thus center of gravity of whole system is not at exactly in the middle of the boat. Boat will move with a constant acceleration in this case (assuming no sinking).

So you have invented a boat propulsion that doesn't require any fuel, just an off center COM?

 

[y33t]

So you have invented a boat propulsion that doesn't require any fuel, just an off center COM?

I didn't invent it and it was there for a long time. Make the math, acceleration is very very small in this system, considering friction etc it might not even be observable from macroscopic point of view. But it is there.

 

[y33t]

So you have invented a boat propulsion that doesn't require any fuel, just an off center COM?

Also I would like to add that here in Turkey we're implementing a fully electric vehicle which has far more better efficiency characteristics than any other vehicle on the planet and the system is based on changing the center of gravity. Segway's also base on same principle, this is a known fact.

 

[Dickfore]

Make the math, acceleration is very very small in this system, considering friction etc it might not even be observable from macroscopic point of view. But it is there.

Sorry to derail, but when you say a physical quantity is small or large, you must compare it to some other physical quantity of the same kind. What exactly is this acceleration small compared to?

 

[A.T.]

So you have invented a boat propulsion that doesn't require any fuel, just an off center COM?

Make the math, acceleration is very very small in this system,

How about you doing the math, and showing us that a boat will accelerate, just because the guy sitting at one end is heavier than the guy on the other end

 

[Doc Al]

THE DOUBT IS WHY DID THE BOAT MOVE WHEN NO EXTERNAL FORCE ACTED ON IT ?

While there's no external force on the system of boat+men, there is certainly an external force on the boat itself. The men push against it when they move.

(Oops... A.T. already made that very point.)

 

[y33t]

How about you doing the math, and showing us that a boat will accelerate, just because the guy sitting at one end is heavier than the guy on the other end

My expression was wrong, the system will have a constant velocity rather than having an acceleration. I will proof and post it here.

While there's no external force on the system of boat+men, there is certainly an external force on the boat itself. The men push against it when they move.

(Oops... A.T. already made that very point.)

That's also another aspect but what I am talking about is the initial imbalance due to weight difference resulting in a motion.

Sorry to derail, but when you say a physical quantity is small or large, you must compare it to some other physical quantity of the same kind. What exactly is this acceleration small compared to?

What I meant by small is we might not perceive the movement of the boat via our sensory organs. Observation of water waves might yield another vectors to work on. On the other hand I realized that my expression was wrong, it doesn't accelerate but it will move with a constant velocity.

 

[Doc Al]

My expression was wrong, the system will have a constant velocity rather than having an acceleration. I will proof and post it here.

So, if the men and boat were initially at rest they'd remain at rest (until they start walking around the boat). I don't think we need a proof of that.

 

[y33t]

So, if the men and boat were initially at rest they'd remain at rest (until they start walking around the boat). I don't think we need a proof of that.

Imagine boat is empty no one abroad. We put heavy man to one end and light man to other end simultaneously. Initially what I mentioned will occur.

Men walking towards each other will also effect the velocity since weights are not equal, change in center of gravity will be towards the heavy guy till they meet each other assuming they are walking at the same speed.

 

[Dickfore]

What I meant by small is we might not perceive the movement of the boat via our sensory organs. Observation of water waves might yield another vectors to work on. On the other hand I realized that my expression was wrong, it doesn't accelerate but it will move with a constant velocity.

The only relevant quantities that I can see are the masses of the two persons $m_{1}, m_{2}$ and the boat $M$, the acceleration due to gravity $g$ and the length of the boat $L$. The only thing I can say by dimensional analysis is that the velocity of the boat must be expressible as:

$$v = f(\frac{m_{1}}{M}, \frac{m_{2}}{M}) \, \sqrt{g \, L}$$

Can you tell us the explicit form of the numerical function $f(x, y)$?

 

[Doc Al]

Imagine boat is empty no one abroad. We put heavy man to one end and light man to other end simultaneously. Initially what I mentioned will occur.

Why? What physics principle are you invoking?

 

[xts]

First of all:
Forum Rules: Do not post your homework/school-type questions here!

Thus I may assume you do not ask how to solve primary school problem of centre of mass, but you expect some deeper view of the problem.

1. Practical answer: the boat will move upside down, and both men fall into water.

2. You've all forgotten about mass of the water displaced by the boat (equal to sum of masses of both men and boat itself), which must be moved as the boat moves, but in opposite direction. But that is just a first order approximation - to make this harder to calculate, the boat is asymetrically submerged at the beginning (men at its ends have different weights), and more symmetrically at the end. In order to compute that you have to know boat profile geometry.

 

[Doc Al]

Thus I may assume you do not ask how to solve primary school problem of centre of mass, but you expect some deeper view of the problem.

The poster's question was a general one and was answered with the very first response.

 

[entphy]

Since there is no external force acting on the system of boat and men, the entire system should remain stationary with respect to an observer on the shore since the boat was stationary in the water to begin with. The total momentum of the boat and men system remains perfectly the same before and after the men move in the boat. All the while assuming ideal conditions of no water and wind resistances, and no stiring of water that carries away some momentum of the boat and men system.

When both men start moving in the boat, the momentum change of the men is transferred to the boat, causing the boat to move with respect to the observer on the shore. Let's analyse the system when the men of mass m₁ and m₂ are moving at constant velocities v₁ and v₂, and the boat of mass M moves at velocities v_b, all velocities are measured with respect to the shore observer.

Due to the conservation of linear momentum without external force,

m₁v₁ + m₂v₂ + Mv_b = 0

Integrate both sides over the same period of time t, and results in the following relationship among the displacements l₁, l₂ and l_b of the men and boat with respect to the shore observer.

m₁l₁ + m₂l₂ + Ml_b = 0

Before the men move, the position of the center of mass of the men and boat system is
X₀ = [m₁l₁₀ + m₂l₂₀ + Ml_b0] / (m₁ + m₂ + M)

After the men move, the position of the center of mass of the men and boat system is
X = [m₁(l₁₀ + l₁) + m₂(l₂₀ + l₂) + M(l_b0 + l_b)] / (m₁ + m₂ + M)

But it has been shown that m₁l₁ + m₂l₂ + Ml_b = 0, thus, X = X₀. Hence the center of mass remain at the same position and stationary with respect to the shore observer.

When both men stopped moving, the momentum change of both men will be transferred to the boat and caused it to stopped as well. The final position of the boat could be different, but the center of mass remain at the same position.

 

[y33t]

The only relevant quantities that I can see are the masses of the two persons $m_{1}, m_{2}$ and the boat $M$, the acceleration due to gravity $g$ and the length of the boat $L$. The only thing I can say by dimensional analysis is that the velocity of the boat must be expressible as:

$$v = f(\frac{m_{1}}{M}, \frac{m_{2}}{M}) \, \sqrt{g \, L}$$

Can you tell us the explicit form of the numerical function $f(x, y)$?

You forgot to take into account buoyancy which will effect the behavior of the system depending on the surface characteristics. Trick of the problem is that it's on water and buoyancy will dynamically respond to stimuli by men aboard. Symbolic solution to problem will be a bit more complicated than that.

Why? What physics principle are you invoking?

Newton's 2nd law, which is apparently not enough to generate a solution in this case.

 

[Dickfore]

You forgot to take into account buoyancy which will effect the behavior of the system depending on the surface characteristics. Trick of the problem is that it's on water and buoyancy will dynamically respond to stimuli by men aboard. Symbolic solution to problem will be a bit more complicated than that.

Do you know the direction of the buoyancy force?

 

[y33t]

Do you know the direction of the buoyancy force?

It is opposite to gravity in sign for Newtonian fluids, but will it have a homogeneous distribution along the boats bottom surface ?

 

[Dickfore]

It is opposite to gravity in sign for Newtonian fluids, but will it have a homogeneous distribution along the boats bottom surface ?

I did not ask you about the distribution, i asked you about the direction. Is opposite of gravity in the direction we are considering?

 

[y33t]

I did not ask you about the distribution, i asked you about the direction. Is opposite of gravity in the direction we are considering?

Doesn't matter if we consider it or not, it will definitely contribute to the behavior of the system.

 

[Dickfore]

Doesn't matter if we consider it or not, it will definitely contribute to the behavior of the system.

how?

 

[y33t]

how?

Masses of two men are not equal thus heavier man's side will sink more than the other man in volume. Thus heavy man will experience buoyancy more than the other man. Buoyancy will also contribute to steady position of COM before men start walking towards.

 

[Dickfore]

Masses of two men are not equal thus heavier man's side will sink more than the other man in volume. Thus heavy man will experience buoyancy more than the other man. Buoyancy will also contribute to steady position of COM before men start walking towards.

Don't bring the motion of the passengers in the discussion, since you had claimed that even if the passengers sit still, the boat would move with a constant velocity. What is so important about the heavier man's side sinking more than the lighter man's side? How does that result into a motion in the horizontal direction?

 

[y33t]

Don't bring the motion of the passengers in the discussion, since you had claimed that even if the passengers sit still, the boat would move with a constant velocity. What is so important about the heavier man's side sinking more than the lighter man's side? How does that result into a motion in the horizontal direction?

I am talking about the period 'before' they walk to each other already. Boat is rigid body, when these two men are initially put to ends of the boat, boat will first align by tending to rotate towards heavy man, and this alignment will be observed as the whole system making a rotational motion from cross-sectional (side) perspective. Very small but definitely a rotational motion till COM shifts too much to heavy mans direction. At this point, boat will take moment from water to shift the COM and this will earn a constant velocity to system. Once the system is in equilibrium (they didn't start walking yet) force applied from boat to water due to derivative of sinked volume with respect to time will become zero.

 

[Dickfore]

I am talking about the period 'before' they walk to each other already. Boat is rigid body, when these two men are initially put to ends of the boat, boat will first align by tending to rotate towards heavy man, and this alignment will be observed as the whole system making a rotational motion from cross-sectional (side) perspective. Very small but definitely a rotational motion till COM shifts too much to heavy mans direction. At this point, boat will take moment from water to shift the COM and this will earn a constant velocity to system. Once the system is in equilibrium (they didn't start walking yet) force applied from boat to water due to derivative of sinked volume with respect to time will become zero.

Rotational motion around what axis?

 

[y33t]

Rotational motion around what axis?

Two men are looking each other from ends of the boat, heavy men opens his arms making 90 degrees with his body, axis his arms symbolize is X, axis from heavy man to other man is Y, axis from heavy men's head to sky is Z.

When no men abroad COM is exactly the middle and current position of COM is origin. Heavy man is at the - region of Y and the other man + region. Between the moment they are abroad and the equilibrium, COM changes from origin to some point towards heavy man. Axis of rotation will shift in Y axis with COM changing and is around X axis. Rotation is always around X axis but with COM changing it will shift rotation axis in Y axis.

Assume rectangular prism boat.

Initial COM(x,y,z); (0,0,0)

Final COM(x,y,z); (0,-3,-2)

Constant velocity I am talking about is caused from COM changing from 0 to -3 in Y axis, which results as rotational motion around X axis. I could elaborate more by defining corners of the rectangular prism boat to compare initial/final positions of the boat.

 

[Dickfore]

Rotation is always around X axis but with COM changing it will shift rotation axis in Y axis.

How did you conclude this? What physical principle did you use to justify your conclusion?

Assume rectangular prism boat.

Initial COM(x,y,z); (0,0,0)

Final COM(x,y,z); (0,-3,-2)

How did you get the final coordinates? What formulas did you use?

Constant velocity I am talking about is caused from COM changing from 0 to -3 in Y axis, which results as rotational motion around X axis.

This is not rotation. It does not even describe uniform motion, since there is no change of coordinates with time.

 

[entphy]

I am just thinking of contributing a little of my 2 cents, hope this does not spark off unncessary argument. Dickfore is right. The fact is there really is no contribution from the vertically pointed buoyancy in this picture that concerns only the horizontal linear momentum and center of mass response to the internal configuration change (men's motion on the boat).

Lets assume there are 2 large cranes that lower the men slowly down to the boat, the base of the cranes are firmly anchored at the shore and remain stationary. The cranes lower the men until the men barely land on the boat, but the full weight of the men are still supported by the tensions in the ropes attached to the cranes. In other words, there is no force exerted on the boat by the weight of the men yet. Suddenly the ropes are cut and the full weight of the men simply added to the boat. Does the boat move?
Consider the system of men + boat shortly before the ropes are cut. The center of the mass of the system is still well defined actually, and the horizontal position of the COM remain the same all the time while the men were being lowered by the cranes. The only change is the vertical position of the COM of the system, which is lowering along the same vertical trajectory while the men are being lowered. So even after the ropes are cut, the horizontal position of the system COM remains firmly the same along the vertical trajectory. Thus, there is no motivation at all for the system of men and boat COM to move across the wafer.

Now consider the moment the ropes are cut. The full weight of the men are suddenly added vertically to the boat. Due to the assymetry distribution of the weight, the boat will temporarily experience a net torque that rotates the boat + men system about the COM. More water is displaced at the heavy men side and the buoyancy catches up shortly, rocking the boat back to try to restore equilibrium. Such a damped oscillation will probably last a while but eventually equilibrium will be reached, and there will be no net vertical and horizontal motion afterwards.

It is just not possible to induce a permanent motion perpendicular to the gravitational field by putting weight onto a platform, no matter how small the effect it is claimed to be. The fact is, if there is however small an effect, a perpetual motion matchine would have been invented and all energy crisis solved by paying no respect to the momentum and energy conservation law. And there is no reason why such an effect should be small at all, if it exists in the 1st place. Because one could always choose to set up a system that magnifies the effect arbitrarily, and the experment is not even hard to perform. Just place a smooth plateform with a very heavy weight at one end on a surface with low friction (lubricants or ice), and vertically place another heavy weight at the other end of the plateform and see if the system move.

The fact is, this is a simple question that is meant to test the concept of linear momentum conservation and the COM. One should be able to grasp the key point and also make sufficient assumptions to remove non-essential components irrelevant to the points of the question. The only reason I could think of that one would like to take buoyancy into account is to see how much the bow and stern of the boat actually submerged into the water, which could affect water resistance that affect dissipation of the kinetic energy of the system. But if this needs to be taken into account, I suppose the question should also make clear the viscosity, temp and density of the water (sea water has higher buoyancy due to higher density), and also the shape of the boat that makes a difference in the water resistance. And one will then venture into the non-linearity of fluid dynamics, but have to give up shortly since too much details are omitted from the question.

 

[y33t]

How did you conclude this? What physical principle did you use to justify your conclusion?


How did you get the final coordinates? What formulas did you use?


This is not rotation. It does not even describe uniform motion, since there is no change of coordinates with time.

I simulated the system using Blender (with real world physics engine and fluid dynamics pack) and system behaved as I guessed it to behave. Who mentioned about uniform motion ? It's not a linear motion over time please read my posts again.

 

[Dickfore]

I simulated the system using Blender (with real world physics engine and fluid dynamics pack) and system behaved as I guessed it to behave. Who mentioned about uniform motion ? It's not a linear motion over time please read my posts again.

lol.