Water bucket on an inclined plane
A bucket of water on small wheels rests on an inclined plane. While at rest, the calm surface of water experiences the same force at every point, and is perpendicular to the force of gravity.
If the bucket were freely rolling down the plane, what would the water surface look like?
Assume the bucket does not slip, and after some initial sloshing, the water comes to rest.
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5 solutions
This is not correct. The problem clearly states that the bucket is rolling freely. This means that there is a constant acceleration down the incline. That acceleration produces component of the for to the rear and the water will accumulate there.
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By that logic an astronaut on the ISS would notice items move away from the Earth
Compared to the first figure where the bucket is at rest, water in figure B has indeed accumulated towards the rear.
That was my thinking! If the bucket is rolling freely, it is accelerating down the hill. Which means it is accelerating in two directions! (Correct answer should be C)
Absolutely correct
Is it not true that some of gravity's force would be converted into the angular momentum of the wheels? The cart would then not be accellerating down the plane at the same rate that it would if it were a frictionless box on a frictionless surface. So as far as the water is concerned the box is still going slower than gravity is trying to pull it, and it would be pooled in the front of the box. It would be closer to even than when the box is at a stand still, but it would still be pooled toward the front. The question says "rolling freely". But for it to roll, the wheels have to absorb energy slowing the box. The water being perpendicular to the inclined plane indicates that the box would roll at the same speed regardless of the inertial mass, size, or distribution of mass of the wheels.
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The problem says that the wheels are small, so the angular momentum of the wheels is small and hardly affects the acceleration of the bucket. If the wheels are massive and we do not ignore the angular momentum of the wheels, then the acceleration of the bucket would be slightly less than g sin θ , and water level would be slightly flatter than the slope of the incline.
You are right that the moment of inertia of the wheels would reduce the acceleration of the car. But I assumed that the mass of the wheels is negligible. I explicitly mentioned this in an earlier version of the problem, but after the problem was revised by Brilliant staff, this hint no longer appears.
This does ignore friction as a force being applied. (I.e. assumes perfect bearings).
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Right, the friction was neglected here. In addition, I have assumed that the moment of inertia of the wheels does not matter. The reality is somewhere between cases A and B.
ok. the first incidence of force on the water is gravity, and as long as nothing else changes the water must distribute the force of gravity equally throughout itself. if another force acts upon the water, and that force is greater than the gravity acting upon it, then the water must equally distribute that force throughout its mass; overcoming the force of gravity. gravity is still there but a lesser force. is this right?
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In principle, you are right. The water surface is always perpendicular to the total force acting on it. Gravity is usually the dominant force, but for small droplets this is the surface tension. The surface tension of the water causes a force directed towards the center of the droplet. Therefore, the water droplet assumes the shape of a sphere because its surface is perpendicular to the radius vector.
Please dumb it down for me.
The water level will always be perpendicular to the acceleration of the cart in its local inertial frame , i.e. when viewed from the perspective of something in free fall.
In that frame, the only force on the cart is the normal force from the inclined plane, perpendicular to the incline. Therefore the water will be parallel to the incline.
So tge water takes on the shape of the incline its going down water forms to whatever it rest upon? That's why if you just got rod of the box and poured the water down the incline it would do the same thing be perpendicular to the incline?
If the bucket had a motor to provide acceleration then, the correct answer would be option 'c'?
What if the motor provided it a constant speed?
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Any force with a component parallel to the surface will affect the direction of the water surface. Thus, a motor "helping" the downhill acceleration will result in situation C, whereas braking, (rolling) friction, or air drag will result in situation A.
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Isn't that the case as it stands? If the cart is left to free fall the acceleration will be increasing and the speed of the cart will be increasing down the incline and thus optin c is correct. If option B were correct the acceleration of the cart would be 0 and would be moving at a constant speed.
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@Jerry Duncan – If the cart were standing still on the ramp, or traveling at constant speed, the situation would be A, as shown in the introduction of the problem.
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@Arjen Vreugdenhil – Thanks a lot for your response @Arjen Vreugdenhil sir.
It is General perception that the water shall move backwards due to inertia during acceleration and then on constant speed it shall be perpendicular to the acceleration of the cart in its local inertial frame, or not, depending on the angle of inclination (as earth gravity shall always pull water towards it and try to make its surface parallel to the surface to the earth), thus the water will be parallel to the incline may not stand true for all conditions.
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Can you suggest the conditions under which the water will not be parallel to the incline?
It is true if and only if the net force on the cart (not counting gravitation) is perpendicular to the incline.
To be parallel to the incline, the cart would have to be moving horizontally at the rate of zero G... 32ft/sec/sec. being an incline, it would have to roll down the incline FASTER than G. The equation leaves out this part of the formula and therefor produces an inaccurate proof.
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If you want to go into the details of the motion, see Markus' proof.
My solution does not make any assumptions about the movement of the cart, except that it accelerates in the direction of the net force acting on it. This acceleration determines the shape of the surface.
You are right that the cart does not accelerate at a rate of g . Rather, the rate of acceleration along the incline is g sin θ , where θ is the incline angle.
I agree with John. I am a professional structural engineer, I believe the assumption is not a fact. The water on inclined surface with take the same surface angle. This case will occurred on level surface, in either the bucket is in stable position or in no acceleration position.
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Do you mean the assumption that "the water level will always be perpendicular to the acceleration of the cart in its local inertial frame "?
Note that this is not the usual inertial frame, which you probably utilize as a structural engineer (unless you design spacecraft).
In particular, when in rest on a level surface on earth, the cart is accelerating upward at rate g in its local inertia frame.
Sir,Is there a proof for the statement "The water level will always be perpendicular to the acceleration of the cart in its local inertial frame, i.e. when viewed from the perspective of something in free fall"!!!
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We assume that the water has reached an constant shape.
Consider a water particle; its acceleration is equal to a , the acceleration of the system. The direction of the force on this particle is, on one hand, the pressure gradient at this point; on the other hand, equal to the direction of the acceleration of the system. (I.e. d F = − ∇ P d V and d F = a d m .) Therefore the pressure gradient is in the opposite direction of the acceleration: ∇ P = − ρ a .
But the surface of a fluid is a surface of equal pressure, and therefore perpendicular to the pressure gradient ∇ P . Therefore, it is perpendicular to the acceleration of the system.
Note
If we do this analysis from the usual perspective with the earth at rest, we have to include a gravitational term d F g = g d m ; the conclusion is that − ∇ P = ρ ( a − g ) . Since for a cart going downhill a is precisely equal to g + F n / m , with F n the normal force, we get − ∇ P = ρ F n / m . Thus the pressure gradient is normal to the incline, and the water surface perpendicular to it, i.e. parallel to the incline.
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Thank u so much!!! But I had a few doubts how did u get dV where P=F/A and the reason for taking an infinitesimal mass element,should mass not change with time and gravity is acting on the opposite direction of the normal force,Shoudn't they have opposite signs???Thanks a lot!!!!
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@Erica Phillips – I added the negative signs for the gradient.
Consider an box of sides d x , d y , d z centered at ( x , y , z ) . The pressures on the left and right face are P left,right = P ± 2 1 d x ∂ x ∂ P . This produces a force equal to F x = ( P left − P right ) A = ( − d x ∂ x ∂ P ) d y d z = − ∂ x ∂ P d x d y d z = − ∂ x ∂ P d V , and similar for the F y and F z components, resulting in F = − ∇ P d V .
How can it be justified as freely falling, doesn't it clearly experiences another force as friction from ground?
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There is a normal force from the incline. The cart is not in free fall. In my analysis, the "camera" is in free fall, and we see the cart accelerating upward and slightly to the right.
If the bucket reaches terminal velocity and therefore stops accelerating, it is correct to say that the water will return to parallel to the horizontal (perpendicular to the force of gravity)?
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That is correct.
In the accelerating reference frame of the bucket, the downhill component of gravity disappears, leaving the component perpendicular to the slope. This causes the water surface to be perpendicular to said component, i.e. parallel to the slope.
The inertia at the beginning of the downhill pushes water backwards .It depends on when you measure the level as the longer it goes it will level out .Similarly the water will slosh up the front of the bucket if it is brought to a quick stop .
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Sorry for my previous post .Somebody asked me that question and I was not told about the water coming to a rest.
If the system initial configuration (A) is at rest due to the blocks, and there is no friction in the system, when the blocks are pulled, I suspect Schrodinger's cat goes into oscillation among all three states. ;-) The only force the wheels apply to the cart & water, is perpendicular to the incline. The remaining component of gravity contributes to the cart & water acceleration parallel to the incline.
Alternative B will be correct initially with some assumptions: -negligible rolling friction -Negligible wheel mass
But after the acceleration phase, the cart will reach a terminal velocity and solution A would best describe the water, assuming the air passing by does not affect the water but only the cart.
My approach was about visualizing how the gravitational and acceleration forces acted on the water bucket.
If gravity was the only acting force then the water will rest perpendicular to the direction of gravity.
If acceleration was the only acting force then the water will accumulate to the back of the bucket in relation to the direction of acceleration.
Now we can align the two images on top of each other and see the differences between the effects. Since we are dealing with water, we can think of this superposition as waves and apply the rules governing wave behavior.
By adding the two "waves", they cancel one another because the crests (highest point) align with troughs (lowest point) leaving the water resting parallel to the base of the bucket. Due to the bucket's orientation, the water is, therefore, also parallel to the slanted plane.
What about the friction against the bucket? Wouldnt that create a force difference between water and the bucket ?
Gravity F g = F g ∥ + F g ⊥ acts on the bucket, which, in relation to the inclined plane, disassembles a parallel and a vertical component. Only the parallel component F g ∥ = m a contributes to the acceleration a of the bucket.
A calm water surface represents an equipotential surface in an external force field. Thus, the external force is in equilibrium perpendicular to the water surface. In the moving reference system of the bucket in addition to gravity, an inertial force F i = − m a = − F g ∥ acts, which is inversely equal to the parallel component of the gravity. The force acting on the water therefore results to F = F g + F i = ( F g ∥ + F g ⊥ ) + ( − F g ∥ ) = F g ⊥ and thus stands perpendicular to the inclined plane. Thus, the water surface must be in equilibrium parallel to the inclined plane.
This situation compares to the weightlessness in free fall. In free fall, the gravitational and inertial forces are completely compensated for in the moving frame, so that one would float freely in a falling elevator, thus experiencing weightlessness. On the inclined plane there is only parallel to the plane a free movement, so that only the parallel component of gravity is canceled. The resulting force therefore points perpendicular to the inclined plane.