A variation on Newton's bucket

LOL...just guessing..HAHAHA

The forces acting on the mass of water are the normal applied by the bucket (which is orthogonal to the water surface) and its weight. Since the bucket was cubic and half full, the normal will make 45 deegres with the horizontal. So, in the accelerated referencial of the mass of water we have: $N \cos 45º = ma (I)$

$N \sin 45º = mg (II)$

Dividing (II) by (I) we get

$\tan 45º = \frac{g}{a} => a = g$

g= 9.8 ms-2 is "almost" constant anywhere on the surface of earth. So water will not spill out if the angular acceleration of the spinning of bucket is more than g. But it will start spilling out if it is equal or less than g.

So the value of a, so that water will start pouring out is g. That is 9.8 ms-2

The cube is uniformly accelerated, so the water suffers a backward acceleration $-\vec{a}$ , and gravitational acceleration $\vec{g}$ , then we have an 'effective acceleration' $\vec{g'} = -\vec{a} + \vec{g}$ .

Under a uniform field of force (without rotation), the surface of water will be plane and is perpendicular to the vector $\vec{g'}$ . Because water level in the cube is half it's height, and if the water is spilled out, it's surface will make $\angle{45}$ degrees to the horizontal direction. On the other hand, $\vec{g'}$ is normal to the water surface, so $\vec{g'}$ will be $-45$ degrees to the horizon, and then we have $|\vec{a}| = g = 9.8 (m/s^{2})$

See http://en.wikipedia.org/wiki/Newton%27s_bucket

Even with some constant acceleration $a$ , the particles of water are essentially in equilibrium. This happens because the sum of the forces acting on a particle, $\mathbf{ma}+\mathbf{mg}$ , is balanced by a normal force, $\mathbf{F_n}$ . (The normal force arises from forces from nearby particles).

We can represent this graphically by the vector $\mathbf{F_n}$ being opposite in direction and equal in magnitude to the sum of the vectors $\mathbf{ma}$ and $\mathbf{mg}$ .

By the argument in the Wikipedia article, the surface of the water is perpendicular to the normal force, $\mathbf{F_n}$ . Since the same $\mathbf{ma}$ and $\mathbf{mg}$ act on every particle in the water, the normal force, $\mathbf{F_n}$ is of the same direction and magnitude for every single particle. Hence, the surface of the water is a straight line.

Obviously when the water starts spilling out the line will be the diagonal from the top left to bottom right corners. Therefore, the normal force, $\mathbf{F_n}$ will be at a 45-degree lookup angle from the bottom of the bucket, and the sum of $\mathbf{ma}$ and $\mathbf{mg}$ will be at a 45-degree lookdown angle. Therefore, $\mathbf{ma}=\mathbf{mg}$ , $a=g$ , and $a=\boxed{9.8}$ $m/s^2$ .

The surface of the water is always perpendicular to the net force acting on it. Seeming as the water surface will be at a 45 degree angle to the horizontal when it is spilling (the height of the cube is equal to the base); the net force will be acting perpendicular to the 45 degree surface. Hence the net force will act at a 45 degree angle in the opposite direction. Taking g as the opposite length and a as the adjacent; using simply trigonometry, $tan(45)=\frac{g}{a}$ which simplifies to $g=a$ .

Hence given that $g=9.8\frac{m}{s^{2}}$ , $a=9.8\frac{m}{s^{2}}$

When the container with the liquid is accelerated to the right the liquid tends to accumulate to the left half of the container due to inertia. Now, to make the liquid overflow the right half of the liquid should occupy the left half of the container along with the previously present other half. To solve any non-inertial frames we use the concept of pseudo-force to make it simpler. Hence the force acting on the right half "imaginative" is (m/2) a. The force required is (m/2) g. Equating this we get the answer to be g=9.8m s^-2

Imagine that our bucket is in free space. We can reproduce gravity by accelerating smoothly upwards at gm/s^2. So we can mimic our bucket by one accelerating g upwards and a forwards in free space. In turn this behaves like a stationary bucket in a gravitational field producing acceleration g down and a backwards. For a stationary bucket the surface is perpendicular to the gravitational acceleration (in our world horizontal). For the water to be on the point of overflowing we want the surface to be 45 degrees to the horizontal. Hence g=a.

When the water is almost spilling out, to see the situation, you draw the diagonal of your container, and the left half is the part that is going to be full with water. So we have an inclination of 45°.

By the formula arctg(angle) = acceleration / g, acceleration = 9.8

Very Simple.

The water will spill out when the water reaches the top surface. This implies that the angle that the water surface makes to the horizontal is 45 degrees.

Considering the forces on the water, there is gravity downwards, and the force due to acceleration to the right (and thus an "fictitious" force to the left). For the angle to be 45 degrees, the force of gravity and the force to the left. This means that they must be equal in magnitude. Therefore, the magnitude of the acceleration is equal. And there fore the answer is 9.8 m/s.

Quando a água está prestes a cair, no topo do círculo, não há normal, apenas o peso. Então a força é igual ao peso. Assim: F=P ma=mg Logo, a=g=9,8m/s2

A fluid cannot resist shear stress ; that is, it cannot resist forces parallel to its free surface .

Consider a hypothetical molecule of mass "m" on the surface of the accelerating liquid in its non inertial frame of reference. The forces acting on it will be

1.It's wight i.e w = mg in vertically downward direction

2.Pseudo force i.e f = ma to the left

Now, let the fluid be about to spill. By using simple geometry we can conclude that the angle that the free surface makes with the horizontal direction is -45 degrees as the box is a hollow cube. On resolving the above mentioned forces in a direction parallel to the free surface of the liquid, we get two components mg $sin 45$ and ma $cos 45$ in exactly opposite directions (angle between them is 180 degrees).

Since the net force along this line is zero (along the line parallel to the free surface), therefore mg $sin 45$ + ma $cos 45$ = 0 vectorially. Considering only magnitude (as we already know the direction of acceleration of the block), we get

mg $sin 45$ = ma $cos 45$

hence, mg = ma

hence, g = a

Therefore magnitude of a is 9.8m/ $s^{2}$ and its direction is to the right.

When θ=45º, water will start to spill out.

$\implies \frac{a}{g}=tan 45º$

$\implies$ at a=g (9.8m/ $s^2$ ), water will start to spill out.

** I apology for some grammar and scientific terms mistakes, because I usually solve these questions in Chinese.

When the bucket moves to its highest position facing down, the water inside the bucket accept 2 distinct forces, which is the gravity $(mg)$ and the supporting force $(F_N)$ from the bottom of the bucket, their resultant force is the water's centripetal force during that position. However, $F_N=0$ because it doesn't act on any effect to the water during that position.

The formula for the centripetal force is $m\frac{v^2}{r}$ . According to Newton's second law of motion derived formula $F=ma$ , a object with a circular motion must have acceleration $(a)$ . The direction of the acceleration is towards the centre of the circle, and so we have another formula which is $a=\frac{v^2}{r}$ .

Now, we arrange and combine our formulas and thinkings, we should have

$mg+F_N=m\frac{v^2}{r}$

$mg+F_N=ma$

Since $(F_N=0)$ ,

$mg=ma$

$a=9.8m/s^2$

Using the Einstein's Principle of Equivalence , we can model the acceleration $a$ like an horizontal gravity. Then, two similars forces act on the water:

An horizontal force: $m\alpha$ and a vertical force: $mg$

The water begin to spill out of the bucket when the horizontal force is greater than the vertical force. So, in the equilibrium, we have:

$m\alpha = mg$ . Therefore, $\alpha = g = 9.8 m/s^2$

Before acceleration the only force acting on the liquid is mg perpendicular to the free surface downwards. For the required condition ( water to spill out) one end of the free surface should be at left top corner and other end should be at right bottom corner. For this to happen the free surface should have turned by 45 degrees. Since the resultant acceleration has to be perpendicular to the free surface directed inwards it should also turn by same angle, this will happen when the pseudo force ma is equal to mg and acting horizontally and directed opposite to the direction of acceleration There fore a=g= 9.8 m/s*2

When the water is about to spill out of the cube, the shape of the water in the cube is a rectangular prism. Consider the body of water, there are 4 forces on it: gravity (down), normal force by floor (up), normal force by the wall (left), force that provides the acceleration (right).

By the symmetry of the problem: Normal force up = Normal force left and Gravity = Force that provides acceleration Hence $a = g = 9.8 m/s^2$

Tilt the axes 45 degrees clockwise and consider the frame of the bucket. The effective gravity in this frame due to the fictitious force and the weight must point to the -y axis. This gives a = 9.8 m/s^2.

When a bucket filled with water accelerates, we can choose the frame of the bucket as our frame. To do this, we will have to add a pseudo force in our bucket system, ma in the opposite direction(to the left).

Also there is the weight acting vertically downwards. The vector sum of these two perpendicular accelerations gives the line along which the surface of water makes the angle with the horizontal.

The resultant of acceleration vectors a and g makes an angle * tanA=a/g * If the angle is 45 degrees, water will start spitting out. So, a=g. a=9.8

The acceleration of the cube is perpendicular to the gravitational force. The angle the water makes with respect to the horizontal is therefore related to the acceleration as $a = g*tan(\theta)$ .

Since the parameters of the problem are constant in the z-axis, the cube can be simplified to a 2-D box with side length $s$ , with the water occupying $A = s^2 / 2$ units. $A$ must be constant before the water spills over the edge.

The angled water in the box is made up of two shapes: a right triangle with height $h_1$ and opposite angle $\theta$ , and a rectangle of height $h_2$ and width $s$ . Trigonometry:

$tan(\theta) = \frac{h_1}{s}$

$A = \frac{s^2}{2} = h_2 * s + \frac{s * h_1}{2} = h_2 * s + \frac{s^2 * tan(\theta)}{2}$

$h_2 = \frac{s(1 - tan(\theta))}{2}$

$h = h_1 + h_2 = \frac{s(1 + tan(\theta))}{2}$

When the water spills over the bucket, $h = s$ :

$1 + tan(\theta) = 2$

$tan(\theta) = 1$

$\theta = atan(1)$

$a = g * tan(\theta) = g * tan(atan(1)) = g$

Looking at the Einstein equivalency principle we can "create" an apparent gravity acceleration. So, we are looking for a triangle of forces to solve this question. In this moment, by the simetry of the cube, when the water begin to spill out of the bucket, the gravity make an 45 angle with the surface of the liquid. Thus, we conclude the fact that the acceleration is equal the gravity.

First, we have a acceleration a and then, a centrifugal force: $Fcent=Ma$ $Fg=Mg$

If we build a graphic between the curve of the, we decompose these forces with a tangent angle that the curve does, it will be exactly:

$tg \theta=\frac{a}{g}$

We know that: $tg \theta=dy/dx$ But with a cube property, the $dy/dx=1$ because the cube has the same edge with the other axis, so we can conclude that the acceleration is: $a=g$ .

Water will spill out when the angle of the surface with respect to the base of the cube is 45º. Thus the vector sum of gravity and acceleration should form a vector 45º from the vertical a/g= tan45º = 1 a = g