what is the weight of a cage with a bird flying inside it?

Question

consider weight of cage is x
weight of the bird is y

Would the weight sum up both

Answer 1

Somewhere between x and x+y

If, instead of a cage, the bird was in a closed box (so a completely closed system) then the weight would be x+y. In order for the bird to fly, there must be a force acting upwards on it to to counteract gravity, casued by the bird flapping. The bird exerts an opposite force downwards on the air (Newton's second law), which in turn exerts a force on the box.

However, the bird is not in a box, but a cage, so presumably the cage is open to the air. This means that some of the air which is forced downwards by the bird's wings can escape from the cage, and interact with the surroundings outside the cage. But , depending on the size of the cage, how open it is, whether it has a solid floor etc, the air from the bird's wings will exert some downwards force on the cage, so the weight will be more than x.

This illustrates the importance of understanding where the boundaries to a system are in physics problems - is there a closed system (like a box) or an open system (like a cage) where air can move in and out.

Incidently, I saw a very good similar problem the other day - can someone sat on a boat make the boat move by blowing into the sail? I've used the ideas from the solution to that problem in answering above.

Answer 2

Weight is a measurement of gravitational force; the classic formula (thanks Newton!) is F = m*a where weight is F, m is mass and a is the acceleration of gravity.

In the example above, the presumption is that the cage is sitting on a scale in a static position and that the bird is flying inside the cage and is not in contact with the cage at all. In this situation, the weight of the bird (i.e. the FORCE that the mass of the bird is accelerated by gravity) is counteracted by the force of the bird's wings in flight. Since the bird is not resting on the cage, it's weight ("y") is not added to the weight of the cage ("x").

For additional information on this type of problem, consider searching Yahoo using the phrase "statics" or "engineering statics".

Update: For all answers that are addressing the turbulence or force of the air as it may incidentally apply force to the cage (which could occur upon any surface of the cage, whether it is closed or not, including the sides of the cage due to turbulent air flow), that does not change the actual weight of the cage if you stick to the definition that the weight of an object is defined by its mass times the acceleration due to gravity. The added force (i.e. acceleration) is additive, but has neither changed the mass of the cage or the acceleration due to gravity.

For example, if you stand on a scale to weigh yourself, but your friend steps on the scale to increase the registered weight, do you actually weigh more? The answer is no.

If you want to know measure the force the cage is applying to the scale, you would include the turbulence created by the bird's wings (which would be neglibile in most conceivable real cases, meaning many orders of magnitude less).

However, considering the information given in this problem, these theories seem to be over thinking it a little bit - you have no knowledge of the downward force of the wings, nor the air density, nor the existing air flow, etc. . If you have to choose between just x or x +y, x is the reasonable choice.

Answer 3

Because the bird is flying you will have just the weight of the cage. The bird weight is supported by the air. (if the bird is flying very close to the bottom of the cage the wind from the wings will give some extra weight but probably only a slight effect).

Answer 4

If the cage is closed one with air inside, then the weight will be x + y, even if the birds fly inside.

If there is an opening, then when the birds fly , the weight will be only x, since the birds weight is balanced due to their flight.

We tie a heavy cylinder to one end of a thread and the other end is tied to the ceiling.
A glass beaker is placed surrounding the cylinder so that the cylinder is well immersed inside the cylinder but without pressing the bottom. Below the glass beaker a weighing machine is arranged to measure the weight of the cylinder and beaker.

Since the weight of the cylinder is balanced by the tension in the string, its weight will not add to glass beaker. If we move up and down the cylinder, due to circulation of air , the balance will fluctuate.
However, we know that the cylinder’s weight is not added to the glass beaker because it is a heavy one and the fluctuation of weight is small.

Similar is the case with flying birds in a cage. Due to the flight there will be fluctuations but the weight y is not added to x. the fluctuation of weight is about the value of x only. And in any time it will not equal to x + y.

Suppose one is standing on a weighing machine and noting his weight. Exactly at the same time, high in the sky, a plane is crossing him. Will he notice in the balance his weight and the plane’s weight? Both are in the vertical line.

Answer 5

It is x, i.e. just the weight of the cage. For a similar experiment, get your baby brother to join you on the same side of a seesaw, put a friend on the other side, and position yourselves so that you all just balance. Then, throw your baby brother in the air and catch him. The seesaw will move up while your baby brother is off the seesaw. By the way, do not try this experiment while your parents are watching.

Answer 6

it is just the weight of the cage. (if the bird has no contact with the cage)

Answer 7

I assume there may be three answers:1.x+y;2.x;3.x-y.
Further, the answer may be in the range of (x-y,x+y).

Answer 8

x + y