you place a sheet of paper on a book, it falls with it rather than being separated & slowed by air resistance. What physics concepts support the observation of the book and paper falling together?

# [Physics] What physics concepts support the observation of the book and paper falling together

accelerationfree fallnewtonian-gravitynewtonian-mechanics

#### Related Solutions

it is because the Force at work here (gravity) is also dependent on the mass

gravity acts on a body with mass m with

$$F = mg$$

you will plug this in to $$F=ma$$ and you get

$$ma = mg$$ $$a = g$$

and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.

This is a peculiar aspect of gravity and underlying this is the equality of inertial mass and gravitational mass (here only the ratio must be the same for this to be true but Einstein later showed that they're really the same, i.e. the ratio is 1)

Suppose the balloon with air (A) weighs $m_a$ and the balloon with concrete (B) weights $m_b$. The force accelerating the balloons downwards is $m_a g$ for A and $m_b g$ for B, where $g$ is the acceleration due to gravity. In the absence of air the acceleration is simply this force divided by the mass, so both balloons accelerate at the same rate of $g$. So far so good.

Now suppose the air resistance is F. We don't need to worry exactly what F is. The force on balloon A is $m_a g - F$, so its acceleration is

$$ a_a = \frac{m_a g - F}{m_a} = g - \frac{F}{m_a} $$

and likewise the acceleration of balloon B is

$$ a_b = \frac{m_b g - F}{m_b} = g - \frac{F}{m_b} $$

So the balloons don't accelerate at the same rate. In fact the difference in the accelerations is simply

$$ \Delta a_{ba} = a_b - a_a = \frac{F}{m_a} - \frac{F}{m_b} $$

Since $m_b \gg m_a$ the difference is positive, i.e. balloon B accelerates at a much greater rate than balloon A.

**Response to comment**

I think there are a couple of possible sources of confusion. Let me attempt to clarify these, hopefully without making things even more confused!

Firstly the air resistance affects both the acceleration and the terminal velocity. I've only discussed acceleration because terminal velocity can get complicated.

Secondly, and I think this is the main source of confusion, the gravitational force on an object depends on its mass while the air resistance doesn't. However the air resistance depends on the velocity of the object while the gravitational force doesn't. That means the two forces change in different ways when you change the mass and speed of the object.

Incidentally, Briguy37 is quite correct that buoyancy has some effect, but to keep life simple let's assume the object is dense enough for buoyancy to be safely ignored.

Anyhow, for a mass $m$ the gravitational force is $F = mg$ so it's proportional to mass and it doesn't change with speed. Since $a = F/m$ the acceleration due to gravity is the same for all masses.

By contrast the air resistance is (to a good approximation) $F = Av^n$, where $A$ is a constant that depends on the object's size and shape and $n$ varies from 1 at low speeds to 2 at high speeds. In your example the size and shape of the ballons is the same so $F$ just depends on the speed and for any given speed will be the same for the two ballons. The deceleration due to air resistance will depend on the object mass: $a = Av^n/m$, so air resistance slows a heavy object less than it slows a light object.

When you first release the balloons their speed is zero so the air resistance is zero and they would start accelerating at the same rate. On the moon there is no air resistance, so the acceleration is independant of speed and the two objects hit the ground at the same speed.

On the Earth the acceleration is initially the same for both balloons, but as soon as they start moving the air resistance builds up. At a given velocity the **force** due to air resistance is the same for both ballons because it only depends on the shape and speed. However because acceleration is force divided by mass the **deceleration** of the two balloons is different. The deceleration of the heavy balloon is much less than the deceleration of the light balloon, and that's why it hits the floor first.

From the way you phrased your question I'm guessing that and answer based on calculus won't be that helpful, but for the record the way we calculate the trajectory of the falling balloons would be to write:

$$ \frac{dv}{dt} = g - \frac{Av^n}{m} $$

This equation turns out to be hard to solve, and we'd normlly solve it numerically. However you can see immediately that the mass of the object appears in the equation, so the change of velocity with time is affected by the mass.

## Best Answer

If you place the book on the desk and a piece of paper on it, if there is no wind turbulence it will stay there until the cleaning lady comes.

Why? because the gravitational force keeps it there, and in addition frictional forces, i.e. electromagnetic interactions between the paper molecules and book molecules . This would be true in vacuum too. Atmospheric pressure on top of the paper and "vacuum", i.e. displacement of air between the book and paper contact surface adds to the stability.

When both fall, unless a turbulence lifts a side of the paper and air enters underneath, they will fall together. If falling from a great enough height, turbulence will separate them inevitably. Just needs some air to intervene between the two surfaces.