- #1
moonman239
- 282
- 0
How much force does air resistance have on a freefalling object on Earth? No equations, please.
Slight definitions problem here, so I'll give two answers...moonman239 said:How much force does air resistance have on a freefalling object on Earth? No equations, please.
Point well taken.russ_watters said:By definition, an object in freefall is not experiencing any drag force at all.
Point well taken (I like to agree with everybody).Stonebridge said:From Wiki: "in nontechnical usage, falling through an atmosphere without a deployed parachute or lifting device is also referred to as free fall."
The request for "no equations" hints at a non-technical interpretation.
moonman239 said:I already understand air resistance. What I wanted, though, was the amount of drag force applied to a freefalling object at sea level (assuming a pressure of 10013 mb)
The point is, it varies. It depends on the shape and size of the object and how fast it's moving (relative to the air, of course). Anything more we could say about it is best expressed as an equation.moonman239 said:I already understand air resistance. What I wanted, though, was the amount of drag force applied to a freefalling object at sea level (assuming a pressure of 10013 mb)
diazona said:The point is, it varies. It depends on the shape and size of the object and how fast it's moving (relative to the air, of course). Anything more we could say about it is best expressed as an equation.
You mean if it's falling at terminal velocity? In that case, no - the air resistance would be simply equal to the weight. But as I understand the OP's question, it wasn't asking about that special case.Phrak said:Would you say this is true if your object is not accelerating?
I'd agree with that... the equation isfawk3s said:The best way for you would be to check the equation. I haven't learned this in school but last time I checked Wikipedia the equation seemed quite easy.
That way you could create your own conditions, such as the shape of the body, density of air which you can get from your preferred air pressure, the speed of the body at an instant and so on. Thats the only way you can get it because you asked for a number...
diazona said:I'd agree with that... the equation is
force = 1/2 × (density of air) × (cross-sectional area of object) × (drag coefficient) × (speed²)
which is written in mathematical notation as
[tex]F = \frac{1}{2}\rho A C v^2[/tex]
It is reasonably simple.
Oh yeah, I missed that, sorry.zeebek said:I was writing the same stuff several posts above.
As a suggestion:
assume your body is smth like a sphere, assume its velocity, prescribe some value for the viscosity and calculate the Reynolds number. Then go here and choose an estimate for the drag coefficient http://en.wikipedia.org/wiki/Drag_coefficient. The force is Cd*density*velocity^2*cross section
I'd agree with that... the equation is
force = 1/2 × (density of air) × (cross-sectional area of object) × (drag coefficient) × (speed²)
which is written in mathematical notation as
LaTeX Code: F = \\frac{1}{2}\\rho A C v^2
It is reasonably simple.
diazona said:Oh yeah, I missed that, sorry.
But the equation is reasonably simple - all it involves is multiplying some numbers together. If one of the parameters in it happens to be hard to calculate, that doesn't make the equation complicated. In practice you usually look up the drag coefficient or experimentally measure it.
Ah, I can see how that would be confusing. The Wikipedia article sort of explains this, if you look down at the "definition" section: the "reference area" is some characteristic area of the object. Usually in simple applications we choose it to be the cross-sectional area, but you can also choose some other area (like surface area, or whatever) if it makes sense for a given object. In fact, for the same object in the same orientation, you could have two different drag coefficients depending on which reference area you choose - so if you look up a drag coefficient, make sure to take note of what kind of area it's associated with! All that really matters in the equation is the product (drag coefficient × reference area).moonman239 said:Oh and Diazona, the first time I looked at the coefficient of drag equation, I didn't know what the heck it meant by "reference area."
Drag forces in air are caused by the interaction between an object and the air molecules surrounding it. When an object moves through the air, it creates a disturbance in the air flow, which results in a force acting on the object in the opposite direction of its motion.
Air density plays a significant role in determining the magnitude of drag forces. Higher air density means there are more air molecules per unit volume, resulting in a higher number of collisions between the object and the air molecules. This leads to a higher drag force acting on the object.
The value of drag forces in air is influenced by several factors, including the shape and size of the object, the speed at which it is moving, the air density, and the surface roughness of the object. Objects with a larger surface area, moving at high speeds through dense air, and with a rough surface experience higher drag forces.
The value of drag forces in air can be calculated using the drag equation, which takes into account the object's velocity, air density, surface area, and a drag coefficient that depends on the object's shape and surface roughness. This equation helps in predicting the value of drag forces for different objects and conditions.
There are several ways to reduce the value of drag forces in air. One way is by changing the shape of the object to make it more streamlined, which reduces the surface area and the disturbance caused in the air flow. Other methods include using smoother surfaces, reducing the speed of the object, and increasing the air density around the object.