# A Science Question About Gravity

#### Tiburon

##### Well-known member
Got it. But when we try to account for the effect of air resistance, do we do so by subtracting r from f or by subtracting r from a? Why is the equation, a=(f-r)/m, rather than a=(f/m)-r?

Suppose, for example, the force wasn't constant, as with gravity, but just an initial push (as with two sleds on a flat surface). The air resistance wouldn't affect the force, then, but it would affect the acceleration. (To maybe make the point clearer, let's say the sleds got the initial push, then went through a vacuum, then encountered air resistance.) Would it still make sense to subtract r from f?
I agree it should be a=(f/m)-r regardless of whether it's travelled through a vacuum.

#### LifeIn

##### Well-known member
Got it. But when we try to account for the effect of air resistance, do we do so by subtracting r from f or by subtracting r from a? Why is the equation, a=(f-r)/m, rather than a=(f/m)-r?
Because that's not how physics works. The effect of air resistance is to exert a force on the object moving through the air. That force is independent of the mass of the object. The effect of that force on the acceleration, however, is dependent on the mass of the object.

Suppose, for example, the force wasn't constant, as with gravity, but just an initial push (as with two sleds on a flat surface). The air resistance wouldn't affect the force, then, but it would affect the acceleration.
In that case f1 = f2 = 0 and the equations still hold, even on a flat surface. Therefore the acceleration would be (0-r)/m, or -r/m, which is a negative number, meaning the velocity is decreasing.

#### The Pixie

##### Well-known member
The way a blade works is the pressure causes the ice the melt (Le Chatelier's principle, as ice is less dense than water), and the water acts as a lubricant.

Is it possible that a greater weight enhances that effect, perhaps by producing more water or producing it faster?