- #31
Elquery
- 66
- 10
Yes M and m are factors, of course. Makes perfect sense and a bit of a duh moment there.
I was also heading where hmmm27 went in thinking about $$F={m}{a}$$ (or g instead of a if you want to be technical, sure
) to describe why without another force (air resistance) acceleration would be equal*. Doubling mass doubles force, yet that yields no change in acceleration: $$a=\frac{F}{m}$$
And of course this is essentially what A.T. said in post #14:
It seems to me that we would then be discussing force subtraction, so relative magnitudes do matter. I.e. the heavier object has, for example, a downward force of 10, the lighter object 5, and the upward (resistive force due to air) is 2 at a given velocity (who cares about units?). For the heavier object, the net downward force is 10-2=8 (80% of the gravitational force). For the smaller object: 5-2=3 (60% of the gravitational force). Thus the smaller object's net force is further reduced proportional to mass at a given velocity.
Wrong, right, confused?
I was also heading where hmmm27 went in thinking about $$F={m}{a}$$ (or g instead of a if you want to be technical, sure
) to describe why without another force (air resistance) acceleration would be equal*. Doubling mass doubles force, yet that yields no change in acceleration: $$a=\frac{F}{m}$$And of course this is essentially what A.T. said in post #14:
*perhaps there's a question still remaining (for me, and perhaps others): is acceleration still equal in the presence of air-resistance up to the point of terminal velocity? Or does air resistance, being an opposing force to gravity, reduce acceleration slightly right 'from the get-go'?A.T. said:
It seems to me that we would then be discussing force subtraction, so relative magnitudes do matter. I.e. the heavier object has, for example, a downward force of 10, the lighter object 5, and the upward (resistive force due to air) is 2 at a given velocity (who cares about units?). For the heavier object, the net downward force is 10-2=8 (80% of the gravitational force). For the smaller object: 5-2=3 (60% of the gravitational force). Thus the smaller object's net force is further reduced proportional to mass at a given velocity.
Wrong, right, confused?
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