- #666
vanesch
Staff Emeritus
Science Advisor
Gold Member
- 5,117
- 20
schroder said:If we use the numbers we were using in the example, 10 m/sec tread, 2 m/sec cart both wrt the floor. I was getting 2 m/sec too low! You were getting 12 m/sec too high!
At the wheel/tread interface, as you and Newton say and I agree, they are working on each other, it is not an all or nothing proposition as we were using (I was giving it nothing while you were giving it all)
So I need to add 5 m/sec to my 2 and the correct velocity is 7 m/sec
You need to subtract 5 m/sec from your 12 and the correct velocity is also 7 m/sec.
But now that I think of it, *even in these frames* you are wrong.
We agree that in the ground frame, the velocity of the cart (with the + sign in the sense opposite to the motion of the tread) is:
+ 2 m/s (it goes the other way)
and the velocity of the wind is 0 m/s
In the frame of the tread we have:
12 m/s for the cart
10 m/s for the wind
In your "half way" frame:
7 m/s for the cart
5 m/s for the wind
In your "vertical" frame (moving up 10 m/s):
the cart: sqrt (2^2 + 10^2) = 10.19 m/s
the wind: 10 m/s
In the "half way vertical" frame (moving up 5 m/s):
the cart: sqrt(2^2 + 5^2) = 5.38
the wind: 5 m/s
DARN. EVEN by picking just ANY of these frames, we STILL have the cart going faster than the wind...