Re: Any reason why this idea wouldn't work?


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Posted by Joel L. Lewis (24.197.38.131) on May 14, 2003 at 13:39:07:

In Reply to: Re: Any reason why this idea wouldn't work? posted by Nick Hall on May 14, 2003 at 13:24:37:

Oh, okay. *Dang!* Thanks for clarrifying that for me!:-)

: Joel wrote:

: : Yes, I considered that. But it's my understanding that objects always fall(and therefore rise) at the same rate.

: Nope.

: Objects FREE FALL 'at the same rate' - or more accurately, they accelerate at the same rate under gravity assuming no other force (like drag) intervenes.

: :As I said, my base assumption is that the initial descent will be at the same rate no matter the 'relative weight' which is being raised on the other side, thus imparting the same momentum to the weight and therefore the same potential to rise. THAT is the vital question to me.

: Sorry - no.

: The speed of the right hand weight as it passes the bottom depends upon its acceleration - which in turn is proportional to the force applied. In your example, the effective acceleratory force is less than that for an object in free fall.

: Imagine your original example WITHOUT the left hand weight. The right hand weight falls through 2 metres, its kinetic energy at the bottom will be 20 joules and its velocity will be 6.32 m/s (ignoring losses and assuming a vacuum!). The acceleratory force was gravity alone acting on the mass.

: But in your original case, the NET, or aggregate acceleratory force on the big weight was reduced by the weight of the left hand mass mediated across the pivot. The net acceleratory force on the system is now less, and the speed as the big weight passes the bottom will be less.

: In this case gravity is 'partially pulling against itself', and so the total force available to accelerate anything is reduced.

: It is easy to verify - get two wheels, one only very, very slightly unbalenced, and one with lots of imbalance - which one goes from horizontal to vertical the fastest??

: Thanks

: Nick





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