At 10:41 AM 11/19/1999 -0500, dg50@daimlerchrysler.com wrote:
>I've been tossing this problem around in my head for the last couple of
days -
>I've got the concept, but I'm having trouble putting an equation on it.
This is the response I got from a friend
---- forwarded message ----
If we idealize the system and ignore shocks, springs
etc..anything that would affect transient behavior,
and replace the turn with the phrase,
"The car must instantly turn 90 degrees to the right"
The the answer is....
The tires need to exert TWICE the FORCE on car b
as they do on car a to accelerate the car in the
new direction (right)
If you break down the original course of the cars
in its vector components, there's no component
going to the right.
But if it turns instantly to the right, then
F=ma applies again, and for equal A, the 2m car
needs 2F vectored to the right.
But what does TWICE the FORCE really mean?
In an idealized system, the force turning the car
to the right is the friction of the tires on the
road, which ideally can be expressed as m*g*(mu)
where mu is the coefficient of friction.
since the heavier car weighs 2m, the heavier car
puts twice the amount of friction to the ground,
and hence twice the force.
It doesn't sound right, but I think it is.
---- end of forwarded message ----
jmb
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