OK, F is applied to m2. Then (let's assume no slippage between the masses):
mu m1 g = m1 a , and F - mu m1 g = m2 a.
Physically, a is determined by the total inertial load which the external
force, F, sees:
a = F/(m1 + m2) . The internal force (mu m1 g) will automatically arise (mu
is numerically adjustable) from internal stresses to make these equations
so.
Perhaps this is more evident if you place m1 behind m2 ( and at the same
vertical level) and let m2 drag m1 through the tension T in a connecting
rope .
Then: T = m1 a , and F-T = m2 a, and a = F/(m1 + m2). Now one can directly
see how T develops to make these so.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
----- Original Message -----
From: "Gene Gordon" <physics@ROCHESTER.RR.COM>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Friday, October 21, 2005 6:09 PM
Subject: Re: Free body diagram misconception
|I am sorry to say that due to the lateness of the hour or just typing, I
| mistyped the problem. Everything in the problem, except for the force
| acting on the system. The force should be acting on block 2. not block
| 1, since that is the accceleration that i was asking about.
|
|
| Sorry,
|
| Gene