Richard,
Adopting the point of view that the particle moves leftward and downward along the wedge while the wedge moves rightward, in falling a distance dy, the particle must, relative to the wedge, move leftward a distance, dx_rel = dy/tan in order to stay on the surface of the wedge. In terms of motion relative to center of mass of entire system, dx_rel=dx+dX where dx is how far left the particle moves and dX is how far right the wedge moves. Substituting and solving for dy yields:
dy = (dx + dX) tan
Divide both sides by dt to get your equation.
Take the time deriv. of the result to get Bob L's equation.
With ay positive downward as in ay = (ax+Ax) tan,
the sum of the vertical forces yields
N cos - mg = - m ay
Bob L. omitted the minus sign on the right, but it must have just been a typo because you need that minus sign to get the (correct) result that Bob L. got.
Jeff Schnick
________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of Richard A. Lindgren
Sent: Fri 5/25/2007 11:04 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] A different normal force. Was: Re: The Normal Force
Bob,
What is the simple logic for writing down the following equation without
getting into LaGrangians.
ay = (ax + Ax) tan