# Block on wedge slides down

1. Oct 14, 2009

### wing_88

1. The problem statement, all variables and given/known data
A block m slides from frictionless inclined (45 with horizontal (a)) of a wedge mass m which on horizontal plane with coefficient of friction u. Prove the wedge moves to right with acc. equal to g(1-3u)/(3-u)

2. Relevant equations

3. The attempt at a solution
block has normal reaction mgcos (a) ...
then...

2. Oct 14, 2009

### Delphi51

I don't see my way through to the end yet, but certainly you must begin by drawing the free body diagrams for the block and the wedge. No friction force for the wedge on the block (or vice versa) so I guess the force of the block on the wedge (and vice versa) will be perpendicular to the wedge surface. Once you have the FBDs, write the vertical and horizontal F = ma for each body and hopefully you'll be able to solve for the acceleration you are looking for! Don't be afraid to start even when the end isn't in sight.

3. Oct 14, 2009

### Delphi51

Oh, it was tough thinking! I didn't have enough equations and started thinking about the fact that the block stays in contact with the wedge as it accelerates down. But the horizontal acceleration of the block is not the same as that of the wedge, because the block is changing its horizontal position with respect to the center of mass of the wedge. Finally ended up with a huge simplification by switching to the direction perpendicular to the wedge when working on the block's forces. The acceleration of the block in that direction must be the same as the component of the wedge's acceleration in that direction - in order to keep them in contact.