Block on wedge slides down

  • Thread starter wing_88
  • Start date
  • #1
4
0

Homework Statement


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)



Homework Equations





The Attempt at a Solution


block has normal reaction mgcos (a) ...
then...
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
11
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
Delphi51
Homework Helper
3,407
11
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.
 

Related Threads on Block on wedge slides down

Replies
3
Views
3K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
8
Views
4K
Replies
5
Views
1K
Replies
4
Views
7K
Replies
3
Views
4K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
14
Views
273
  • Last Post
Replies
12
Views
5K
Top