Newtonian Mechanics : Block a wedge with no friction. How to solve for accelerations?

Oct 2017
1
0
A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge . The base of the wedge is at an angle of A degrees to the horizontal. There is no friction between the block and the wedge. The system is released from rest.

Determine:
1) The horizontal acceleration of the wedge
2) The horizontal acceleration of the block
3) The vertical acceleration of the block

I don't know how to solve any of these.. I've applied Newton's laws, but I get only 3 equations. I'm guessing there's some additional constraint which gives me the fourth one, but I haven't been able to find it. Should I try Newton's laws in laboratory coordinates?

This is what I know so far :
1) The wedge accelerates horizontally to the left ( by the normal b/w wedge,block)
2) The block accelerates horizontally to the right ( by the normal b/w wedge,block)
3) The block accelerates downwards vertically ( I don't know how to reason this one.. perhaps because if it didn't accelerate downward vertically, then it would lose contact with the wedge, but I'm not sure)
4) The centre of mass of the system remains stationary horizontally as all the horizontal forces are internal, so they're cancelled out by Newton's 3rd law.
5) If the block accelerates vertically, and the wedge stays on the same vertical level, then the centre of mass of the system accelerates vertically.

Please help.
 
Jun 2016
1,239
588
England
To get things to move you need to supply energy
As you indicate (in point 5 in your original post) the centre of mass of the system moves downwards.
This supplies the energy for the lateral movements.

Note that the wedge cannot move downwards (it rests on the table)
so the only way to get the centre of mass of the system to go down, is if the centre of mass of the Block goes down.

You are correct that, while the centre of mass of the block and the centre of mass of the wedge both move (in opposite directions) the horizontal location of the centre of mass of the system stays the same.
so P.M + p.m stays constant
where P is the position of the Centre of Mass of the Wedge
and p is the position of the Centre of Mass of the Block.
 
Last edited:
Apr 2015
1,084
249
Somerset, England
The base of the wedge is at an angle of A degrees to the horizontal.
What does this mean?

Do you have a diagram?