Motion of particle on wedge on smooth surface

[mybrainhurts]

Homework Statement

a particle of mass m moves down the inclined face (angle A) of a wedge of mass M which is free to move on a smooth fixed horizontal table.

by determining the forces acting on i) the particle, ii)the wedge and iii) the system of wedge + particle, show that the acceleration f of the wedge = mgcosAsinA/(M+m(sinA)^2) and f' of the particle with respect to the wedge = (m+M)gsinA/(M+m(sinA)^2)

Homework Equations

f=ma?

The Attempt at a Solution

System as a whole I would assume the only force would be gravitational force acting downwards, which would be completely counteracted by the normal force of the table, leaving no net force, right?

particle: only forces acting on it are gravitational force, straight down, and the normal force of the wedge acting on it, which would be equal to mgcosA, leading to an overall force F1 of mgsinA parallel to the wedge right?

wedge:
forces; gravitational force on the wedge, completely counteracted by normal force of table
force of particle on wedge, equal to mgcosA perpendicular to the surface of the wedge, downwards component would be counteracted by the reaction force of the table, resulting in a total force F2 in the direction of f of mgsinAcosA

this gives f as F2/M = mgsinAcosA/M
and f' as F1/m =gsinA
which is obviously not the solution I was expected to find.

I'm guessing I overlooked something obvious, probably to do with the reaction forces between the particle and the wedge, but can't work out what.

 

[Doc Al]

System as a whole I would assume the only force would be gravitational force acting downwards, which would be completely counteracted by the normal force of the table, leaving no net force, right?

The net force would be zero if there were no vertical acceleration of the system, but is that true? After all, the particle is accelerating down the incline. Hint: Focus on horizontal forces.

particle: only forces acting on it are gravitational force, straight down, and the normal force of the wedge acting on it, which would be equal to mgcosA, leading to an overall force F1 of mgsinA parallel to the wedge right?

No. You can't assume the normal force is mgcosA--that's what it would be if the wedge were fixed.

I'm guessing I overlooked something obvious, probably to do with the reaction forces between the particle and the wedge, but can't work out what.

Don't assume an answer for the normal force--call that force F_n and continue.

Consider the horizontal accelerations of the particle and wedge. What is their relationship? What horizontal forces act on each?

(You may find it helpful to view things from the accelerating frame of the wedge.)

 

[ank_gl]

System as a whole I would assume the only force would be gravitational force acting downwards, which would be completely counteracted by the normal force of the table, leaving no net force, right?

if the net force in vertical dir is 0, n there is no net force in horizontal direction, why would the system change its state at the first place?

the body on the wedge has a reaction perpendicular to the inclined surface, horizontal component of this force is the force which makes the system to change the state

 

[Doc Al]

if the net force in vertical dir is 0, n there is no net force in horizontal direction, why would the system change its state at the first place?

Careful with this reasoning: It is perfectly possible for there to be no net force on a system as a whole yet have its pieces accelerate due to internal forces.