# Mass on a wedge. Relative acceleration.

1. Jul 16, 2011

### Darth Frodo

1. The problem statement, all variables and given/known data

A particle of mass M is on a wedge of mass 2M. The wedge is smooth, and is inclined at 30* to the horizontal. The system is released from rest. Find the acceleration of the wedge, and find the acceleration of the particle relative to the wedge.

2. Relevant equations

F=ma

3. The attempt at a solution

Particles motion parallel to the wedge

F=ma
(mg)sin30 = m[f - (a)cos30]
mg = 2mf - ma$\sqrt{3}$

Particles motion perpendicular to the wedge

F = ma
(mg)cos30 - R = m(asin30)
mg$\sqrt{3}$ -2R = ma
R = $\frac{m(g\sqrt{3} - a)}{2}$

Wedges Motion

F = ma
Rsin30 = 2ma
R = 4ma

Substitution

$\frac{m(g\sqrt{3} - a)}{2}$ = 4ma

mg$\sqrt{3}$ - ma 8ma

g$\sqrt{3}$ - a = 8a

g$\sqrt{3}$ = 9a

a =$\frac{g\sqrt{3}}{9}$

The answer at the back of the book is $\frac{g}{3\sqrt{3}}$

2. Jul 16, 2011

### ehild

Your solution is the same as that of the book. (Write 9 as 3*√3*√3 and simplify by √3.)

ehild