Mass on a wedge. Relative acceleration.

[Darth Frodo]

Homework Statement

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.

Homework Equations

F=ma

The Attempt at a Solution

Particles motion parallel to the wedge

F=ma
(mg)sin30 = m[f - (a)cos30]
mg = 2mf - ma[itex]\sqrt{3}[/itex]Particles motion perpendicular to the wedge

F = ma
(mg)cos30 - R = m(asin30)
mg[itex]\sqrt{3}[/itex] -2R = ma
R = [itex]\frac{m(g\sqrt{3} - a)}{2}[/itex]Wedges Motion

F = ma
Rsin30 = 2ma
R = 4ma

Substitution

[itex]\frac{m(g\sqrt{3} - a)}{2}[/itex] = 4ma

mg[itex]\sqrt{3}[/itex] - ma 8ma

g[itex]\sqrt{3}[/itex] - a = 8a

g[itex]\sqrt{3}[/itex] = 9a

a =[itex]\frac{g\sqrt{3}}{9}[/itex]

The answer at the back of the book is [itex]\frac{g}{3\sqrt{3}}[/itex]

 

[ehild]

a =[itex]\frac{g\sqrt{3}}{9}[/itex]

The answer at the back of the book is [itex]\frac{g}{3\sqrt{3}}[/itex]

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

ehild