How to find the acceleration of a particle on a freely moving wedge

[gnits]

Could anyone show me the error of my ways with his question please? I must be setting up the equations wrongly.

The question is to find the acceleration of the particle A.

All surfaces are frictionless and the wedge is free to move.

Here's a picture of the setup. The top diagram is from the text book. The lower one is with my additions. Thick lines are forces, thin are accelerations. Black is for the particle. Red is for the wedge.



Hopefully I've labelled all forces and accelerations correctly.

(The small black arrow on A is meant to be labelled N, same size as red arrow N)

(I'll write $\displaystyle \frac{1}{\sqrt{2}}$ for $\displaystyle cos(45)$)

using $\displaystyle F=ma$

Resolving forces for A parallel to the slope:

$\displaystyle 8g\frac{1}{\sqrt{2}} = 8(a_2-\frac{a_1}{\sqrt{2}})$

gives:

$\displaystyle a_1=\sqrt{2}a_2-g$

Resolving forces for A vertically:

$\displaystyle 8g-\frac{N}{\sqrt{2}}=8\frac{a_2}{\sqrt{2}}$

gives:

$\displaystyle 8\sqrt{2}g-N=8a_2$

Resolving forces for B horizontally:

$\displaystyle \frac{N}{\sqrt{2}}=10a_1$

giving:

$\displaystyle N=10\sqrt{2}a_1$

Are these equations correct?

When I solve for $\displaystyle a_2$ I get:

$\displaystyle a_2=\frac{9\sqrt{2}g}{14}$and so to find the acceleration of A need to calculate $\displaystyle a_2-\frac{a_1}{\sqrt{2}}$ for which I get:

$\displaystyle \frac{g}{\sqrt{2}}$Book answer is $\displaystyle \frac{g}{14}\sqrt{106}$

(my answer re-written is $\displaystyle \frac{g}{14}\sqrt{98}$)

Thanks for any help,
Mitch.

 

[gnits]

Solved

No worries. Found my error.