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

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

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.

Wedge2.jpg

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 $\frac{1}{\sqrt{2}}$ for $cos(45)$)

using $F=ma$

Resolving forces for A parallel to the slope:

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

gives:

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

Resolving forces for A vertically:

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

gives:

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

Resolving forces for B horizontally:

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

giving:

$N=10\sqrt{2}a_1$

Are these equations correct?

When I solve for $a_2$ I get:

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

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

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

Thanks for any help,
Mitch.

2. ## Solved

No worries. Found my error.