# Homework Help: Hard inclined plane problem

1. Oct 30, 2009

### bon

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

so a block of mass m slides on a frictionless surface of an inclined plane of angle theta which itself has mass M and can slide on a horizontal surface. Assuming no friction, find the accel of the block and the inclined plane..

3. The attempt at a solution

well i see that there are two components to the accel of the block and one component to the accel of plane, giving 3 equations...but how do i solve to find the accel of the block and the plane? is there another constraint equation i'm missing?

2. Oct 30, 2009

### tiny-tim

Welcome to PF!

Hi bon! Welcome to PF!
The geometrical constraint is that the block stays on the plane …

how would you write that?

(oh, and I'd go for conservation of energy)

3. Oct 30, 2009

### bon

Re: Welcome to PF!

ahh thanks i think ive got it now..

just working it through..stuck on the last line..

the bottom of the fraction should be sin^2(theta) + M/m

but i have M/m + M/m (sin^2(theta)) - sin^2(theta)...

i know these are equivalent..but how do i simplify my expression to sin^2(theta) + M/m

4. Oct 30, 2009

### bon

actually i think i've made a mistake..

i can't see why though in my working!

the constraint is: tan theta = ay (block) / ax (block) + ax (plane), yes?

5. Oct 30, 2009

### tiny-tim

Hi bon!

(have a theta: θ and try using the X2 tag just above the Reply box )
Nooo …

try it with x and y, rather than ax and ay,

and concentrate on the position of the block relative to a fixed point on the plane.

6. Oct 30, 2009

### bon

thanks.. :)

so so tanθ = y / xblock - xplane

or tanθ = y / xplane - xblock??
which one?

also then could you differentiate twice to get ratio of accel's? what would the ratio be?

Many thanks again

7. Oct 30, 2009

### tiny-tim

(hmm … brackets would help )

you decide!
A linear combination of x and y will be the same linear combination of ax and ay

8. Oct 30, 2009

### bon

i really don't know which one it is?! how can i decide?

thanks

9. Oct 31, 2009

### rl.bhat

Such problems can be solved by identifying the forces acting on each objects, and noting down their relations.
In this problem, the forces acting on the block are
mg*sinθ..along the inclined plane
mg*cosθ ....perpendicular to the inclined plane
Normal reaction N due to the wedge. Since wedge is moving horizontally, mgcosθ is not equal to N. The block is moving along the wedge.
So (mg*cosθ - N) = m*ay, where ay is the acceleration of m along the normal,say y.
Hence N = (mg*cosθ - m*ay)....(1)
When M moves horizontally through X, block moves through y along normal such that
y = X*sinθ. Οr ay = Ax*sinθ ..(2) where Ax is the acceleration of the wedge.
Forces acting on wedge are
Weight Mg downward, normal reaction N1 from the ground upward and normal force N due to block. Only component which moves the wedge horizontally is N*sinθ
So N*sinθ = M*Ax ...(3)
Using equations (1) and (2) solve for Ax.