# Block on an accelerating wedge

Hello,

## Homework Statement

A 45o wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find its acceleration in y direction. (Gravity directed down, acceleration due to gravity is g).
[PLAIN]http://img210.imageshack.us/img210/6958/wedge45.jpg [Broken]

## Homework Equations

if i make x' be a horizontal coordinate moving with wedge, then this is in not inertial frame so -mA term adds:
$$m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA$$ (1)
y coordinate is inertial, and Newton second law is:
$$m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg$$ (2)
The only forces as seen by inertial observer are mg and normal force N.
To associate accelerations of x and y in inertial frame I make X (acceleration A) to be position of wedge and x that of block, then:
$$-\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta)$$ (3)
sins theta = 45, tan(45)=1

## The Attempt at a Solution

Solving equations (1) and (2) for N, and dividing one by another, i get:
$$tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g)$$
Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:
$$\frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g$$
Combining this with equation (3), i get:
$$\frac{d^2 y}{dt^2} = (A - g)/2$$
Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.

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dude, the answer is ay= |g-A| /2

it can be brought by simply using the pseudo force on the block. bikramjit das is right ... just use pseudo force ... pseudo forces come into action when tyou need to apply newton's laws in non inertial frame ... F(pseudo) = -ma
m is mass of body on which newton's laws are to be applied and a is acc. of frame ((the bigger block)

note F is opposote to a

well i made few mistakes in my solution, forgot to divide by two and mixed + and - sign in one place. So my final answer is ay = (A-g)/2. What is pseudo i don't know ;] if it's F = -mA, this is kinda way i did this problem. Just i have no idea how to bring this in a simple way.

wat u did is conventional. but pseudo force takes place when the system is in non- inertial frame of ref. Refer google with the term "pseudo force", u'll definitely get what I am trying to say

ok, thank you both for help.