Is the Solution for the Block's Acceleration on an Accelerating Wedge Correct?

AI Thread Summary
The discussion revolves around solving a physics problem involving a block sliding on a wedge that is accelerating. The user derives equations of motion for the block using both inertial and non-inertial frames, ultimately arriving at an expression for the block's acceleration in the y-direction. There is confusion regarding the conditions under which the block moves upward when the wedge's acceleration exceeds gravitational acceleration. The concept of pseudo forces is introduced as a necessary component when applying Newton's laws in non-inertial frames. The final correct expression for the block's acceleration is confirmed to be ay = (A - g) / 2.
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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

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. :approve:
 
is my answer correct??
 
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.
 

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