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Block on an accelerating wedge

  • #1
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:
[tex]m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA[/tex] (1)
y coordinate is inertial, and Newton second law is:
[tex]m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg [/tex] (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:
[tex] -\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta) [/tex] (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:
[tex] tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g) [/tex]
Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:
[tex] \frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g [/tex]
Combining this with equation (3), i get:
[tex] \frac{d^2 y}{dt^2} = (A - g)/2 [/tex]
Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.
 
Last edited by a moderator:

Answers and Replies

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

it can be brought by simply using the pseudo force on the block. :approve:
 
  • #3
is my answer correct??
 
  • #4
1,137
0
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
 
  • #5
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.
 
  • #6
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
 
  • #7
ok, thank you both for help.
 

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