Trouble interpreting fictitious forces for block on wedge problem

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Homework Statement





This is a standard block on wedge problem - we have an incline of angle [tex]\alpha[/tex], a block of mass m on wedge of mass M. The block is released from the inclined surface. The wedge is not fixed and can accelerate. The question is typical - find the horizontal accelerations of the block and wedge a1 and a2 respectively, and the normal foces N1 and N2 (in terms of the masses the angles and g and stuff). I know this is a standard problem and I in fact know how to solve it using the constraint invoking the geometry of the incline.

However, I really want to do this problem using fictious forces. I want to use a reference frame in which the wedge is fixed.

Homework Equations



Let inertial frame be S, accelerated one be S' (not rotated).

ma = F + Ffictitious

The Attempt at a Solution



I thought the above equation wold make this problem fall out very quickly using the above equation, because the acceleration of the wedge with respect to the inertial frame (and hence the fictitious force) is easy to calculate once we know the normal force N1.

Now, in the frame fixed with respect to the wedge, I thought we would have N1=mgcos[tex]\alpha[/tex] (and consequently an acceleration of the block down the incline of gsin[tex]\alpha[/tex]). But this isn't the case, right? Why not, though? And how would we solve for N1 using our fictitious force framework?
 

Answers and Replies

  • #2
Now, in the frame fixed with respect to the wedge, I thought we would have N1=mgcos[tex]\alpha[/tex] (and consequently an acceleration of the block down the incline of gsin[tex]\alpha[/tex]). But this isn't the case, right? Why not, though? And how would we solve for N1 using our fictitious force framework?

It is not so because the fictitious force is also acting. So you should take that into account as well. Show complete working preferably with a diagram so we could tell where you are wrong.
 
  • #3
Well I haven't been able to use fictitious forces, so there is no point in me writing down the working out. Fictitious forces act IN addition to the forces present in inertial frames, like N1. Shouldn't N1 have the same value in the inertial frame and the frame in which the block is fixed? So how do we go about finding N1 (without solving the problem entirely in the inertial frame, which defeats the entire point/advantage of using the noninertial frame of reference)?
 
  • #4
Doc Al
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Shouldn't N1 have the same value in the inertial frame and the frame in which the block is fixed?
It does; the value of a 'real' force such as N does not depend on the reference frame. But that value is different from what it would be if the wedge were fixed. (Which is an entirely different problem.)
So how do we go about finding N1 (without solving the problem entirely in the inertial frame, which defeats the entire point/advantage of using the noninertial frame of reference)?
Call the acceleration of the wedge 'a'. Then write your force equations for the block in the accelerating frame of the wedge. (How is the block's motion constrained?)
 

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