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Newton's Laws: Wedge and block magnitude, acceleration

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A.) A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest. Calculate:
i) the acceleration of the wedge
ii) the horizontal and vertical components of the acceleration of the block. (Do these answers reduce to the correct results when M is very large?)

b) The wedge and block now have a horizontal force P applied to the wedge. What magnitude must P have if the block is to remain at a constant height above the tabletop?

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a.) let the slope of wedge with horizontal = Q
see freebody diagram in attached jpg file.
Let the acceleration of the wedge is a and the acceleration of block with respect to wedge along the slope is a'.
N = mg.cos(Q) - m.a.sin(Q)
horizontal component of N on wedge:
Nh = {mg.cos(Q) - ma.sin(Q)}.sin(Q)= mgcos(Q).sin(Q) -ma.sin^2(Q)
vertical component of N:
Nv = {mg.cos(Q) - ma.sin(Q)}.cos(Q) = mg.cos^2(Q) - ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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