From the diagram below, a block of mass lies on the surface of mass . The wedge is connected to another mass hanging over a pulley of mass . What must be the hanging mass such that the smaller mass on the wedge does not slip up or down relative to the wedge?
Assumptions:
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1) The horizontal accelerations of M ′ and m are equal, and are the same as the vertical acceleration of M
2) The vertical acceleration of m is zero
Let T be the rope tension and N be the magnitude of the normal reaction force between m and the ramp. Translating into math yields:
M M g − T = M ′ T − N s i n θ = m N s i n θ N c o s θ = m g
Substituting in for the normal force magnitude gives:
M M g − T = g t a n θ M ′ T − m g t a n θ = g t a n θ
Further substitution:
M ( g − g t a n θ ) = T = M ′ g t a n θ + m g t a n θ M = g − g t a n θ M ′ g t a n θ + m g t a n θ = c o t θ − 1 M ′ + m