Wedge Hedge

From the diagram below, a block of mass m m lies on the surface of mass M M' . The wedge is connected to another mass hanging over a pulley of mass M M . What must be the hanging mass M M such that the smaller mass m m on the wedge does not slip up or down relative to the wedge?

Assumptions:

  • All surfaces are frictionless
  • Pulleys and rope are weightless
  • g = 9.81 m / s 2 9.81 m/s^2
  • Force of tension is the same throughout the whole rope.
( m + M ) / ( c o t θ 1 ) (m + M') / (cotθ - 1) ( m + M ) g t a n θ (m + M')gtanθ 2 m + M c o t θ + 1 2m + M' - cotθ + 1 ( m + M ) / ( t a n θ + 1 (m + M') / (tanθ + 1

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1 solution

Steven Chase
Mar 4, 2019

1) The horizontal accelerations of M M' and m m are equal, and are the same as the vertical acceleration of M M
2) The vertical acceleration of m m is zero

Let T T be the rope tension and N N be the magnitude of the normal reaction force between m m and the ramp. Translating into math yields:

M g T M = T N s i n θ M = N s i n θ m N c o s θ = m g \large{\frac{M g - T}{M} = \frac{T - N sin\theta}{M'} = \frac{N sin \theta}{m} \\ N cos \theta = m g}

Substituting in for the normal force magnitude gives:

M g T M = g t a n θ T m g t a n θ M = g t a n θ \large{\frac{M g - T}{M} =g \, tan \theta \\ \frac{T - m g \, tan\theta}{M'} = g \, tan \theta }

Further substitution:

M ( g g t a n θ ) = T = M g t a n θ + m g t a n θ M = M g t a n θ + m g t a n θ g g t a n θ = M + m c o t θ 1 \large{M ( g - g \, tan \theta) = T = M' g \, tan \theta + m g \, tan \theta \\ M = \frac{M' g \, tan \theta + m g \, tan \theta}{g - g \, tan \theta} = \frac{M' + m }{cot \theta - 1} }

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