Problem Statement:
A wedge M and a block N are subjected to forces P and Q as shown in the figure. If force P is sufficiently large, then the block N can be raised. The weights of the wedge and the block are negligible compared to the forces P and Q. The coefficient of friction (μ) along the inclined surface between the wedge and the block is 0.2. All other surfaces are frictionless. The wedge angle is 30°.
The limiting force P, in terms of Q, required for the impending motion of block N to just move it in the upward direction is given as P = αQ . The value of the
coefficient ‘α’ (round off to one decimal place) is
Solution: (D) 0.9
Reaction Forces and the frictional forces are named appropriately, as shown in the FBDs.
Friction Force, Ff = μ*Rn1 = 0.20Rn1 —(1)
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For the overall system’s FBD, ∑Fy =0, hence Rn = Q —(2)
Consider the FBD of the wedge M,
As ∑Fx = 0 => P = Rn1*Sin(30˚) + Ff*Cos(30˚)
or P = 0.5Rn1 + 0.20*Rn1*Cos(30˚) or Rn1 = 1.485P —(3)
Also, ∑Fy = 0, => Rn = Rn1*Cos(30˚) – Ff*Sin(30˚)
Put values from eq(1), eq(2) and eq(3)
or Q = 1.485P*Cos(30˚) – 0.20*1.485P*Sin(30˚)
or Q = 1.137P
or P = 0.878Q ≈ 0.9P
hence α ≈ 0.9