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A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip. The acceleration is:
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Q: A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip. The acceleration is:

Question Analysis & Solution

Step-by-Step Solution

  1. Draw a free body diagram: Draw the free body diagram of the block in the accelerating frame of reference. The forces acting on the block are:
    • Weight (mg) acting vertically downwards.
    • Normal reaction (N) from the wedge, acting perpendicular to the inclined surface.
    • Pseudo force (ma) acting horizontally opposite to the direction of acceleration of the wedge.
  2. Resolve the forces: Resolve the normal reaction N into its horizontal and vertical components:
    • Vertical component: N cos θ
    • Horizontal component: N sin θ
  3. Apply Newton's Second Law: Since the block does not slip, the net force on the block in both horizontal and vertical directions must be zero.
    • Vertical equilibrium: N cos θ = mg
    • Horizontal equilibrium: N sin θ = ma
  4. Solve for acceleration (a):
    • From the vertical equilibrium equation, we have: N = mg / cos θ
    • Substitute this value of N into the horizontal equilibrium equation: (mg / cos θ) * sin θ = ma
    • Simplify to find the acceleration: a = g tan θ

Correct Answer: g tan θ

AI Suggestion: Option D
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