Q: A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of resultant downward force along the inclined plane, then find the angle between the inclined plane and the horizontal
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Sol: μ = 0.5 , N = mg cosθ
N = 2F , F = mg(sin θ – μ cos θ)
N = 2mg (sinθ – μ cosθ)
mg cosθ = 2mg (sinθ – μ cosθ)
cosθ = 2 cos θ(tan θ – μ)
1/2 = tan θ – 0.5
⇒ tan θ = 1
⇒ θ = 45°