Question

Problem 4 A 2 kg block is released at point A on an inclined plane that is tangent to a circular arc. The plane is tilted 15° form horizontal, and the coefficient of friction is 0.4, Which point is the equilibrium position of the block? 0.4 curve radius 30 cm 25 cm 15 cm (A) A (B) B (C) C (D) D

Answer 1

Given that: - When the block is at point A, its weight mg will act in vertically downward direction. Since The plane is inclined, The component of weight parallel and perpendicular to the plane will be mg sin theta and mg cos theta respectively If block has tendency to move downward due to the mg sin theta component of weight, Frictional Force f will act on the block in opposite or backward direction as shown. f = mu N = mu mg cos theta The block will remain steady i.e., not move until mg sin theta greaterthanorequalto f implies mg sin theta greaterthanorequalto mu mg cos theta i.e., tan theta greaterthanorequalto mu So, for block to move from point A, Tan 15 degree should be greater than mu but (tan 15 degree = 0.27) & (mu = 8.4) implies Hence block will not move and be at possible A itself upon release