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A block is placed on an inclined plane whose angle of inclination is greater than the angle of repose. If the coefficient of friction is mu (a) find force P applied parallel to incline that will keep it in equilibrium (minimum and maximum value of force). (b) Find the range of P if it is applied horizontally on the block. ( c) What is the minimum value of P if it is applied perpendicular to the inclined plane? |
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Answer» SOLUTION :(a) (i) If the block has tendency to move down , friction will act in the upward direction `_|_^(AR)` to the PLANE : `N = mg cos theta , f = mu N = mu mg cos theta` Along the plane : `P + f = mg sin theta rarr P = mg sin theta - f` `P_(min) = mg sin theta - mu mg cos theta` `= mg ( sin theta - mu cos theta)` (ii) If the block has tendency to move up , friction will act in the downward direction `P = mg sin theta + f` `P_(max) = mg ( sin theta + mu cos theta)` (b) (i) If the block has tendency to move down , friction will act in the upward direction `_|_^(ar)` to the plane : `N = mg cos theta + P sin theta` `f = mu N = mu (mg cos theta + P sin theta)` Along the plane : `P cos theta + f = mg sin theta` `P cos theta + mu(P sin theta + mg cos theta) = mg sin theta` `P (cos theta + mu sin theta) = mg ( sin theta - mu cos theta)` `P = (mg (sin theta - mu cos theta))/((cos theta + mu sin theta))` `P_(min) = (mg (tan theta - mu))/((1 + mu tan theta))` (ii) If the block has tendency to move up , friction will act downwards `P_(max) = (mg (tan theta + mu))/(( 1 - mu tan theta))` ( c) `_|_^(ar)` to the plane : `N = P + mg cos theta , f = mu N = mu (P + mg cos theta)` Along the plane : `mg sin theta = f = mu(P + mg cos theta)` `P_(min) = mg((sin theta)/(mu) - cos theta)`    
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