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Derive relation between angle of friction and angle of repose
Dear Mahesh Suppose angle of friction = alpha and angle of repose = theta Let us suppose a body is placed on an inclined plane as in the above figure.Various forces involved are:- 1. weight,mg of the body,acting vertically downwards.2.normal reaction,R,acting perpendicular to inclined plane.3.Force of friction,F, acting up the plane.Now, mg can be resolved in two components:- mgcos theta opposite to Randmgsin theta opposite to FIn equilibrium, F = mgsin theta -------- eq.1R = mgcos theta ---------eq.2 On dividing eq.1 by eq.2we get, F/R = mgsin theta/mgcos thetamu = tan theta Where mu = coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact. Now, since mu = tan alpha i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.Thus, Theta = alpha i.e angle of friction is equal to angle of repose.
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