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        relationship between angle of friction and angle of repose
one year ago

							 I.e. angle of friction is equal to angle of repose.= i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.Thus,  = tan= coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact.Now, since  Where  = tan/mgcos ---------eq.2On dividing eq.1 by eq.2 we get,F/R = mgsin -------- eq.1R = mgcos opposite to FIn equilibrium,F = mgsin opposite to Rand mgsin 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 and angle of repose = Suppose angle of friction = one year ago
							 I.e. angle of friction is equal to angle of repose.= i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them. Thus,  = tan= coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact. Now, since  Where  = tan/mgcos ---------eq.2 On dividing eq.1 by eq.2 we get, F/R = mgsin -------- eq.1 R = mgcos opposite to F In equilibrium, F = mgsin opposite to R and mgsin 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 and angle of repose = Suppose angle of friction = one year ago
							Sorry about the previous answers.... Solution: ,since coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.Thus, =coefficient of limiting friction)We know, (Therefore,   ---------eq.2On dividing eq.1 by eq.2 we get, -------- eq.1 opposite to FIn equilibrium, opposite to Rand Let us consider 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:- and angle of repose = Let angle of friction = one year ago
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