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Grade: 11
        
relationship between angle of friction and angle of repose
one year ago

Answers : (3)

Sujit Kumar
111 Points
							
 I.e. angle of friction is equal to angle of repose.\alpha=\theta i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.Thus, \alpha = tan\mu= coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact.Now, since \mu Where \theta = tan\mu\theta/mgcos\theta ---------eq.2On dividing eq.1 by eq.2 we get,F/R = mgsin\theta -------- eq.1R = mgcos\theta opposite to FIn equilibrium,F = mgsin\theta opposite to Rand mgsin\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 and angle of repose = \alphaSuppose angle of friction = 213-1012_MASS.JPG
one year ago
Sujit Kumar
111 Points
							
 I.e. angle of friction is equal to angle of repose.\alpha=\theta i.e coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them. Thus, \alpha = tan\mu= coefficient of limiting friction.This is equal to the tangent of the angle of repose between two surfaces in contact. Now, since \mu Where \theta = tan\mu\theta/mgcos\theta ---------eq.2 On dividing eq.1 by eq.2 we get, F/R = mgsin\theta -------- eq.1 R = mgcos\theta opposite to F In equilibrium, F = mgsin\theta opposite to R and mgsin\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 and angle of repose = \alphaSuppose angle of friction = 213-1012_MASS.JPG
one year ago
Sujit Kumar
111 Points
							
Sorry about the previous answers.... 
Solution:
\theta=\alpha ,since coefficient of limiting friction between any two surfaces in contact is equal to tangent of the angle of friction between them.Thus, \mu = tan\alpha=coefficient of limiting friction)We know, \mu(\mu=tan\thetaTherefore,  \frac{F}{R}=\frac{mgsin\theta}{mgcos\theta}=\mu ---------eq.2On dividing eq.1 by eq.2 we get,R=mgcos\theta -------- eq.1F=mgsin\theta opposite to FIn equilibrium,mgsin\theta opposite to Rand mgcos\thetaLet 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:-\theta and angle of repose = \alphaLet angle of friction = 213-1012_MASS.JPG
one year ago
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