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relationship between angle of friction and angle of repose

Arka , 6 Years ago

Grade 11

3 Answers

Sujit Kumar

Last Activity: 6 Years ago

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 = $\alpha$ Suppose angle of friction =

Sujit Kumar

Last Activity: 6 Years ago

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 = $\alpha$ Suppose angle of friction =

Sujit Kumar

Last Activity: 6 Years ago

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, $\mu$ ( $\mu=tan\theta$ Therefore, $\frac{F}{R}=\frac{mgsin\theta}{mgcos\theta}=\mu$ ---------eq.2On dividing eq.1 by eq.2 we get, $R=mgcos\theta$ -------- eq.1 $F=mgsin\theta$ opposite to FIn equilibrium, $mgsin\theta$ opposite to Rand $mgcos\theta$ 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:- $\theta$ and angle of repose = $\alpha$ Let angle of friction =