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        relationship between angle of friction and angle of repose
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.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 =

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 = $\alpha$Suppose angle of friction =

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, $\mu$($\mu=tan\theta$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 =

one year ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions