Question

explain relation between cofficent of friction and cofficent of repose


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Answer 1

Answer:

Because of the angle of inclination, this weight can be resolved into two components; mgcosθ and mgsinθ as shown in fig. Thus, coefficient of limiting friction is equal to the tangent of the angle of repose.

I hope it's helpful for you

Answer 2

Explanation:

if u is coefficient of static friction and @ is the angle of repose then relation between is ------

@ = tan^-1 (u)

angle of repose :- it is the minimum angle that an inclined plane makes with the horizontal when a body pleced on it just begins to slide down .

let any body which mass is m placed in inclined which inclination with horizontal is @ .

then ,

you resolve the mg along inclination and perpendicular to inclination.

you see .

in equilibrium condition.

mg.sin@ = friction.

but we know,

friction =uN

where N is normal reaction.

also N =mg.cos@

then,

mg.sin@ = umg.cos@

tan@ = u

hence,

@ =tan^-1 (u)

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Hope it helps you !☺✨

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