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Question

(c) Maximum force by surface when friction works

$F = \sqrt {{f^2} + {R^2}} = \sqrt {{{(\mu R)}^2} + {R^2}} = R\sqrt {{\mu ^2} + 1} $

Minimum fo

Vedclass · Accepted Answer

$Mg \le F \le Mg\sqrt {1 + {\mu ^2}} $

Vedclass · Answer

(c) Maximum force by surface when friction works

$F = \sqrt {{f^2} + {R^2}} = \sqrt {{{(\mu R)}^2} + {R^2}} = R\sqrt {{\mu ^2} + 1} $

Minimum force $ = R$ when there is no friction

Hence ranging from $R$ to $R\sqrt {{\mu ^2} + 1} $

We get, $Mg \le F \le Mg\sqrt {{\mu ^2} + 1} $

A body of mass M is kept on a rough horizontal surface (friction coefficient $\mu $). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is $F$, where

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