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Correct options are C) and D)

$N_{1}=μN_{2}$

Balancing forces in vertical direction,

$N_{2}+μN_{1}=mg$

Solving we get,

$N_{2}=1+μ_{1}μ_{2}mg $

$N_{1}=1+μ_{1}μ_{2}μ_{2}mg $

Applying rotational equilibrium,

$mg2l cosθ=N_{1}lsinθ+μN_{1}lcosθ$

Solving, $tanθ=2μ_{2}1−μ_{1}μ_{2} $

If $μ_{1}=0,μ_{2}=0$

$tanθ=2μ_{2}1 $

$N_{1}=μ_{2}mg$

or $N_{1}tanθ=2mg $

Solve any question of Laws of Motion with:-

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