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Physics-

General

Easy

Question

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle null and that between the floor and the ladder is $mu subscript 2 end subscript$ . The normal reaction of the wall on the ladder is $N subscript 1 end subscript$ and that of the floor is $N subscript 2 end subscript$ . If the ladder is about to slip, then

$μ_{1} = 0 μ_{2} \neq 0 and N_{2} t a n θ = \frac{m g}{2}$
$μ_{1} \neq 0 μ_{2} = 0 and N_{1} t a n θ = \frac{m g}{2}$
$μ_{1} \neq 0 μ_{2} \neq 0 and N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$
$μ_{1} = 0 μ_{2} \neq 0 and N_{1} t a n θ = \frac{m g}{2}$

The correct answer is: $mu subscript 1 end subscript not equal to 0 mu subscript 2 end subscript not equal to 0 text and end text N subscript 2 end subscript equals fraction numerator m g over denominator 1 plus mu subscript 1 end subscript mu subscript 2 end subscript end fraction$

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