In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle \theta with the horizontal floor. The coefficient of friction between the wall and the ladder is \mu_1 and that between the floor and the ladder is \mu_2.The normal reaction of the wall on the ladder is N_1 and that of the floor is N_2.If the ladder is about to slip, then [JEE Adv 2014 P1]

\mu_1=0 \mu_2\neq0 and N_2 tan \theta={mg}/2

\mu_1\neq0 \mu_2=0 and N_1 tan \theta={mg}/2

\mu_1\neq0 \mu_2\neq0 and N_2=\frac{mg}{1+\mu_1\mu_2}

\mu_1=0 \mu_2\neq and N_1 tan \theta={mg}/2

Laws of Motion

Section: Physics

Question : 11 of 20

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In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ1 and that between the floor and the ladder is μ2.The normal reaction of the wall on the ladder is N1 and that of the floor is N2.If the ladder is about to slip, then

[JEE Adv 2014 P1]

μ1=0‌μ2≠0 and N2 tan θ=
mg
2
μ1≠0‌μ2=0 and N1 tan θ=
mg
2
μ1≠0‌μ2≠0 and N2=
mg
1+μ1μ2
μ1=0‌μ2≠ and N1 tan θ=
mg
2

Validate

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