A ladder of mass m is 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 μ_1 and that between the floor and the ladder is μ_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 (1) μ_1 = 0, μ_2 ≠ 0 and N_2 tan θ = mg/2 (2) μ_1 ≠ 0, μ_2 = 0 and N_1 tan θ = mg/2 (3) μ_1 ≠ 0, μ_2 ≠ 0 and N_2 = ((mg)/(1 + μ_1 μ_2 )) (4) μ_1 = 0, μ_2 ≠ 0 and N


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Question Number 22479 by Tinkutara last updated on 19/Oct/17

$A ladder of mass m is leaning against a$ $wall . It is in static equilibrium making$ $an angle θ with the horizontal floor .$ $The coefficient of friction between the$ $wall and the ladder is μ_{1} and that$ $between the floor and the ladder is μ_{2} .$ $The normal reaction of the wall on the$ $ladder is N_{1} and that of the floor is N_{2} .$ $If the ladder is about to slip, then$ $(1) μ_{1} = 0, μ_{2} \neq 0 and N_{2} \tan θ = m g / 2$ $(2) μ_{1} \neq 0, μ_{2} = 0 and N_{1} \tan θ = m g / 2$ $(3) μ_{1} \neq 0, μ_{2} \neq 0 and N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}$ $(4) μ_{1} = 0, μ_{2} \neq 0 and N_{1} \tan θ = \frac{m g}{2}$
Commented by Tinkutara last updated on 19/Oct/17

Commented by ajfour last updated on 19/Oct/17

$(3) a n d (4) a r e c o r r e c t .$ $S e e Q . 22481 f o r s o l u t i o n .$
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