In the figure- a ladder of mass m is shown leaning against a wall- It is in static equilibrium making an angle - with the horizontal floor- The coefficient of friction between the wall and the ladder is - and that between the floor and the ladder is -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 - 0- -2- 0- and -N-2- tan- - mg-2-1- 0- -2 - 0 -and- N-1- tan- - mg-2-1-0- -2- 0 -and- N-2 - mg-1-1-2-1 - 0- -2- 0- and- N-1- tan- - mg-2

The correct option is D μ_1 = 0 μ _2≠ 0 and N_1 tan θ = mg/2Consider of translational equilibrium N_1 = μ_2N_2 (i) ...

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The Complete Equilibrium

In the figure...

Question

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 $μ$ 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

μ_{1} = 0 μ_{2} \neq 0 a n d N_{2} t a n θ = m g / 2

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μ_{1} \neq 0 μ_{2} = 0 a n d N_{1} t a n θ = m g / 2

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μ_{1} \neq 0 μ_{2} \neq 0 a n d N_{2} = \frac{m g}{1 + μ_{1} μ_{2}}

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μ_{1} = 0 μ_{2} \neq 0 a n d N_{1} t a n θ = m g / 2

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