JEE Advance - Physics (2014 - Paper 1 Offline - No. 13)
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 N1 and that of the floor is N2. If the ladder is about to slip, then
$$\mu$$1 = 0, $$\mu$$2 $$\ne$$ 0 and $${N_2}\tan \theta = {{mg} \over 2}$$
$$\mu$$1 $$\ne$$ 0, $$\mu$$2 = 0 and $${N_1}\tan \theta = {{mg} \over 2}$$
$$\mu$$1 $$\ne$$ 0, $$\mu$$2 $$\ne$$ 0 and $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$
$$\mu$$1 = 0, $$\mu$$2 $$\ne$$ 0 and $${N_1}\tan \theta = {{mg} \over 2}$$
Explanation
As the rod is about to slip, wall and floor exert limiting friction on the ladder.
Case 1 : If $$\mu$$1 = 0, $$\sum {{{\overrightarrow \tau }_A} = \overrightarrow 0 } $$
$$mg\left( {{l \over 2}\cos \theta } \right) = {N_1}(l\sin \theta )$$
$${N_1} = {{mg\cot \theta } \over 2}$$
or $${N_1}\tan \theta = {{mg} \over 2}$$
and $${N_2} = mg$$
Case 2 : If $$\mu$$2 = 0, N1 remain unbalanced and rod can never be in equilibrium.
Case 3 : If $$\mu$$1 $$\ne$$ 0, $$\mu$$2 $$\ne$$ 0
$${N_1} = {f_2} = {\mu _2}{N_2}$$
$${N_2} + {f_1} = mg$$
or $${N_2} + {\mu _1}{N_1} = mg$$
or $${N_2} + {\mu _1}({\mu _2}{N_2}) = mg$$
or $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$
Case 1 : If $$\mu$$1 = 0, $$\sum {{{\overrightarrow \tau }_A} = \overrightarrow 0 } $$
$$mg\left( {{l \over 2}\cos \theta } \right) = {N_1}(l\sin \theta )$$
$${N_1} = {{mg\cot \theta } \over 2}$$
or $${N_1}\tan \theta = {{mg} \over 2}$$
and $${N_2} = mg$$
Case 2 : If $$\mu$$2 = 0, N1 remain unbalanced and rod can never be in equilibrium.
Case 3 : If $$\mu$$1 $$\ne$$ 0, $$\mu$$2 $$\ne$$ 0
$${N_1} = {f_2} = {\mu _2}{N_2}$$
$${N_2} + {f_1} = mg$$
or $${N_2} + {\mu _1}{N_1} = mg$$
or $${N_2} + {\mu _1}({\mu _2}{N_2}) = mg$$
or $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$
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