Question 290: Laws of Motion

Question

In the figure, a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle $$	heta$$ 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

Accepted Answer

As the rod is about to slip, wall and floor exert limiting friction on the ladder.
Case 1 : If $$\mu$$
1
= 0, $$\sum {{{\overrightarrow 	au  }_A} = \overrightarrow 0 } $$
$$mg\left( {{l \over 2}\cos 	heta } ight) = {N_1}(l\sin 	heta )$$
$${N_1} = {{mg\cot 	heta } \over 2}$$
or $${N_1}	an 	heta  = {{mg} \over 2}$$
and $${N_2} = mg$$
Case 2 : If $$\mu$$
2
= 0, N
1
remain unbalanced and rod can never be in equilibrium.
Case 3 : If $$\mu$$
1
$$
e$$ 0, $$\mu$$
2
$$
e$$ 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 $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$

Answer

$$\mu$$
1
= 0, $$\mu$$
2
$$
e$$ 0 and $${N_2}	an 	heta  = {{mg} \over 2}$$

Answer

$$\mu$$
1
$$
e$$ 0, $$\mu$$
2
= 0 and $${N_1}	an 	heta  = {{mg} \over 2}$$

Answer

$$\mu$$
1
$$
e$$ 0, $$\mu$$
2
$$
e$$ 0 and $${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$$

Answer

$$\mu$$
1
= 0, $$\mu$$
2
$$
e$$ 0 and $${N_1}	an 	heta  = {{mg} \over 2}$$

Laws of Motion

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