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
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4k4dae/15dee390-ba62-444c-a7ef-a236dd067a41/01e1ee60-6fcf-11ec-905a-a9d085ead7cd/file-1ky4k4daf.png?format=png)

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 ${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1ky4k75qg/d7f51591-cd7a-4201-8c7a-a4faf4eb9fd5/4f798480-6fcf-11ec-905a-a9d085ead7cd/file-1ky4k75qh.png?format=png)

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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