Laws of Motion 3 Question 6
6. 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 $N_{1}$ and that of the floor is $N_{2}$. If the ladder is about to slip, then
(2014 Adv.)
(a) $\mu_{1}=0, \mu_{2} \neq 0$ and $N_{2} \tan \theta=\frac{m g}{2}$
(b) $\mu_{1} \neq 0, \mu_{2}=0$ and $N_{1} \tan \theta=\frac{m g}{2}$
(c) $\mu_{1} \neq 0, \mu_{2} \neq 0$ and $N_{2}=\frac{m g}{1+\mu_{1} \mu_{2}}$
(d) $\mu_{1}=0, \mu_{2} \neq 0$ and $N_{1} \tan \theta=\frac{m g}{2}$
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Answer:
Correct Answer: 6. (b)
Solution:
- NOTE
It is not given in the question, best assuming that both blocks are in equilibrium. The free body diagram of two blocks is as shown below,
Reaction force, $R$ =applied force $F$
For vertical equilibrium of $A$;
$f_{1}=$ friction between two blocks $=W_{A}=20 \mathrm{~N}$
For vertical equilibrium of $B$;
$f_{2}=$ friction between block $B$ and wall $=W_{B}+f_{1}=100+20=120 \mathrm{~N}$
