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 $N _1 \tan \theta=\frac{m g}{2}$
Analytical & Descriptive Questions
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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 $=W _A=20 N$
For vertical equilibrium of $B$;
$f _2=$ friction between block $B$ and wall $=W _B+f _1=100+20=120 N$
