System Of Particles And Rotational Motion Ques 48
48. A weightless ladder $20$ $ ft$ long rests against a frictionless wall at an angle of $60^{\circ}$ from the horizontal. A $150$ pound man is $4$ $ ft$ from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude of the force from the following
[1998]
(a) $175 $ $lb$
(b) $100 $ $lb$
(c) $120 $ $lb$
(d) $69.2 $ $lb$
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Answer:
Correct Answer: 48.(d)
Solution:
- (d) $A B$ is the ladder, let $F$ be the horizontal force and $W$ is the weight of man. Let $N_1$ and $N_2$ be normal reactions of ground and wall, respectively. Then for vertical equilibrium
$W=N_1$ $\quad$ …….(1)
For horizontal equilibrium, $N_2=F$ $\quad$ …….(2)
Taking moments about $A$,
$N_2(A B \sin 60^{\circ})-W(A C \cos 60^{\circ})=0$ $\quad$ …….(3)
Using (2) and $A B=20 ft, B C=4 ft$, we get
BETA
