SATHEE: System of Particles and Rotational Motion

System of Particles and Rotational Motion - Result Question 46

49. A weightless ladder $20 f t$ long rests against a frictionless wall at an angle of $60^{\circ}$ from the horizontal. A 150 pound man is $4 f t$ 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

(a) $175 l b$

(b) $100 l b$

(d) $69.2 l b$

[1998]

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Answer:

Correct Answer: 49. (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 $W = N_{1}$

For horizontal equilibrium, $N_{2} = F$

Taking moments about $A$ , $N_{2} (A B \sin 60^{\circ}) - W (A C \cos 60^{\circ}) = 0$

Using (2) and $A B = 20 f t, B C = 4 f t$ , we get

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System of Particles and Rotational Motion - Result Question 46

49. A weightless ladder 20ft long rests against a frictionless wall at an angle of 60∘ from the horizontal. A 150 pound man is 4ft 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

Answer:

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49. A weightless ladder $20 f t$ long rests against a frictionless wall at an angle of $60^{\circ}$ from the horizontal. A 150 pound man is $4 f t$ 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