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A weightless ladder 20 ft long rests against a frictionless wall at an angle of 60° 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

1. (a) 175 lb
2. (b) 100 lb
3. (c) 120 lb
4. (d) 69.2 lb

Correct answer: (d) 69.2 lb

Solution

To solve, we apply the torque equilibrium condition about the base of the ladder. The torque due to the man's weight and the ladder's reaction force must balance the horizontal force. After calculation, the horizontal force required is approximately 69.2 lb.

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