Question

A uniform ladder 15 ft long is leaning against a frictionless wall at an angle of 53 above the horizontal. the weight of the ladder is 30 pounds. a 75-lb boy climbs 6.0-ft up the ladder. what is the magnitude of the friction force exerted on the ladder by the floor

Answer 1

Answer;

33.9 pounds

Explanation;

In order for the ladder to be in equilibrium, the net torque should be equal to zero. Therefore, the torque in the opposite directions should equal each other:

Clockwise torque = Counter clockwise torque

Torque is the product of the applied force and the distance between that force and the axis of rotation.

Wι (7.5 ft) cos 53° + Wb (6 ft) cos 53° = F (15 ft) sin 53 °

Substitute the values for the weights of the ladder and the boy, respectively.

(20 lb) (7.5 ft)cos 53° + (75 lb) (6 ft) cos 53° = F(15 ft) sin 53°

Solving for F;

F = ((30 ×7.5 × cos 53°) + (75 × 6 × cos 53°))/ (15 × sin 53°)

  = 33.9 lb

  = 33.9 Pounds