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19 August, 19:17

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

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  1. 19 August, 22:51
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    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
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