Auniform 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

33.9 pounds

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

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6

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information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

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