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 magnitude of the friction force exerted on the ladder by the floor is mathematically given as
x= 33.9 Pounds
Question Parameter(s):
A uniform ladder 15 ft long is leaning against a frictionless wall at an angle of 53
The weight of the ladder is 30 pounds. a 75-lb boy climbs 6.0-ft up the ladder.
Where Clockwise torque = Counterclockwise torque
Generally, the equation for the statement is mathematically given as
W (7.5 ) cos 53° + Wb (6 ) cos 53° = x (15 ) sin 53 °
(20 lb) (7.5 )cos 53° + (75 lb) (6 ) cos 53° = x(15 ) sin 53°
Therefore
x = ((30 *7.5 * cos 53°) + (75 * 6 * cos 53°))/ (15 * sin 53°)
x= 33.9 Pounds
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