Question

if the 52 lb bucket is released from rest, determine its velocity after it has fallen a distance of 12 ft . the windlass a can be considered as a 31 lb cylinder, while the spokes are slender rods, each having a weight of 2 lb . neglect the pulley's weight.(figure 1)

Answer 1

The velocity of the bucket after falling a distance of 12 ft is 7.54 m/s.

To determine the velocity of the bucket after it has fallen a distance of 12 ft, we need to use the equation of motion for an object falling under the influence of gravity:

v = √(2gh)

where v is the velocity of the object, g is the acceleration due to gravity (9.8 m/s²), h is the height the object has fallen, and h is the distance the object has fallen.

In this case, h = 12 ft = 3.66 m, so the velocity of the bucket after falling a distance of 12 ft is:

v = √(2 × 9.8 m/s² × 3.66 m) = 7.54 m/s

Note that this is the velocity of the bucket relative to the ground, and does not take into account any effects of the windlass or the spokes.

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