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A crank handle with a length of 30 cm is attached to an axle with a radius of 5 cm and is used to lift a bucket of water from a deep well. If the bucket of water weighs 120 N and friction is neglible, (a) how much force is required to turn the crank? (b) find the number of turns of the crank required to raise the bucket to the surface if the well is 510 m deep.
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(a) To calculate the force required to turn the crank, we can use the concept of torque. Torque is equal to the force applied multiplied by the distance from the axis of rotation. In this case, the force required to lift the bucket is equal to its weight, which is 120 N. The distance from the axis of rotation (the axle) to the point where the force is applied (the crank handle) is 30 cm (or 0.3 m). The torque required to lift the bucket is then:
Torque = Force x Distance
Torque = 120 N x 0.3 m
Torque = 36 Nm
The torque required to lift the bucket is 36 Nm.
(b) To find the number of turns of the crank required to raise the bucket to the surface, we can use the work-energy principle. The work done in lifting the bucket is equal to the change in potential energy of the bucket. The potential energy stored in the bucket at the surface is equal to its weight multiplied by the height it is lifted. The height the bucket is lifted is equal to the depth of the well, which is 510 m. The work done in lifting the bucket is then:
Work = Force x Distance
Work = 120 N x 510 m
Work = 61200 J
The number of turns of the crank required to raise the bucket to the surface can be found by dividing the work done by the torque required to lift the bucket:
Number of turns = Work / Torque
Number of turns = 61200 J / 36 Nm
Number of turns ≈ 1700
Therefore, approximately 1700 turns of the crank are required to raise the bucket to the surface.