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Zorluk: ZorWheel and Axle

A bucket is being raised from a well using a windlass mechanism (a type of wheel and axle). The cylindrical drum (axle) around which the rope winds has a diameter of 8 inches8 \text{ inches}. The crank handle used to turn the drum has a length (radius) from the center of rotation of 20 inches20 \text{ inches}. If the bucket is raised vertically by exactly 15 feet15 \text{ feet}, what is the total distance, in feet, that the crank handle must travel along its circular path? (Assume the rope does not overlap itself on the drum).

Cevap: 75 feet

Cevap

75
In any simple machine, assuming 100% efficiency, the work put in equals the work put out. The Ideal Mechanical Advantage (IMA) for a wheel and axle is the ratio of the effort radius to the load radius. Here, the load radius (the drum) is 4 inches4 \text{ inches} (half of the 8 inch8 \text{ inch} diameter) and the effort radius (the crank) is 20 inches20 \text{ inches}. Thus, the IMA is 20/4=520 / 4 = 5. This geometric relationship dictates that the effort force must travel 55 times further than the load to conserve work. If the load is lifted 15 feet15 \text{ feet}, the effort distance is 15×5=75 feet15 \times 5 = 75 \text{ feet}.

Adım Adım Çözüm

1
Calculate the drum's radius.
4 inches4 \text{ inches}
The Ideal Mechanical Advantage compares radii, but the drum's dimension was provided as a diameter.
2
Calculate the Ideal Mechanical Advantage (IMA).
IMA = 55
Dividing the effort radius (20 inches20 \text{ inches}) by the load radius (4 inches4 \text{ inches}) gives the mechanical advantage of the system.
3
Calculate the total distance traveled by the crank handle.
75 feet75 \text{ feet}
By the principle of conservation of energy, the effort distance must be equal to the load distance (15 feet15 \text{ feet}) multiplied by the IMA (55).

Anahtar Kavram

Ideal Mechanical Advantage (IMA) and Force-Distance Tradeoff
Tahmini Süre:1m 30s
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Konular

Wheel and Axle

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