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Physics - Current Mine Rescue - Can U Solve?
The mine rescue operation in Chile, currently ongoing, divulges data thru the news media and TV that affords some simple physics calculation: Here's the data:
We're told the depth of the mine is 2040 ft.
The weight of the rescue capsule + average weight of (each) man = 1200 lbs.
Time capsule takes from mine floor to ground = 15 min. {assume constant speed}
Time capsule takes from ground to mine floor = 10 min. (assume constant speed}
The large pulley wheel {seen on TV at the top of the lift rig is divided into ten(10) equal angular segments (by spokes). The lift cable subtends approx 3.5 of these segments.
Questions:
1) What is the minimum horsepower(hp) reqd. of lift motor/engine to rescue a miner?
{1.0 hp = 550 ft-lb/s}
2) What is the minimum hp reqd. to lower capsule from ground to mine floor?
3) How many revolutions of the big pulley atop the rig to rescue a miner?
Add missing data to 3): when lifting capsule the large pulley wheel period = 5 sec.
1 Answer
- 1 decade agoFavorite Answer
1) Since this question involves only lifting the weight of the capsule + the weight of a man to the ground from the mine floor, we can easily substitute the data given.
If 1 hp = 550 ft-lb/s, then by dimensional analysis, we will compute power (in hp) by multiplying the distance traveled by the weight of the lifted things and then divided by the time (in seconds) it took to do that job.
P = dw/t
d = 2040 ft.
w = 1200 lbs.
t = 15 min (60 s / min) = 900 s
P = (2040)(1200)/900 = 2720 ft-lb/s
To express in terms of horse power,
P = 2720 ft-lb/s ( 1 hp / 550 ft-lb/s) = 4.9 hp
2) The answer in this question is 0 hp. This is because we will just let the Earth do work on the rescue capsule. If we use the *lift* motor, we will just be doing some negative work on it, thus wasting energy. And also, the individual weight of the rescue capsule was not given so I assume that we will never be able to give an answer to this question assuming that we try to waste some energy.
3) Since the distance that will be traveled by the capsule + miner will be 2040 ft., the velocity will be
v = d/t = 2040 ft / 900 s = 2.3 ft/s
It was said that large pulley wheel is divided into ten equal angular segments, this means,
360/10 = 36 degrees or pi/5 radians per angular segment
The lift cable subtends approx 3.5 of these segments, that means it subtends 3.5 x pi/5 = 7pi/10 radians. This information is insufficient, however. This is because even if we are given the velocity of 2.3 ft/s, we can never know the angular velocity of the pulley: v = rw given that r = radius of the pulley and w is the angular velocity. The angular velocity of the pulley is necessary for the computation of number of revolutions(because we will effectively calculate the total angular displacement of the wheel). We can never know that because w = v/r (it varies indirectly proportional to its radius).
Or we can look at the brighter side.
We can assume that the lift cable is 2040 ft. long and it subtends (as a whole) the large pulley wheel approximately 3.5 of those segments. That means, the wheel will only turn with a total angular displacement of 3.5 of the segments. Which means, 3.5 (pi/5) = 7pi/10 = (7/20) 2pi where 2pi is the angle traversed in one revolution. Therefore, it will only revolve at 7/20 of a single revolution.
