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Answer :
Final answer:
Given that the crane's hoist motor is rated at 159 hp and it can use only 72.0% of this power for safety reasons, the shortest time it can lift a 5550 kg load over a distance of 89.0 m is approximately 570 seconds.
Explanation:
To solve this problem, one needs to first convert the hoist motor power from horsepower to watts because the internationally recognized unit of power is the watt, not the horsepower. Power is equal to work divided by time. After finding the maximum power of the crane in watts, we calculate how much of this power the crane can really use, considering its safety limit. Then, we can calculate the shortest time it will take to lift a given load to a certain height.
Here are the calculations:
- Convert horsepower to watts: 1 horsepower is approximately 746 watts, so 159 hp * 746 = 118734 watts.
- Calculate the actual power the crane can use: 118734 watts * 72.0% = 85449.28 watts.
- Calculate the work done in lifting the load: work = mass * gravity * height = 5550 kg * 9.8 m/s^2 * 89 m = 48658200 J.
- Calculate the time: time = work / power = 48658200 J / 85449.28 watts = 569.51 seconds.
So, the shortest time in which the crane can lift a 5550 kg load over a distance of 89.0 m is approximately 570 seconds, considering that it can use only 72.0% of its maximum hoisting power for safety reasons.
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Answer:
The value is [tex]t = 56.68 \ s [/tex]
Explanation:
From the question we are told that
The rating of the hoist motor is [tex]k = 159hp = 159 *746 =118614 \ W[/tex]
The percentage of it power used is [tex]z = 0.72 * 118614=85402.08 \ W[/tex]
The mass of the load is m = 5550 kg
The distance is h = 89.0 m
The potential energy required to lift the load through that distance is
[tex]E = m * g * h[/tex]
=> [tex]E = 5550 * 9.8 * 89.0[/tex]
=> [tex]E = 4840710 \ J[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{E}{ z}[/tex]
=> [tex]t = \frac{4840710}{ 85402.08}[/tex]
=> [tex]t = 56.68 \ s [/tex]
