EMT 1220 Horsepower calculation for Elevator Motor- Homework The elevator shown
ID: 1570845 • Letter: E
Question
EMT 1220 Horsepower calculation for Elevator Motor- Homework The elevator shown weighs 4000lb when fully to a 3000 lb eounterwe C and is powered by an electro motor ight Determine the power required when the elevator (a) is moving upward at a constant speed of 25 ft/s, (b) 3000 upward and an upward acceleration of 4 ft/s. Find Force, "F" on the Motor Cable. I) Apply the Formula: Power (Force) x (Velocity) solution. sinoe Fandv have the same direction, the powerla equal to Fo We must first determine the foree F exerted by cable AB on the elevator In each of the two given situations. Force The forces acting on the elevator and on the counter- weight are shown the sketches. a, Uniform Motion. We have a 0, both bodies are in equilib- rium. Free Body +12F, Free Body E: +12F 0: F+ T -(.000) o 4000 lb 3000 Eliminating T we land. F (1,000) b Power. Substituting the given values of o and the values found for Finto the expresslon for the power, we have a. Fo (1.000) IbX25 ft/) (25,000) ft lb/s 1 b Accelerated Motion. We have a 4 ft/s, The equations of motion are +NEF, 3,000 T (4) Free Body E: +13F (4) Eliminating T F (1,000) 4 (1870)lb Power substituting the given values of o and the values found for Finto the expression for the power, we have Power A) 45.45 HP B) 84.90 HPExplanation / Answer
As the solution here is given I am going to explain how he gets 32.2 and how he cancels ft/s^2.
SI units of acceleration is m/s^2 but here the units of acceleration given is ft/s^2. Now 1m= 3.28ft.
Now g (acceleration due to gravity) =9.8m/s^2 = 9.8*3.28ft/s^2 (Putting the value of m in terms of feet) = 32.144ft/s^2 which is here approximately taken as 32.2ft/s^2.
Now as the acceleration is given in ft/s^2 and g is in ft/s^2, they have cancelled the terms of both of them as they have the same units because in denominator they have g and in numerator they have a (acceleration).
This question is set up in a unit system where mass unit is pound (lb), distance unit is feet (ft) so all the calculations are done according to that rest the way of solving the problem will be same in any unit system.
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