p0414

# p0414 - 4.14. CHAPTER 4, PROBLEM 14 275 4.14 Chapter 4,...

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Unformatted text preview: 4.14. CHAPTER 4, PROBLEM 14 275 4.14 Chapter 4, Problem 14 Problem: An electric motor is pulling a block of mass, m, at constant speed, V , up an incline that is at an angle φ to the horizontal. The coefficient of kinetic friction between the block and the incline is µ. (a) Find the average output power, Pp , delivered by the motor. Express your answer as a function of µ, φ, V , m, and gravitational acceleration, g = |g|. (b) If µ = 1 , find the angle φo for which Pp is a maximum. 4 (c) For µ = 1 , φ = φo , m = 10 kg and V = 30 m/sec, determine the efficiency of the motor if the 4 average power required to run it is Pe = 5 kW. . ... .......... .............. ...................... ....................... ...... ............ ... ..... ........................... ...................................... .. .. . . .. ............................ ... ............................. ... ............................. .. ... ...................... ... ........................ ... ....................... . .. ............ .................................................... .. . . .. ... .......... ... .......................................................................... .. .. .. ............................................................................................... . .............................. ... ........................ ............ .. .......... .. ... .................................................................................................................................... ... ...................................................................................................................................... . ... . . ..... ... ...................................................................... . ... ......................................................................................... .... ......................... . . . .................. .................. . ... ............................. ... .................................. . . ... ... ................................ .. . ....... ... .................................. . . . ... .......... ........ ... ....................... ... ..... ... ......................... ... .......................... ... .. ............................ . ............... . . ... ....................... . ...................... ... ......................... . . .. .. . . .. .... ..... ....................... . .... ..... ......................... . ...................... . . . . ...... . . . . ... .. ...................... .... ........... . ... ... .......................... . . ... ..... ....................... . .. . . ..................... ... ..... .......................... ................ . . . . . . .. ... .................. ... ... ....... .... ....................... . ..................... . ..... . . .......... ... . ..... .... ............ . ............... ..................... .................. ..................... .. . . . . . .... .......... .... . ..... ...... ...................... ........................ ............................... ........................ .. . ......................................... .............................. . . .. . .. . .. . ... .................................... .. ....... .. . ............................................................... ............................................................... .. . ... .. .................................................. . ............................................... ... . .. .......... ... . .............................................. .............. ............... ................................... ................................. . ... .................. .. . . ........................... . ............... . ..... . . ...................... . ..... ... . ................................................................................ . . ..... . .... ..... .. . . . . . ... .. .. .. ...... .. .. .. ........................................................................................ ........................................................ .......................................................................................... . ................................................................. ............................................................................................... . ...................................................................................... .. . . . . . . . ... ..... . . . . . . . .............................................................. . . . ................................... .. .. .. .. ..... .. .................................................................................... ..... • V g m • φ Solution: To solve, we first balance forces acting on the block. Next, we compute the average output power, Pp , required to pull the block up the incline. Then, we use the result to determine the incline angle for which Pp is a maximum and the efficiency of the motor for this angle. (a) The forces acting on the block when it’s on the incline are as shown below. The block’s acceleration is zero because it is not moving in the direction normal to the chute, and its velocity is constant. .. ... .... .... .. ... .... ... ... .... ... ................. .. .................. .............. .. ... .................. . ............. .. ... ....................... . . .. . .............. ... ................ ..... ..... ... ....... ......... . .................... .......... .... ................ .............. ................... ...... ... .. ... .. . .... .. ...................... ................. ........................ ................. ............... .. ......................... .................. . . .. .. ........ .. ....... ... .. .................................................... ................................................... .................... ... ..... ....................................................... ................................................... .... . ... .. ................ ...................................... ... .. .. ... . ......................................... .. . ................................... ..... .. ........ . ...... .................... ..... . ...... ................. ....... ....... .. ... ................... ... . ... ....................... .. . .. . .. .... .... .................... .... ... .......... .. . .. ... ............... ..... .. ... ..... .................. ... . .. .... ....................... ............... ... ... .... ....... ............... .. . .. ... ... .... ....... . .. .. .. ................ . ... .............. . ................ . . ... .... ... ... .............. . ................ . ............... . ... . .. . ... .............. . .................. ....... .... ... . . .................. ... . . ....... .. ... . .............. ... ....... .... .............. .. .. ... ... .. .......... . .............. . .. . ............. .............. ..... .. .. . .. . ..... ... . ............. ... .. ............... .................... . . ..... ..... ...... ... ................. .. ................. .. .. . ... .. . .............. .............. ............... . . . . ...... . .............. . . ..... ..... . ................ . .. ....... ............. .... .. . .. .............. ........................... ................................ ............... ............................................................. . F µN N mg φ So, the forces must sum to zero. Thus, we have F − mg sin φ − µN = 0 Solving for F , we have F = mg (sin φ + µ cos φ) Thus, the motor’s average output power is Pp = mgV (sin φ + µ cos φ) and N − mg cos φ = 0 276 CHAPTER 4. WORK AND ENERGY (b) To find the angle for which Pp is a maximum, φo , we compute the first and second derivatives of Pp , and determine what is needed for the following conditions to hold. dPp = mgV (cos φ − µ sin φ) = 0 dφ d2 Pp = −mgV (sin φ + µ cos φ) < 0 dφ2 Since µ > 0, The second derivative of Pp is obviously negative for 0o ≤ φ ≤ 90o . The first derivative vanishes when 1 tan φo = µ Hence, for µ = 1 , the angle at which Pp is a maximum is 4 φo = tan−1 (4) = 76.0o (c) For the given values of µ = 1 , m = 10 kg and V = 30 m/sec, the average output power when φ = 76o 4 is 1 Pp = (10 kg) 9.807 m/sec2 (30 m/sec) sin 76o + cos 76o = 3.033 kW 4 Finally, since the power required to run the motor is Pe = 5 kW, the motor’s efficiency is η= 3.033 kW Pp = = 60.7% Pe 5 kW ...
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## This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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