PS4-1 - Problem Set 4: Problem 1. Problem: A conveyer belt...

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Unformatted text preview: Problem Set 4: Problem 1. Problem: A conveyer belt moving with speed V carries boxes to a stationary incline at Point A where they slide and fall off at Point B. The coefficient of sliding friction on the incline is . Determine the speed of the boxes, VB , as they leave the incline at Point B. Express your answer as a function of V , L, , the angle of the incline, , and gravitational acceleration, g. Solution: The first thing we must do is identify the forces acting on each box. Clearly, as shown in the diagram below, there are three forces, viz., weight, mg, the reaction force from the incline, N , and the friction force, N . B.............. B.. ... B............ ......................... . .. ........................... .. . ........................ ..... . .. .. .... . . . . ..................................... ........ . ............................................... ...B.......................................... ....... .................................................... ... . . . . .................................................................. . .................................................................. . . . . mg ... . . . .. . . .......................................................................... ..................................................................................... ........................................................................................................... ........................................................................................................................ . ... ...................................................................................... . ...................................................................................... . . ... .................................................................................... . ....... .................................................................................... ................................................................................ . ..................................................................................... ... .............. .. .. ..................................................................................... ............... ................................................................................. . . . . .. ................................................................................ . . . . ................................................................................... ................................................................................. . . ................................................................................. . . . .. . . . . . . ................................................................................. ......... ............ . ...... . .... ............................................................................. . ................................................................................ ............... . .. ... . ..... . . .. .. . ................................................................................... . ........................................................................................ ... . .. . .. .. ..................................................................................... . .. .. ..... .................................................................................... ........... . ... . . ............................................................................................... . ....................................................................................... ................................................................................ ........................................................................... .. .. ................................................................ .. . ........................................................ . .. ..................................................... .... . .............................................. . .. ....................................... . ............................ .. ........................ ... . ... ......... .. ........ .... .... ... . . . . . . . . . . . . . . . . . . . . . . . N N The kinetic energy at Points A and B is given by 1 and mV 2 2 Only the forces tangential to the surface do work so that TA = TB = 1 mv 2 2 B Also, note that balancing forces normal to the surface tells us that N = mg cos Thus, we conclude that UA-B = (mg sin - N )L Now, the Principle of Work and Energy is UA-B = TB - TA so that mgL(sin - cos ) = Therefore, vB = UA-B = (mg sin - mg cos )L = mgL(sin - cos ) 1 1 mv 2 - mV 2 2 B 2 V 2 + 2mgL(sin - cos ) ...
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This note was uploaded on 10/12/2009 for the course AME 301 at USC.

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