p0236 - 2.36. CHAPTER 2, PROBLEM 36 139 2.36 Chapter 2,...

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Unformatted text preview: 2.36. CHAPTER 2, PROBLEM 36 139 2.36 Chapter 2, Problem 36 Problem: A man is pulling a block up an incline using an inextensible cable wraps around two pulleys as shown. Due to fatigue, the man’s speed decreases exponentially, i.e., v = vo e−t/τ , where vo is his initial speed and τ is a constant of dimensions time. At what time is the block’s deceleration vo /(40τ )? . . . ........... .. .. .. . .. ............ .. .. ... . ............ 3 ............ .. . . ..... . ... ... .............. .............. . . .. . ... ................................... ................................... ... . .. . . . ... ... .. . .•. ... ... .. . . . .. ... ..... ... ... .......... . .. 2 ..... ... ............................................................................................................... ... .... ........................................................................................................... .... . . ...................... .. .. .. ...................................... .. .. .. . .. ..... .. . . .... . . . .. .. . ... ... .. .............................................................................................................................. ...... ... .. ................................................................................................................................... ... ... . ... ... ....................................................................................................................................... . . ... ... ......................................................................................................................................... ... .... . . ... ... ... ... .................... . ..... ... ... .......................... ... . ....... .. ........ .. ..... ... ... ..................... . ... ... ............................ ... ....... .. ... ............................ ..... ............... ... ................................ ..... . . . .... .. .................. . . . . ................. . .. ... .... ...................... ......... .. .. .. ..... . .... ... ............... .. .•... ..................... . ...... ... . ... ........ ...... ...................... ........... ..... ................ ... 1 ................. ................. ... ... ..... ... . . . . .............. . ............. .......................... . ............. ............................... ................................................. .................................................... . .... ........................................................... ..... ... . .. .. .. .................. ... .................................. . .... .. . .................................. . .. .. ............... ........................... ... ................... . ... ... ............... . . .. . . . ........ ............... .................... ............... ...................... ..... .. . .. ... ............ ........ . ................. . . ........... . ............................... .............................. ............ . ... ... .. ..... ........ . x v x z t x m θ Solution: The total cable length is xtot = x1 + x2 + x3 + constant where the constant represents the portion of the cable that makes contact with the pulleys. Since the cable is inextensible, its total length cannot change so that x1 + x2 + x3 = 0 ˙ ˙ ˙ Clearly, since x1 = x2 , we must have x1 + x2 = 0. Thus, ˙ ˙ 2x1 + x3 = 0 ˙ ˙ Solving for x1 yields ˙ 1 1 x1 = − x3 = − v ˙ ˙ 2 2 ˙ But, the block’s speed, vblock , is −x1 . Therefore, vblock = The block’s acceleration is ablock = vo 1 1d v= ˙ vo e−t/τ = − e−t/τ 2 2 dt 2τ 1 v 2 So, the block’s deceleration is equal to vo /(40τ ) when − Solving for t yields t = 3τ vo vo −t/τ =− e 2τ 40τ =⇒ e−t/τ = 1 20 ...
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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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