p0242

# p0242 - 146 CHAPTER 2. PARTICLE KINEMATICS 2.42 Chapter 2,...

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Unformatted text preview: 146 CHAPTER 2. PARTICLE KINEMATICS 2.42 Chapter 2, Problem 42 Problem: A box of gold bars is connected to a four pulley system as shown. A classic 1964 Ford Mustang is initially at rest directly below Pulley 4. The Mustang begins moving with constant acceleration, a, at time t = 0. What is the acceleration of the box, ab , at time t = τ ? Express your answer as a function of a, τ and the distance H . Compute ab for a = 1 m/sec2 , τ = 4 sec and H = 5 m. 2 ................................................................................................................................. . . .. . . . . .. . . ............................................................................................................................................................. ..... . . . . .. . . . ..... . . . .. .. . . . . . . .. ... .. .... .. . . .... ...... . ..................................................................................................................................................................... .. ..................................... ............. .. ........... .. .. .. .. .• .. ........... .. ......... . .... .............................................................................................. ................. . .. . . . . ..........................................•................................................................................................... . • . . . . .. .. . . . .... ... ... .... . . ... .. .. . .. .. . . . . • . . • ... . . ... . .. . . . ...... ......... . . . . ....... ....... .. . . .... .. . . ... .. . . . . . . .2. .4 .. . .... . .. . .... . .... . . .... .. . . .. .... . . .... . . . .. . .. .... . .... . .... .. .... . .. . . .... . .... .. .... .. . .... . . .... . .. . .... .. . .... . . .... . . .. .... . .. .... . . .1. .3. . . .. .... . .. . . ... . . ... . . ....... ....... .. . .. .. .. . . . . .. ... . .... . .... . .. . •. .. . .. • . .. . . . ...... ..... . .. . .. . . .. ... . .. . ....... ............................. . ... .. . . . ...•..... . .. ......... . . .. ......................................... ................ .....•.... .................................. . .. ...... . . . . . ............................................. . .. ............................................. . ............................................. . .. . ............................................. . . . . . . .. ............................................. . . ............................................. .. . . . . ............................................. . . . .. ............................................. . ............................................. .. . ............................................. . . ..... .. .. .. .. .... .. ..... . . . . . . . . . ... .. .. . .................................. .. .... . . . . .. ... .. . . . . .. .. . . . . . . . .. . .. .. . . ................ . ................. . .. .. . .................. .................. . . . ............... .............. . . .. . .. . ....... ...... . . . .. . .... ... ............................................................................................................................................................................................................................................................................................................... . . . . . . . . . . . . . . . . .... ...... ...... ...... ................................................. . .. ........... . . . ....... . . . . . .. ... ... ........................... .. ........... .... .. .... .. .. .. .. .. .... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .... .. . .... .. .. .. .. .......... .. .. ............ .. . ... ..... ............. ........ ..... . ..................................................................................................................................................................................................................................................................................................................................................... .............................................................................................................................................................................................................................................................................................................................................................................................. ............................................................................................................................................................................................................................................................................................................................................................................................... xx h H xx Gold s a Solution: The total cable length is tot = 4h + + 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 ˙ 4h + ˙ = 0 ˙ But, the box’s speed is v = −h. Therefore, Now, inspection of the geometry tells us that s2 + H 2 = 2 =⇒ v= 1˙ 4 1 ˙ h=− ˙ 4 =⇒ = s2 + H 2 Since the height H is constant, there follows ˙ ˙ = √ ss 2 + H2 s =⇒ v= 1 ss ˙ √ 4 s2 + H 2 Since the Mustang’s acceleration is constant and equal to a, the distance it travels in time t is s= So, the speed at which the box moves is v= Simplifying, the speed at time t is v= 1 a2 t3 √ 4 a2 t4 + 4H 2 1 4 12 2 at 1 24 4a t 12 at 2 =⇒ s = at ˙ · at + H2 2.42. CHAPTER 2, PROBLEM 42 Differentiating with respect to t again yields the acceleration of the box, ab . Thus, ab = = = Therefore, when t = τ , we have ab = For the given values of a = 1 ab = 4 1 2 1 2 1 2 2 1 23 a t 4a2 t3 1 3a2 t2 √ −2 4 a2 t4 + 4H 2 (a2 t4 + 4H 2 )3/2 147 1 3a2 t2 a2 t4 + 4H 2 − 2a4 t6 4 (a2 t4 + 4H 2 )3/2 1 a2 t4 + 12H 2 2 2 at 4 (a2 t4 + 4H 2 )3/2 1 a2 τ 4 + 12H 2 2 2 aτ 4 (a2 τ 4 + 4H 2 )3/2 m/sec2 , τ = 4 sec and H = 5 m, we find 2 m/sec m/sec (4 sec)4 + 12(5 m)2 3/2 2 2 (4 sec)4 + 4(5 m)2 1 m/sec2 2 2 4 sec2 2 = 0.173 m sec2 ...
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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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