p0237 - 118 CHAPTER 2. PARTICLE KINEMATICS 2.37 Chapter 2,...

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Unformatted text preview: 118 CHAPTER 2. PARTICLE KINEMATICS 2.37 Chapter 2, Problem 37 Problem: In the system shown, there are two pulleys and three weights connected as shown. The upper pulley is fixed in space while the lower pulley can move. (a) Determine the velocity, v3 , and acceleration, a3 , of Weight 3 as a function of x1 , x2 , x1 and x2 . ˙˙¨ ¨ (b) How many degrees of freedom does this system have? (c) Compute the velocity and acceleration of Weight 3 for x1 = −4 cm/sec, x2 = −3 cm/sec, ˙ ˙ x1 = −2.5 cm/sec2 and x2 = −2 cm/sec2 . ¨ ¨ ................................................................................................................................................................................................................................................................................................... . ................................................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................................................... ................................................................................................................................................................................................................................................................................................ ...................................................................................................................................................... . . . . ................................................................................................................................................................................................................................................................................................. .. . .. . . .. .. .. .. . . . . .. ... .. .. ... . . . ....... . .......... . . . ........... . . . . ............... . ............... .. . . . . . . . .................. .... . . . . . . . ................... . . ................... . . . . . .. . . . ............. . . . . . . ........... ..... . . . . ................. . ................ . . . . .. ..... .. . ............... . . . .............. . . . . . ........... . . .. ..... . . . ............... .. . . . . ........... . . . . . .... .... . . . . . . . .. .. . . . 1 . .... . . . . . . . . . . . . . . . . . . . . . p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................ . . . . . . ........................ . . . . ........................ . .. . . . ........................ . . . . . ........................ . . . ........................ ........ ....... . . . . ........................ . . .. 2 ........................ . . .. ........................ .......... . . . ..... . ........................ .............. . . ........... . ........................ ............... . . . .......... . . ............... . . . .. . ......... . . ....... .. .. . .. .. . . . ................ ..... . . ................. .... . ....... ........ .... .... . .. . . .. ................ . 3 .. .. . . .................. . . .. . . . .................. . . ............. . . . . .............. . .... . . . . ........ ... . . .. . . . ..... ... . . . . ........... . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ......................... . . . .. ......................... . . . . ......................... . .. . . . ......................... ..... . . .... ......................... . . . . .... .... ......................... . . . ......................... . . ......................... . . . . . . ......................... . . . ......................... . . . ......................... . . . . . . . . . . . . . . .. . . .. . . .. . . .. . . .. . . . .. . . . ............ . . . . . ......................... . . . ............ . ......................... .. ......................... .. ......................... ........ ....... . ......................... ......................... ......................... ......................... ......................... ......................... • • x x 1 x • x i 2 3 Solution: (a) Conservation of cable length tells us that x1 + xp = constant and (x2 − xp ) + (x3 − xp ) = constant and x2 + x3 − 2xp = 0 ˙ ˙ ˙ Thus, the speeds in this system are given by x1 + xp = 0 ˙ ˙ Therefore, the speed of Weight 3 is Similarly, the acceleration of Weight 3 is x3 = 2xp − x2 = −2x1 − x2 ˙ ˙ ˙ ˙ ˙ x3 = −2¨1 − x2 ¨ x ¨ and a3 = − (2¨1 + x2 ) i x ¨ So, the velocity and acceleration vectors for Weight 3 are v3 = − (2x1 + x2 ) i ˙ ˙ (b) Because any two quantities can be prescribed arbitrarily, this system has 2 degrees of freedom. (c) For the given values of x1 = −4 cm/sec, x2 = −3 cm/sec, x1 = −2.5 cm/sec2 and x2 = −2 cm/sec2 , ˙ ˙ ¨ ¨ we have v3 a3 = − [2(−4) + (−3)] i = 11 i (cm/sec) (cm/sec2 ) = −[2 (−2.5) + (−2)] i = 7 i ...
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This note was uploaded on 10/15/2010 for the course AME 301 at USC.

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