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ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9)

ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9). Objectives : To relate the positions, velocities, and accelerations of particles undergoing dependent motion. APPLICATIONS.

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ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9)

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  1. ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES (Section 12.9) Objectives: To relate the positions, velocities, and accelerations of particles undergoing dependent motion.

  2. APPLICATIONS The cable and pulley modify the speed of B relative to the speed of the motor. It is important to relate the various motions to determine the power requirements for the motor and the tension in the cable. If the speed of the A is known, how can we determine the speed of block B?

  3. APPLICATIONS (continued) Rope and pulley arrangements are used to assist in lifting heavy objects. The total lifting force required from the truck depends on the acceleration of the cabinet A. How can we determine the acceleration and velocity of A if the acceleration of B is known?

  4. DEPENDENT MOTION COMPATABILITY EQUATIONS In many kinematics problems, the motion of one object will depend on the motion of another. The blocks are connected by an inextensible cord wrapped around a pulley. If block A moves downward, block B will move up. sA and sBdefine motion of blocks. Each starts from a fixed point, positive in the direction of motion of block.

  5. DEPENDENT MOTION (continued) sA and sB are defined from the center of the pulley to blocks A and B. If the cord has a fixed length, then: sA + lCD + sB = lT lT is total cord length and lCD is the length of cord passing over arc CD on the pulley.

  6. DEPENDENT MOTION (continued) Velocities can be found by differentiating the position equation. Since lCD and lT remain constant, so dlCD/dt = dlT/dt = 0 dsA/dt + dsB/dt = 0 => vB = -vA The negative sign indicates that as A moves down (positive sA), B moves up (negative sB direction). Accelerations can be found by differentiating the velocity expression. Prove: aB = -aA .

  7. DEPENDENT MOTION EXAMPLE sA and sB are defined from fixed datum lines, measured along the direction of motion of each block. sB is defined to the center of the pulley above block B, since this block moves with the pulley. The red colored segments of the cord and h remain constant in length

  8. DEPENDENT MOTION EXAMPLE (continued) The position coordinates are related by 2sB + h + sA = l Where l is the total cord length minus the lengths of the red segments. Velocities and accelerations can be related by two successive time derivatives: 2vB = -vA and 2aB = -aA When block B moves downward (+sB), block A moves to the left (-sA).

  9. DEPENDENT MOTION EXAMPLE (continued) The example can also be worked by defining the sB from the bottom pulley instead of the top pulley. The position, velocity, and acceleration relations become 2(h – sB) + h + sA = l and 2vB = vA 2aB = aA Prove that the results are the same, even if the sign conventions are different than the previous formulation.

  10. EXAMPLE PROBLEM Given: In the figure the cord at A is pulled down with a speed of 8 m/s. Find: The speed of block B.

  11. DATUM sA sC sB EXAMPLE (Solution) : Define the position coordinates one for point A (sA), one for block B (sB), and one relating positions on the two cords (pulley C). Coordinates are defined as +ve down and along the direction of motion of each object.

  12. DATUM sA sC sB EXAMPLE (continued) If l1=length of the first cord, minus any segments of constant length and l2 for the second: Cord 1: 2sA + 2sC = l1 Cord 2: sB + (sB – sC) = l2 Eliminating sC, 2sA + 4sB = l1 + 2l2 Velocities are found by differentiating: (l1 and l2 constants): 2vA + 4vB = 0 => vB = - 0.5vA = - 0.5(8) = - 4 m/s

  13. GROUP PROBLEM SOLVING Given: In this system, block A is moving downward with a speed of 4 m/s while block C is moving up at 2 m/s. Find: The speed of block B.

  14. GROUP PROBLEM SOLVING (Solution) A datum line is drawn through the upper, fixed, pulleys. Define sA, sB, and sC Differentiate to relate velocities:

  15. End of 12.9 Let Learning Continue

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