Lecture 19

# Lecture 19 - Lecture 19 ABSOLUTE DEPENDENT MOTION ANALYSIS...

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Dynamics: Lecture Notes for Sections 12.9-12.10 1 Lecture 19 ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES RELATIVE-MOTION ANALYSIS OF TWO PARTICLES USING TRANSLATING AXES Section 12.9-12.10 Ehab Zalok 2 ABSOLUTE DEPENDENT MOTION ANALYSIS OF TWO PARTICLES Objectives : Students will be able to: 1. Relate the positions, velocities, and accelerations of particles undergoing dependent motion.

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Dynamics: Lecture Notes for Sections 12.9-12.10 2 3 APPLICATIONS (continued) Rope and pulley arrangements are often used to assist in lifting heavy objects. The total lifting force required from the truck depends on the acceleration of the cabinet. How can we determine the acceleration and velocity of the cabinet if the acceleration of the truck is known? 4 DEPENDENT MOTION (Section 12.9) In many kinematics problems, the motion of one object will depend on the motion of another object. The blocks in this figure are connected by an inextensible cord wrapped around a pulley. If block A moves downward along the inclined plane, block B will move up the other incline. The motion of each block can be related mathematically by defining position coordinates , s A and s B . Each coordinate axis is defined from a fixed point or datum line , measured positive along each plane in the direction of motion of each block.
Dynamics: Lecture Notes for Sections 12.9-12.10 3 5 DEPENDENT MOTION (continued) Position coordinates s A and s B can be defined from fixed datum lines extending from the center of the pulley along each incline to blocks A and B. If the cord has a fixed length , the position coordinates s A and s B are related mathematically by the equation s A + l CD + s B = l T Here l T is the total cord length and l CD is the length of cord passing over arc CD on the pulley. 6 DEPENDENT MOTION (continued) The negative sign indicates that as A moves down the incline (positive s A direction), B moves up the incline (negative s B direction). Accelerations can be found by differentiating the velocity expression. Prove to yourself that a B = -a A . The velocities of blocks A and B can be related by differentiating the position equation. Note that l CD and l T remain constant , so dl CD /dt = dl T /dt = 0 ds A /dt + ds B /dt = 0 => v B = -v A

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Dynamics: Lecture Notes for Sections 12.9-12.10 4 7 EXAMPLE Consider a more complicated example. Position coordinates (s A and s B ) are defined from fixed datum lines, measured along the direction of motion of each block. Note that s B is only defined to the center of the pulley above block B, since this block moves with the pulley. Also, h is a constant.
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