hw4solution - Homework 4 Solution CHANG 1 The motion of the...

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Homework 4 Solution CHANG 1 The motion of the man is 1-D vertical motion downward. Since the platform also travels downward along with the man, they move as one unit and we can take the man and the platform together as the free body. Let the total mass for the man and the platform be M 80 20 100 Kg, and the FBD is shown here: Mg T TT +y a Note that the positive y axis is downward, same as the direction of the acceleration. Apply Newton’s 2nd law: F y ma y Mg 3 T Ma T M g a 3 100 9.8 .2 3 320 N From Newton’s 3rd law the force exerted by the man equals the tension in the cable, so F T 320 N. CHANG 2 Since the marble settles at the same elevation on the inclined surface and rides along with the spinning cone, its motion is a horizontal uniform circular motion. The FBD of the marble is shown here: w=mg N a n y n Set up n y coordinate system with n axis pointing to the left toward the center of the circular motion. Apply Newton’s 2nd law along both y and n axes, with a y 0(motionis horizontal) and radius R L cos from geometry: F y ma y N cos mg 0 N mg cos F n ma n N sin m 4 2 L cos T 2 L gT 2 sin 4 2 cos 2 1.5 2 sin30 4 2 cos 2 30 0.372 m Y&F 5.52 The motion of the rider on the giant swing is a horizontal uniform circular motion, and the FBD of the rider (and the seat) is shown here:
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mg Tension in cable P y n (a) From geometry the radius of the circular motion is R 3 5sin30 5 .5m .Use n y coordinate system (shown in the figure) for horizontal circular motion with n -axis pointing to the left toward the center of the circle. Now apply Newton’s 2nd law in the
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This note was uploaded on 10/17/2009 for the course PHYS 2305 taught by Professor Tschang during the Spring '08 term at Virginia Tech.

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hw4solution - Homework 4 Solution CHANG 1 The motion of the...

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