Final S06 sln'

Final S06 sln' - 2 2 2 1 1 = − − − h N L N h N L N k...

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6. The key issue of this problem is that when there’s linear acceleration, the only choice of pivot point is the center of mass (CM). In a statistics problem, we have the freedom to choose a pivot point, but here because of the deceleration, we lose that freedom. In other words, we have to decompose the motion of the system to two parts: the translational motion of CM and the rotational motion around CM. This is important because we need to know the pivot point before we calculate all the torques. a) After the brakes are applied, the force diagram is as following: There’s no acceleration in the vertical directions, so I have Mg N N = + 2 1 (1) There’s no rotational acceleration around CM, which gives rise to 0 * 2
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Unformatted text preview: * * 2 * 2 2 1 1 = − − − h N L N h N L N k k µ (2) Here I’ve already used the relation 1 1 N f k = and 2 2 N f k = . From equation (1), 1 2 N Mg N − = . Then plug it into equation (2), ( ) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − h L N Mg h L N h L N k k k 2 * 2 * 2 * 1 2 1 Therefore, Mg L h L N k 2 2 1 + = & Mg L h L N k 2 2 2 − = . At the same time, Mg L h L f k k 2 2 1 + = & Mg L h L f k 2 2 2 − = b) The critical condition for the overturning to happen is the normal force N 2 becomes zero. 2 2 2 = − = Mg L h L N k , Then k h L = 2 . The condition to prevent that from happening is k h L > 2 ....
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This note was uploaded on 03/09/2010 for the course PHYSICS 7A taught by Professor Lanzara during the Fall '08 term at Berkeley.

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Final S06 sln' - 2 2 2 1 1 = − − − h N L N h N L N k...

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