06_P40InstructorSolution

06_P40InstructorSolution - 6.40. Model: The steel block...

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6.40. Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of friction s 0.80 μ = and a kinetic coefficient of friction k 0.60. = Visualize: Solve: (a) While the block is at rest, Newton’s second law is () ( ) net s s net G G 0 N xy FT f T f Fn F n F m g =− = ⇒= =− ⇒= = The static friction force has a maximum value ( ) ss max . f mg = The string tension that will cause the block to slip and start to move is ( ) ( ) 2 s 0.80 2.0 kg 9.80 m/s 15.7 N Tm g == = Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for motion. (b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the x - component of Newton’s second law as follows:
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at UGA.

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