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amyvincent12

  • 2 years ago

A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is μs= 0.350, and the coefficient of kinetic friction is μk= 0.160. The angle of elevation θ is increased slowly from the horizontal. At what value of θ does the block begin to slide (in degrees)?

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  1. TuringTest
    • 2 years ago
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    I proved this for you in the other question you asked

  2. TuringTest
    • 2 years ago
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    http://openstudy.com/study#/updates/4f4d6419e4b0acf2d9fe245e

  3. hosein
    • 2 years ago
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    from 2nd newtonian law we have\[mgsin \theta-mg \mu _{s}\cos \theta=0\]

  4. hosein
    • 2 years ago
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    first, block start to motion with a=0

  5. amyvincent12
    • 2 years ago
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    so how would i find acceleration once i found my angle to be 19.3 x 10^1 deg?

  6. sunjeet95
    • 2 years ago
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    \[mgsin \theta - mg \cos \theta \mu k =0\] \[mgsin \theta = mgcos \theta \mu k \] \[\sin \theta = \cos \theta \mu k\] \[\tan \theta = \mu k\] \[\theta = \tan^{-1} \mu k\]

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