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amyvincent12
 3 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)?
amyvincent12
 3 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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TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I proved this for you in the other question you asked

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/study#/updates/4f4d6419e4b0acf2d9fe245e

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1from 2nd newtonian law we have\[mgsin \thetamg \mu _{s}\cos \theta=0\]

hosein
 3 years ago
Best ResponseYou've already chosen the best response.1first, block start to motion with a=0

amyvincent12
 3 years ago
Best ResponseYou've already chosen the best response.0so how would i find acceleration once i found my angle to be 19.3 x 10^1 deg?

sunjeet95
 3 years ago
Best ResponseYou've already chosen the best response.0\[mgsin \theta  mg \cos \theta \mu k =0\] \[mgsin \theta = mgcos \theta \mu k \] \[\sin \theta = \cos \theta \mu k\] \[\tan \theta = \mu k\] \[\theta = \tan^{1} \mu k\]
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