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amyvincent12
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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)?
 2 years ago
 2 years ago
amyvincent12 Group Title
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)?
 2 years ago
 2 years ago

This Question is Closed

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I proved this for you in the other question you asked
 2 years ago

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

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

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

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

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