Question

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)?

Answer 1

I proved this for you in the other question you asked

Answer 2

http://openstudy.com/study#/updates/4f4d6419e4b0acf2d9fe245e

Answer 3

from 2nd newtonian law we have\[mgsin \theta-mg \mu _{s}\cos \theta=0\]

Answer 4

first, block start to motion with a=0

Answer 5

so how would i find acceleration once i found my angle to be 19.3 x 10^1 deg?

Answer 6

\[mgsin \theta - mg \cos \theta \mu k =0\]
\[mgsin \theta = mgcos \theta \mu k \]
\[\sin \theta = \cos \theta \mu k\]
\[\tan \theta = \mu k\]
\[\theta = \tan^{-1} \mu k\]