## anonymous 4 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)?

1. TuringTest

I proved this for you in the other question you asked

2. TuringTest
3. anonymous

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

4. anonymous

first, block start to motion with a=0

5. anonymous

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

6. anonymous

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