- #1
gracy
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Homework Statement
:[/B]block M slides down on frictionless incline .Find the minimum co efficient of friction so that m does not slide with respect to M.
Homework Equations
:acceleration of the system=total driving force/total mass:[/B]
The Attempt at a Solution
m does not slide with respect to M means there should not be kinetic friction rather static friction.both small and big block will have same acceleration.
acceleration of the system=total driving force/total mass:total driving force=(m+M)g sin theta
acceleration of the system=(m+M)g sin theta/m+M
=g sin theta
force responsible for acceleration of small block would be force of static friction between two blocks.
Hence force of static friction between two blocks=mass of small block i.e "m" multiplied by g sin theta
but this force of static friction between two blocks is also equal to or smaller than μ multiplied by normal force on small block i. mg
force of static friction ≤ μmg force of static friction/mg≤ μ
m× g sin theta= force of static friction
Hence m× g sin theta/mg ≤ μ
theta= 37 degrees
sin theta ≤ μ
sin 37 degrees ≤ μ
0.601 ≤ μ
but the correct answer should be 0.75 ≤ μ
where am I going wrong?[/B]