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a thin cylindrical rod of uniform density resists against a smooth wall , find the coefficient of friction between ground and rod ,if the minimum angle that the rod can make with the horizontal is theta  
Asked by nikitha532002 | 31 Jul, 2019, 20:25: PM
answered-by-expert Expert Answer
Figure shows a rod AB of length l and mass m is resting on a smooth wall without slipping and rotating.
 
Figure shows all forces acting on the rod  i.e., weight mg acting at centre of gravity, Reaction force N1 at A ,
Reaction force N2 at B and friction force f acting on the point of contact of rod on the floor.
 
Since rotational motion is absent, sum of moments of all forces is zero.
 
If we take moment of all forces about about point A, we have,  
 
(mg)(l/2)cosθ = N2 l sinθ   or     N2 = (1/2) mg cotθ  ...............(1)
 
For equilibrium, sum of vertical forces is zero,  hence we have,  N1 = mg  .............(2)

For equilibrium, sum of horizontal forces is zero,  hence we have,  N2 = f = μ N1   .............(3)
 
where μ is friction coefficient between ground and rod.
 
from eqn.(3), we get,  μ  = ( N2 / N1 )  ..................(4)
 
using eqn.(1) and eqn.(2),  coefficient of friction μ is obtained from eqn.(4) as, 
 
μ = (1/2) cotθ  ........................(5)
 
 
Answered by Thiyagarajan K | 31 Jul, 2019, 23:24: PM

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