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Grade: 12
        
Block M slides down on a frictionless incline as shown. Find the minimum friction coefficient so that m does not slide with rest to M.
one month ago

Answers : (2)

Arun
22330 Points
							
Let the slope of the inclination is Ф
Hence component of weight along the slide = mgsinФ
Normal component of weight , N = mgcos
=> force of friction = F = μN = μmgcosΦ
For the block does not slide
F = mgsinΦ
=> μmgcosΦ = mgsinΦ
=> μcosΦ = sinΦ
=> μ = tanΦ
So the block does not slide if the co-efficient of the friction will be tanΦ = tan 37 = 0.75
 
one month ago
Khimraj
3008 Points
							
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

 
one month ago
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