A thin hoop of weight 500N and radius 1m rests on a rough inclined plane as shown in the figure. The minimum coefficient of friction needed for this configuration is :
A
13√3
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B
1√3
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C
12
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D
12√3
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Solution
The correct option is D12√3 Let the coefficient of friction be μ. The force by upper rope be F; Fricition force be Ff The normal force on the hoop is mgcosθ Therefore maximum friction force can be μmgcosθ As the hoop is in rest. Therefore moment balance on hoop about its bottom point(where friction is acting) is: ∑→r×→F=02rF^j−rmgsinθ^j=0 F=mgsinθ2 Also force balance along the incline: Ff+F=mgsinθ∴Ff=mgsinθ2 For μ to be minimum. Ff=μmgcosθ=mgsinθ2∴μ=tanθ2=12√3