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`        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.`
7 months ago

Arun
23513 Points
```							Let the slope of the inclination is ФHence component of weight along the slide = mgsinФNormal component of weight , N = mgcos=> force of friction = F = μN = μmgcosΦFor the block does not slideF = mgsinΦ=> μmgcosΦ = mgsinΦ=> μcosΦ = sinΦ=> μ = tanΦSo the block does not slide if the co-efficient of the friction will be tanΦ = tan 37 = 0.75
```
7 months 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
```
7 months ago
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Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions