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sphere of mass M is held between 2 smooth inclined planes for Sin 37 is equal to 3 by 5 the normal reaction of the wall 2 to is equal to

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2 years ago

## Answers : (1)

bramashini
19 Points
```							Mg=2Rsin37R=Mg/2sin37R=Mg / 2*3/5    ( sin 37=3/5 is given) therefore , R=5Mg/6this sphere is lying  between  2 inclined planes so there are 2 normal reactions on the sphere by the plane. if we resolve these reactions the vertical forces will be Rsin37*2 this system is under  equllibrium so verical forces 2Rsin37 is equal to weight of the sphere Mg (here assume normal reaction= R)
```
2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions