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```        Ques1) A particle is placed at rest inside a hollow hemishere of radius R. The coefficient of friction b/w the particle and hemisphere is 1 / (3) 1/2. Show  the maximum height upto which the particle can remain stationary [1- (3) 1/2/2]R.
Ques2) A block of mass m  is paled at rest on a horizontal rough surface with the angle of friction @. The block is pulled with a force F at an angle theta with the horizontal.Show that the mmin value of F required to move the block is [mgsin@ ] / cos( theta - @).
Ques3)  A horizontal force F = mg/3 is applied on the upper surface of a uniform cube of mass m and side a which is resting on  a rough horizontal surface having coefficient of friction = 1/2.Show that the distance b/w the line of action of mg and normal reaction is a/3.```
8 years ago

Ramesh V
70 Points
```										(1)

here OM = R , to find= AM
on balancing forces we have
normal rxn. N = mg.cosx
f = mg.sinx
where f = friction coeff. * N
i.e.,
f = mgcox /31/2
so tan x = 1/31/2
so height at which particle remains stationary is Am = OM - OA = ( 1 - 31/2 /2 )R
(2)
Its solved in my previous posts asked by you
(3)

on balancing moments about centre of gravity gives:
mg*a/2 + mga/3 = mg (a/2 + x)
on solving for x gives
x=a/3
Hence,the distance b/w the line of action of mg and normal reaction is a/3.
---
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

```
8 years ago
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