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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 t he 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.

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.

Grade:12

1 Answers

Ramesh V
70 Points
12 years ago

(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

http://askiitians.com/forums/Mechanics/10/3377/Laws-of-Motion.htm

(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.

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IIT Kgp - 2005 batch


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