IIT JEE , AIEEE Forum - Physics , Mechanics - The culprit is ma imagination of the diagram

asagwal

ohhhh.......
wonderful question, i must say
my answer is r/2.
lets see if i can give a satisfactory solution.
it goes this way..............
consider the cross-section u gave, assume angle @ to be the angle between the horizontal and the tip of the sledge(as seen from the centre of one of the quarters)
now N be the normal reaction from each
2Ncos@=mg cos pi/6
(balancing perpen. to inclined)(cos@ because the two sin components cancel from the 2 surfaces)
friction = u*(mg(cospi/6)/cos@)
putting friction=mg sin pi/6
we get cos@=3/4
and trigo gives the width=r/2
uhhhhh.......that was long
but i think if @ increases, it stops sliding
therefore width is maximum that can slide down, not the minimum
please check the question again

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