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![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 Apr 2007 23:05:48 IST
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An inclined slide of inclination pi/6 has a cross section as shown ( two quarter circular cylinders in contact).The minimum width w of a sledge that can slide down is [ coeff of fric = (sqrt 3)/4]..............................
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SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
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30 Apr 2007 23:26:59 IST
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hope this is ryt.... the ans to this Q is, w = (4 -  7)R/2 do rate me!(if found satisfactory) cheers!
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
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1 May 2007 00:27:09 IST
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sorry this doesnt feel satisfactory n the diagram is still ambiguous......
neone else??????
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SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
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1 May 2007 00:50:20 IST
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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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1 May 2007 13:26:31 IST
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i shd thank asagwal n vasanth for trying  but i'm still blindfolded in this problem bcoz  I SIMPLY CAN'T VISUALIZE THE DIAGRAM  WHERE ARE THE EXPERTS?
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SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
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1 May 2007 14:11:05 IST
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The diagram given by vasanth is correct. Try to visualize it. I will tell you how. First consider two cylinders with their cross sections gradually decreasing from one end to the other. Now cut each cylinders into four equal quarters just as a birthday-boy/girl cuts down the cake. Take one quarter from each cylinder and rest each of these on any of their cut edges. bring them close so that they touch each other. Now it is found that the top surface of these quarter-cylinders is pi/6. put the slide in the narrow grove that is created. While uploading the picture files, the serial order of the files was mixed up. So, I am giving diagram numbers on the pictures.
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Let us build a new world with love, peace, happiness and engineering! (DON'T CHOOSE THE ODD ONE OUT)
          
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1 May 2007 14:41:04 IST
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so frm evrybody's explanation i get this: THE PLANE IS NOT A CONVENTIONAL ONE AS SHOWN FIG 1 now tell me whether ma understanding is correct in fig 2
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SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
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1 May 2007 14:59:06 IST
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Ya, your second diagram says the truth, for_succes
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Let us build a new world with love, peace, happiness and engineering! (DON'T CHOOSE THE ODD ONE OUT)
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1 May 2007 15:00:47 IST
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thanx evrybody..........................
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SOME SAY SKY IS THE LIMIT BUT TO ME
THERE'S SOMETHING BEYOND THE SKY
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