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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 12:59:20 IST
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The wedge moves to the left withh acc=a. Find acc of mass m wrt wedge. Surfaces are all frictionless.. . . Salutes assured!
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16 Mar 2008 13:09:36 IST
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the small block will have acceleration in acceleration in x as well
as y direction .....
let acceleration of block is A...
and  is angle of inclination of wedge...
then
(Ax + a)tan = Ay..
I can give u explain u full solution if u want
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I DONOT FOLLOW THE RULES I MAKE THEM TO FOLLOW ME. |
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16 Mar 2008 13:12:29 IST
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Ans:2(M+m)a/m
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16 Mar 2008 13:18:04 IST
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I WAS BORN INTELLIGENT!!!!!!!
EDUCATION RUINED ME
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
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16 Mar 2008 13:23:04 IST
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Extremely sorry people-i forgot to mention that the inclination of the wedge is 60degrees.. sorry!!
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21 Mar 2008 10:22:03 IST
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Please help! Thanks!
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21 Mar 2008 12:01:15 IST
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let a1 be the accln of the m relative a , down the incline , therefore its horizantal componenet is a1cos ,therefore its original accln in the horizantal directn =a1cos60 - a now let N be the normal reaction of m on M therefore we get the eqn Nsin = Ma = m(a1cos60 - a) sust the value we get 2(m+M)a/m sorry unable to attach diagram
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Moderation Team
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