My bad. Guess my decription was horrible. Both the wedges are facing the same direction, lets say east. thread starts from pulley on left wedge, to mass m1, then mass M1 is attched to pulley of wedge A, which is again connected to Mass M2. It is basically one long thread starting from pulley in wedge B, to mass M1, to pulley on wedge A, to mass M2.
the question has nothing much in it so long as you leave pen and paper alone, but most go wrong as soon as u star using all formulae. its a nicely framed question w=if ure given the options. lures u into a false belief that finally a simple question after several tough ones. the source is last year AITS.
the left string will have no tension... and by length constraint....accelerations of the blocks in their respective directions are equal in magnitude... now apply II law mgsin30-T=ma T+mgsin60=ma
My bad. Guess my decription was horrible.
ReplyDeleteBoth the wedges are facing the same direction, lets say east.
thread starts from pulley on left wedge, to mass m1, then mass M1 is attched to pulley of wedge A, which is again connected to Mass M2. It is basically one long thread starting from pulley in wedge B, to mass M1, to pulley on wedge A, to mass M2.
and wen i say theyre both facing east i mean that the slanting side is towards the east of both the pulleys.
ReplyDeletethe question has nothing much in it so long as you leave pen and paper alone, but most go wrong as soon as u star using all formulae. its a nicely framed question w=if ure given the options. lures u into a false belief that finally a simple question after several tough ones.
ReplyDeletethe source is last year AITS.
vaibhav u yourself give the diagram
ReplyDeletei dunno how
ReplyDeletehttp://vibhav-question5.blogspot.com/
ReplyDeletethe left string will have no tension...
ReplyDeleteand by length constraint....accelerations of the blocks in their respective directions are equal in magnitude...
now apply II law
mgsin30-T=ma
T+mgsin60=ma
add the two equations
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