tension

hey u got the correct ans dats why u r awarded 10 marks

yayyyyyyyyyyyyyyyyyyyyyyyyyy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

lemme celebrate for 5 mins!!!!!!!!!!1

then i will post the ans in next 5 mins .. [3] [3]

hey sky wont the tension will be different at diff points since it is an uniform chain

let at some time 't',

B falls by a distance 'x'.

vel of point B = √2gx. (free fall)

now, ocnsiedr the free body diag of right part...

mg(l/2 - x) downwards n m/L(v²) upwards.

now consider the free body diag of left part ..

T upwards, mg(L/2+x/2)/L downwards and 1/2vdm/dt.

vdm/dt = m/L (v²)

so, T=mgx/L + mg(L/2 + x/2) = mg/2 (3x/L +1) .

where is the string not tight ??

it is always tight....

see, if the tip B comes down x dist, the lowermost point P comes dwn by x/2 ... jus think physicaly........

can u give the fbd for the strings

pls rep sooooooooon

mom is angry,,,, telling me to log off......

actually i was confused sky . so far i had seen such relations when a

massless pulley is used ..

clueless sky abt that dist moved by P = 1/2 the dist moved by B

see, B is moving x dist down,

there is a curve .... one min i will draw

is it alr8 wid you now ???

if there any other doubt in this question i will ask u tomorrow
will u come tomorrow..

i am also feeling sleepy shall we start tomorrow...

great idea...... i scrapped you [3]...

delete it afetr reading :P :P

ok sky i got the constraint relation

and wat abt the tension. and how to include the variation of the tension along the string..............

sky i cant get u....

heywhy do u hide these chats

[7] why is this thread highlighted
is Mg the correct answer??

philip u r rong