62
Lokesh Verma
·2009-03-24 22:08:49
aragon.. you are partly right.
but if you "ignore rotatin", which you have to in this cae then only you will get the constraint equation as what is done by asish
sorry for the typo
I think virang has made some mistake.. i overlooked that one
11
virang1 Jhaveri
·2009-03-24 22:04:39
let M move a distance x.
Therefore first the longest string will move by x distance.
Pulling thestring on the otherside of the pulley by x
This the second pulley is pulled up by x . The second string joining M will have to move 2x to keep the M at x level. Therefore the string between the second and first pulley will also move by 2x. This will pull the first puley up by 2x. Therefore now the first(smallest) string attached M will move 3x to keep the M at x level . So will m move 3x
11
Anirudh Narayanan
·2009-03-24 22:03:41
Constraint:
if M moves down by l, then 2nd pulley moves up by l and the string in the second pulley moves to the right by l/2......similarly the string attached to m, moves up by l/4
Thus if acc. of M is a, acc. of m is a/4
(sorry for the bad english but pls adjust)
62
Lokesh Verma
·2009-03-24 22:02:58
aragon.. you are partly right.
but if you ginorerotatin, which you have to in this cae then only you will get the constraint equation as what is done by asish and virang.
106
Asish Mahapatra
·2009-03-24 21:59:18
So, Mg - 7T = Ma ... (i) a = acceleration of M
T-mg = 7ma .. (ii)
Solving them,
Mg-7mg = Ma + 49ma
==> a = (M-7m)g/(m+49m) = aM
so, am = 7(M-7m)g/(m+49m)
106
Asish Mahapatra
·2009-03-24 21:57:25
Constraint:
Let M move dx down and let m move dy up..
Work done by tension = 0 on system
So, T*dy - 7T*dx =0
==> dy=7dx
==> am = 7aM
11
virang1 Jhaveri
·2009-03-24 21:55:11
Acc of m is 3a while acc of M is a .
I feel but may be wrong
11
Anirudh Narayanan
·2009-03-24 21:50:28
If acc of M is 4a, won't acc of m be a?? (applying constraints)
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Anirudh Narayanan
·2009-03-24 21:49:37
Oi...i've had my names distorted before this,......but Strider---> stringer is the worst distortion!!! [16]
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virang1 Jhaveri
·2009-03-24 21:45:44
Stringer do u have the answer
11
virang1 Jhaveri
·2009-03-24 21:42:27
First m
T - mg = 3ma
Therefore tension in the string between the first and second is 2T
Tension in the string between the second and third is 4T
Therefore the total force trying to pull M is T+2T+4T = 7T
Mg - 7T = Ma
T = ma+mg
Mg - 21ma - 7mg = Ma
g(M - 7m) = Ma + 21ma
g(M-7m) = a ( M + 21m)
a = g(M-7m)/( M + 21m)
11
Anirudh Narayanan
·2009-03-24 21:35:24
So we have to consider the rotation as well??
But as i said nothing else is given!!!