**1**
dimensions (dimentime)
**·**2008-12-17 04:40:55
i think i got the ans

acc of M = (M-5m)g/(M+15m)...........

**1**
dimensions (dimentime)
**·**2008-12-17 06:07:57
let acceleration M be a & of m be b

see, by equation of constraint u will get

b = 3a

by newtons motion

Mg-2.5T = Ma

2.5T-5mg = mb

where T is tension in rope connecting roof & M

solve & u will get the ans,

one more thing

can u tell me how to use latex in this forum..............

**341**
Hari Shankar
**·**2008-12-17 06:33:04
Are you sure its not g(M-5m)/(M+25m)?

Thats the answer I get.

Logically the minus sign has to be there.

**1**
greatvishal swami
**·**2008-12-17 06:47:50
i m getin same as dimen but prophet frm where dat M+25m come isnt it M+15m

**341**
Hari Shankar
**·**2008-12-17 07:11:12
If the acceleration of M is a, what is the acceleration of m?

You seem to have taken it as 3a. Its 5a.

**1**
greatvishal swami
**·**2008-12-17 07:14:10
********edited

its 5a ......correction

& -ve sign has to be there

**3**
msp
**·**2008-12-17 10:10:43
great vishal please post me the soln with your constraint relations

**3**
msp
**·**2008-12-17 10:24:18
hey dimensions I think u have went wrong with your equations

**3**
msp
**·**2008-12-17 10:35:41
yes the acceleration of m is 5 times that of the acc of M

the eqns are

Mg-(T_{1}+T)=Ma

T-mg=m3a

where T=T_{1}/4

**1**
skygirl
**·**2008-12-17 20:56:32
yes the ans given is correct only..

the constraint eqn is : 5A + a=0 (A= acc of M, a= acc of m)

the force eqns are:

5T-Mg = - MA --- 1

T-mg= -ma ----2

I) now find a and A in terms of T.

II) put in the constarint eq. ... so u will get T.

III) then find a and A from the force equations.. :)

**1**
aaru s
**·**2008-12-17 22:10:04
5T-Mg=Ma-----------1

mg-T = 5ma---------2

multiplying eqn 2 wid 5 and thn adding them....we get....5mg- Mg=Ma +25 ma

a=g(5m-M)/M+25m