1. ## pulley system

a light inextensible string passes over a fixed pulley A, under a movable pulley B of mass M and then over a second fixed pulley C. A mass m is attached to one end of the string and a mass 3m is attached to the other end. the system is released from rest. (i) prove that the tension t in the string is given by the equation t(1/m + 1/3m)=g (g stands for gravity) i cant prove this even though i feel like i am doing all the the right equations. the 3 equations i get by using f=ma and looking at all the particles seperately are t-mg=ma t-3mg=3mb Mg-2t=M(a/2 + b/2) a stands for the acceleration the particle of mass m. b stands for the acceleration of the particle of mass 3m. a/2 + b/2 stands for the acceleration . when i get a and b on there own from the first 2 equations and put this in for a and b in the last equation i do not get the answer required. i can post a picture of the diagram in the question if anyone needs it. any help much appreciated

2. ## Re: pulley system

Post a diagram of the system. Do pulleys A and C have mass? Since pulley B has mass, I suspect a net torque equation will be involved.

3. ## Re: pulley system

in this questions you are not supposed to take into account the pulleys weight. it is question 10

4. ## Re: pulley system

can you post the text part of the question as an image?

5. ## Re: pulley system

i can also post pictures of a similar worked example where they go about the question the same way i did in my first post. i follow of the example but for some reason i cant get the answer. the only difference in the worked example was they were working with acctual masses but it shouldnt make a difference

6. ## Re: pulley system

t(1/m + 1/3m)=g
... this is why I asked for an image of the text, which states $T\left(\dfrac{1}{\color{blue}{M}}+\dfrac{1}{3m} \right)=g$. You had a lower case m instead of the upper case M

three scalar equations ...

$T - mg = ma \implies a = \dfrac{T}{m} - g$

$T - 3mg = 3mb \implies b = \dfrac{T}{3m} - g$

note ... $a+b=\dfrac{4T}{3m}-2g$

$Mg - 2T = \dfrac{M}{2}(a+b)$

--------------------------------------------------------------

sub in for $a+b$ ...

$Mg - 2T = \dfrac{M}{2}\left(\dfrac{4T}{3m}-2g \right)$

$Mg - 2T = \dfrac{2MT}{3m} - Mg$

$2Mg = \dfrac{2MT}{3m} + 2T$

$Mg = \dfrac{MT}{3m} + T$

$g = \dfrac{T}{3m} + \dfrac{T}{M}$

$g = T\left(\dfrac{1}{3m}+\dfrac{1}{M}\right)$

thank you