Accelerating pulleys (3 pulleys and 3 masses)

[Yossi33]

Homework Statement

newtons second law and multiple pulley problem.

Relevant Equations

f=ma

hi,i have this question :
m1=3kg m2=6kg m3=20kg
there is no friction between m3 and the floor.
what is the acceleration of each block?

my attempt is :
the pulley that moves is moving downward at the acceleration of m3.
so the system of m1,m2 is moving downward at the acceleration of m3, then i exerted a "ghost force" on that system that works upward and it works on m1 with magnitude of m1a* and on m2 with magnitude m2a*

then i solved their acceleration as they are not in accelerating system:
m2g-T-m2a*=m2a (1)
T+m1a*-m1g=m1a (2)
the moving pulley is 2T=T* (3)
and of m3 ----> T*=m3a* (6)

if i add 1 +2 --> m2g-m2a*+m1a*-m1g=(m1+m2)a
(m2-m1)g-(m2-m1)a*=(m1+m2)a / divide by (m2-m1)
g-a*=(m1+m2)/(m2-m1) multiply a
g-a*=3a (4)

then i subtitue (4) in (2) ----> T=m1(a-a*+g)
T=3(g/3 -a*/3 -a* +g)
T=4g-4a* (5)
subtitue (5) and (6) in (4)
8g-8a*=m3a*
8g=28a*
a*=8g/28 (7)

subtitue (7) in (4) ---> g/3-8g/78=a

i don't have answers, so please tell me if I am wrong and if i am, then how should i solve this. thank you

 

[TSny]

I think your calculations are correct, except I believe the 78 in the last line of the calculation should be 84. Note that $a$ is the magnitude of the acceleration of $m_1$ and $m_2$ in the frame moving with the falling pulley. You still need to find the accelerations of $m_1$ and $m_2$ relative to the lab frame.

The "ghost" force is often called a "fictitious" force.

 

[Chestermiller]

It seems to me your equations 1 and 2 are correct.

 

[Yossi33]

The magnitude i got is a*=2.8=a3
And a=2.33=a13=a23(a1 and a2 relative to m3)

So acceleration of m1 relative to the ground is a1=-a13 + a3 ------》 a1=-2.8+2.33=0.47
And a2=a23+a3 ----》a2=2.8+2.33=5.133
?

Thanks for your help

 

[TSny]

Those numbers agree with what I got. You should include units and indicate the direction of each acceleration.

Nice work.

 

[Yossi33]

Thanks for your help