Three Pulleys 3 masses, find tension

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The discussion revolves around calculating the tension in a system of three masses (m1 = 6 kg, m2 = 2 kg, m3 = 3 kg) suspended from massless, frictionless pulleys. The user attempts to derive the tension using equations of motion but consistently arrives at 15.68 N, which does not match any provided answer choices. Other participants confirm the user's equations are correct but suggest a misinterpretation of the problem. The conversation highlights the importance of careful reading and verification in solving physics problems. Ultimately, the tension is noted to be around 16 N, indicating a possible error in the answer options provided.
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Homework Statement



Three objects mI. m2 and m3 are suspended
from three massless and frictionless pulleys as
shown in the diagram. ml is 6 kg, m2 is 2 kg
and m3 is 3 kg. Find the tension of the string
when the system is set in motion (i.e. all
masses are moving). Answer in N

Options are:
(a) 11.3 (b) 12.1 (c) 13.5 (d) 14.2

..O...O... Top 2 pulleys attached to ceiling, m1 hangs from 1st pulley, m2 is attached
.| |..| |... to the middle pulley, m3 hangs from third pulley.
.|..O...|
.|..[]..|
[] m2 []
m1...m3

Homework Equations


Fg=mg
F=ma

The Attempt at a Solution



I made 3 equations, tried to relate them but I am getting 15.68N instead of any of the options (this was an old exam question). It's too far from 14.2 to be right. Here were the four equations I made:

a1=(m1g-T)/m1
a2=(2T-m2g)/m2
a3=(m3g-T)/m3

2a2=a1+a3

I keep getting none of the possible answers but I think I'm doing it right. Where did I go wrong?
 
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Your equations are right but your answer is wrong. Please post your working.
 
a1=(m1g-T)/m1=g-T/m1=g-T/3
a2=(2T-m2g)/m2=2T/m2-g=T-g
a3=(m3g-T)/m3=g-T/m3=g-T/6


From 4th eqn:

2T-2g=g-T/3+g-T/6

2T+T/2=4g

5T/2=4(9.8)

T=15.68
 
Oh, sorry, I misread the end of your first post. I thought you were getting 14.2. So, I agree with you, it's something around 16.
 
k thanks
 
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