Simple Atwood apparatus question

[mcshadypl]

I had this question on the MCAT earlier, and I'm not sure if I did it properly.
There is an Atwood machine such that a person of mass m is holding onto both ends of the string. What is the tension in the string? According to my calculations, since a=0 since the string is not moving, T=0.5m*g. I think if you would cut the person in half, each half, m1 and m2, would have a mass of 0.5m. Thus:
T-m1g= ma
m2g-T=ma
So a=0 since there is no acceleration and m1=m2=0.5m. This gives us T=0.5mg Did I approach this correctly? I've enclosed a picture of the problem:

http://img217.imageshack.us/i/68744608.jpg/

 

[Doc Al]

Your answer is correct. Rather than cut the person in half, just realize that the ropes pull up on the person twice:
2T - mg = 0, thus T = mg/2.

 

[mcshadypl]

Thanks for your help! That makes more sense.

 

[GrizzlyBat]

Yeah, pulleys can be counter-intuitive if you are not careful. So I tend to be extra careful when dealing with them :).