What is the acceleration of each mass in a pulley system with two masses?

[devanlevin]

2 masses are tied in the followin configuaration, the 1st with a mass of m1 the 2nd mass of m2, pulleys and ropes have no mass or friction.

what is the acceleration of each mass, using only parameters g, m1, and m2

http://picasaweb.google.com/devanlevin/DropBox?authkey=sbH95pBl_D8#5273036312921418818

not too much i can see from this, i know that |a1|=2|a2|,
i think that
1 F=m1a=m1g-T
a1=g-T/m1

2 F=m2a=m2g-2T
a2=g-2t/m2

not sure about those though,

 

[ak1948]

looks right to me.

 

[Doc Al]

So far, so good. But now you need to combine all three equations to solve for a1 & a2 in terms of the given parameters. Hint: Careful with signs. If a1 is down, a2 must be up.

 

[devanlevin]

a1=-2*a2,

a1=g-T/m1

a2=g-2t/m2
-------------------
T=m1(g-a1)=m1(g+2a2)

a2=g-[2m1(g+2a2)]/m2
m2a2=g-(2m1g)-(4m1a2)

a2(m2+4*m1)=g(1-2*m1)

a2=g*(1-2m1)/(m2+4m1)

is that correct,?? the answer in my textbook is similar, either a mistake on their behalf or I am missing something,, their answer is

a2=g*(m2-2m1)/(m2+4m1)

 

[Doc Al]

a2=g-[2m1(g+2a2)]/m2
m2a2=g-(2m1g)-(4m1a2)

You made an error in this step. Note how the terms in your last equation all have the same units--except one. (When you multiplied by m2, you did not do it completely.)

 

[devanlevin]

thanks